Answer

The definition of sufficiency tells us that if the conditional distribution of \(X_1, X_2, \ldots, X_n\), given the statistic \(Y\), does not depend on \(p\), then \(Y\) is a sufficient statistic for \(p\). The conditional distribution of \(X_1, X_2, \ldots, X_n\), given \(Y\), is by definition:

\(P(X_1 = x_1, ... , X_n = x_n |Y = y) = \dfrac{P(X_1 = x_1, ... , X_n = x_n, Y = y)}{P(Y=y)}\) (**)

Now, for the sake of concreteness, suppose we were to observe a random sample of size \(n=3\) in which \(x_1=1, x_2=0, \text{ and }x_3=1\). In this case:

\( P(X_1 = 1, X_2 = 0, X_3 =1, Y=1)=0\)

because the sum of the data values, \( \sum_{i=1}^{n}X_i \), is 1 + 0 + 1 = 2, but \(Y\), which is defined to be the sum of the \(X_i\)'s is 1. That is, because \(2\ne 1\), the event in the numerator of the starred (**) equation is an impossible event and therefore its probability is 0.

Now, let's consider an event that is possible, namely ( \(X_1=1, X_2=0, X_3=1, Y=2\)). In that case, we have, by independence:

\( P(X_1 = 1, X_2 = 0, X_3 =1, Y=2) = p(1-p) p=p^2(1-p)\)

So, in general:

\(P(X_1 = x_1, X_2 = x_2, ... , X_n = x_n, Y = y) = 0 \text{ if } \sum_{i=1}^{n}x_i \ne y \)

and:

\(P(X_1 = x_1, X_2 = x_2, ... , X_n = x_n, Y = y) = p^y(1-p)^{n-y} \text{ if } \sum_{i=1}^{n}x_i = y \)

Now, the denominator in the starred (**) equation above is the binomial probability of getting exactly \(y\) successes in \(n\) trials with a probability of success \(p\). That is, the denominator is:

\( P(Y=y) = \binom{n}{y} p^y(1-p)^{n-y}\)

for \(y = 0, 1, 2,\ldots, n\). Putting the numerator and denominator together, we get, if \(y=0, 1, 2, \ldots, n\), that the conditional probability is:

\(P(X_1 = x_1, ... , X_n = x_n |Y = y) = \dfrac{p^y(1-p)^{n-y}}{\binom{n}{y} p^y(1-p)^{n-y}} =\dfrac{1}{\binom{n}{y}} \text{ if } \sum_{i=1}^{n}x_i = y\)

and:

\(P(X_1 = x_1, ... , X_n = x_n |Y = y) = 0 \text{ if } \sum_{i=1}^{n}x_i \ne y \)

Aha! We have just shown that the conditional distribution of \(X_1, X_2, \ldots, X_n\) given \(Y\) does not depend on \(p\). Therefore, \(Y\) is indeed sufficient for \(p\). That is, once the value of \(Y\) is known, no other function of \(X_1, X_2, \ldots, X_n\) will provide any additional information about the possible value of \(p\).

Answer

Because \(X_1, X_2, \ldots, X_n\) is a random sample, the joint probability mass function of \(X_1, X_2, \ldots, X_n\) is, by independence:

\(f(x_1, x_2, ... , x_n;\lambda) = f(x_1;\lambda) \times f(x_2;\lambda) \times ... \times f(x_n;\lambda)\)

Inserting what we know to be the probability mass function of a Poisson random variable with parameter \(\lambda\), the joint p.m.f. is therefore:

\(f(x_1, x_2, ... , x_n;\lambda) = \dfrac{e^{-\lambda}\lambda^{x_1}}{x_1!} \times\dfrac{e^{-\lambda}\lambda^{x_2}}{x_2!} \times ... \times \dfrac{e^{-\lambda}\lambda^{x_n}}{x_n!}\)

Now, simplifying, by adding up all \(n\) of the \(\lambda\)s in the exponents, as well as all \(n\) of the \(x_i\)'s in the exponents, we get:

\(f(x_1, x_2, ... , x_n;\lambda) = \left(e^{-n\lambda}\lambda^{\Sigma x_i} \right) \times \left( \dfrac{1}{x_1! x_2! ... x_n!} \right)\)

Hey, look at that! We just factored the joint p.m.f. into two functions, one (\(\phi\)) being only a function of the statistic \(Y=\sum_{i=1}^{n}X_i\) and the other (h) not depending on the parameter \(\lambda\):

Therefore, the Factorization Theorem tells us that \(Y=\sum_{i=1}^{n}X_i\) is a sufficient statistic for \(\lambda\). But, wait a second! We can also write the joint p.m.f. as:

\(f(x_1, x_2, ... , x_n;\lambda) = \left(e^{-n\lambda}\lambda^{n\bar{x}} \right) \times \left( \dfrac{1}{x_1! x_2! ... x_n!} \right)\)

Therefore, the Factorization Theorem tells us that \(Y = \bar{X}\) is also a sufficient statistic for \(\lambda\)!

If you think about it, it makes sense that \(Y = \bar{X}\) and \(Y=\sum_{i=1}^{n}X_i\) are both sufficient statistics, because if we know \(Y = \bar{X}\), we can easily find \(Y=\sum_{i=1}^{n}X_i\). And, if we know \(Y=\sum_{i=1}^{n}X_i\), we can easily find \(Y = \bar{X}\).

Answer

Because \(X_1, X_2, \ldots, X_n\) is a random sample, the joint probability density function of \(X_1, X_2, \ldots, X_n\) is, by independence:

\(f(x_1, x_2, ... , x_n;\mu) = f(x_1;\mu) \times f(x_2;\mu) \times ... \times f(x_n;\mu)\)

Inserting what we know to be the probability density function of a normal random variable with mean \(\mu\) and variance 1, the joint p.d.f. is:

\(f(x_1, x_2, ... , x_n;\mu) = \dfrac{1}{(2\pi)^{1/2}} exp \left[ -\dfrac{1}{2}(x_1 - \mu)^2 \right] \times \dfrac{1}{(2\pi)^{1/2}} exp \left[ -\dfrac{1}{2}(x_2 - \mu)^2 \right] \times ... \times \dfrac{1}{(2\pi)^{1/2}} exp \left[ -\dfrac{1}{2}(x_n - \mu)^2 \right] \)

Collecting like terms, we get:

\(f(x_1, x_2, ... , x_n;\mu) = \dfrac{1}{(2\pi)^{n/2}} exp \left[ -\dfrac{1}{2}\sum_{i=1}^{n}(x_i - \mu)^2 \right]\)

A trick to making the factoring of the joint p.d.f. an easier task is to add 0 to the quantity in parentheses in the summation. That is:

Now, squaring the quantity in parentheses, we get:

\(f(x_1, x_2, ... , x_n;\mu) = \dfrac{1}{(2\pi)^{n/2}} exp \left[ -\dfrac{1}{2}\sum_{i=1}^{n}\left[ (x_i - \bar{x})^2 +2(x_i - \bar{x}) (\bar{x}-\mu)+ (\bar{x}-\mu)^2\right] \right]\)

And then distributing the summation, we get:

\(f(x_1, x_2, ... , x_n;\mu) = \dfrac{1}{(2\pi)^{n/2}} exp \left[ -\dfrac{1}{2}\sum_{i=1}^{n} (x_i - \bar{x})^2 - (\bar{x}-\mu) \sum_{i=1}^{n}(x_i - \bar{x}) -\dfrac{1}{2}\sum_{i=1}^{n}(\bar{x}-\mu)^2\right] \)

But, the middle term in the exponent is 0, and the last term, because it doesn't depend on the index \(i\), can be added up \(n\) times:

So, simplifying, we get:

\(f(x_1, x_2, ... , x_n;\mu) = \left\{ exp \left[ -\dfrac{n}{2} (\bar{x}-\mu)^2 \right] \right\} \times \left\{ \dfrac{1}{(2\pi)^{n/2}} exp \left[ -\dfrac{1}{2}\sum_{i=1}^{n} (x_i - \bar{x})^2 \right] \right\} \)

In summary, we have factored the joint p.d.f. into two functions, one (\(\phi\)) being only a function of the statistic \(Y = \bar{X}\) and the other (h) not depending on the parameter \(\mu\):

Therefore, the Factorization Theorem tells us that \(Y = \bar{X}\) is a sufficient statistic for \(\mu\). Now, \(Y = \bar{X}^3\) is also sufficient for \(\mu\), because if we are given the value of \( \bar{X}^3\), we can easily get the value of \(\bar{X}\) through the one-to-one function \(w=y^{1/3}\). That is:

\( W=(\bar{X}^3)^{1/3}=\bar{X} \)

On the other hand, \(Y = \bar{X}^2\) is not a sufficient statistic for \(\mu\), because it is not a one-to-one function. That is, if we are given the value of \(\bar{X}^2\), using the inverse function:

\(w=y^{1/2}\)

we get two possible values, namely:

\(-\bar{X}\) and \(+\bar{X}\)

Answer

Because \(X_1, X_2, \ldots, X_n\) is a random sample, the joint probability density function of \(X_1, X_2, \ldots, X_n\) is, by independence:

\(f(x_1, x_2, ... , x_n;\theta) = f(x_1;\theta) \times f(x_2;\theta) \times ... \times f(x_n;\theta)\)

Inserting what we know to be the probability density function of an exponential random variable with parameter \(\theta\), the joint p.d.f. is:

\(f(x_1, x_2, ... , x_n;\theta) =\dfrac{1}{\theta}exp\left( \dfrac{-x_1}{\theta}\right) \times \dfrac{1}{\theta}exp\left( \dfrac{-x_2}{\theta}\right) \times ... \times \dfrac{1}{\theta}exp\left( \dfrac{-x_n}{\theta} \right) \)

Now, simplifying, by adding up all \(n\) of the \(\theta\)s and the \(n\) \(x_i\)'s in the exponents, we get:

\(f(x_1, x_2, ... , x_n;\theta) =\dfrac{1}{\theta^n}exp\left( - \dfrac{1}{\theta} \sum_{i=1}^{n} x_i\right) \)

We have again factored the joint p.d.f. into two functions, one (\(\phi\)) being only a function of the statistic \(Y=\sum_{i=1}^{n}X_i\) and the other (h) not depending on the parameter \(\theta\):

Therefore, the Factorization Theorem tells us that \(Y=\sum_{i=1}^{n}X_i\) is a sufficient statistic for \(\theta\). And, since \(Y = \bar{X}\) is a one-to-one function of \(Y=\sum_{i=1}^{n}X_i\), it implies that \(Y = \bar{X}\) is also a sufficient statistic for \(\theta\).

Answer

The probability mass function of a geometric random variable is:

\(f(x;p) = (1-p)^{x-1}p\)

for \(x=1, 2, 3, \ldots\) The p.m.f. can be written in exponential form as:

\(f(x;p) = \text{exp}\left[ x\text{log}(1-p)+\text{log}(1)+\text{log}\left( \frac{p}{1-p} \right)\right] \)

Therefore, \(Y=\sum_{i=1}^{n}X_i\) is sufficient for \(p\). Easy as pie!

Answer

Because \(X_1, X_2, \ldots, X_n\) is a random sample, the joint probability density function of \(X_1, X_2, \ldots, X_n\) is, by independence:

\(f(x_1, x_2, ... , x_n;\theta_1, \theta_2) = f(x_1;\theta_1, \theta_2) \times f(x_2;\theta_1, \theta_2) \times ... \times f(x_n;\theta_1, \theta_2) \times \)

Inserting what we know to be the probability density function of a normal random variable with mean \(\theta_1\) and variance \(\theta_2\), the joint p.d.f. is:

\(f(x_1, x_2, ... , x_n;\theta_1, \theta_2) = \dfrac{1}{\sqrt{2\pi\theta_2}} \text{exp} \left[-\dfrac{1}{2}\dfrac{(x_1-\theta_1)^2}{\theta_2} \right] \times ... \times = \dfrac{1}{\sqrt{2\pi\theta_2}} \text{exp} \left[-\dfrac{1}{2}\dfrac{(x_n-\theta_1)^2}{\theta_2} \right] \)

Simplifying by collecting like terms, we get:

\(f(x_1, x_2, ... , x_n;\theta_1, \theta_2) = \left(\dfrac{1}{\sqrt{2\pi\theta_2}}\right)^n \text{exp} \left[-\dfrac{1}{2}\dfrac{\sum_{i=1}^{n}(x_i-\theta_1)^2}{\theta_2} \right] \)

Rewriting the first factor, and squaring the quantity in parentheses, and distributing the summation, in the second factor, we get:

\(f(x_1, x_2, ... , x_n;\theta_1, \theta_2) = \text{exp} \left[\text{log}\left(\dfrac{1}{\sqrt{2\pi\theta_2}}\right)^n\right] \text{exp} \left[-\dfrac{1}{2\theta_2}\left\{ \sum_{i=1}^{n}x_{i}^{2} -2\theta_1\sum_{i=1}^{n}x_{i} +\sum_{i=1}^{n}\theta_{1}^{2} \right\}\right] \)

Simplifying yet more, we get:

\(f(x_1, x_2, ... , x_n;\theta_1, \theta_2) = \text{exp} \left[ -\dfrac{1}{2\theta_2}\sum_{i=1}^{n}x_{i}^{2}+\dfrac{\theta_1}{\theta_2}\sum_{i=1}^{n}x_{i} -\dfrac{n\theta_{1}^{2}}{2\theta_2}-n\text{log}\sqrt{2\pi\theta_2} \right]\)

Look at that! We have factored the joint p.d.f. into two functions, one (\(\phi\)) being only a function of the statistics \(Y_1=\sum_{i=1}^{n}X^{2}_{i}\) and \(Y_2=\sum_{i=1}^{n}X_i\), and the other (h) not depending on the parameters \(\theta_1\) and \(\theta_2\):

Therefore, the Factorization Theorem tells us that \(Y_1=\sum_{i=1}^{n}X^{2}_{i}\) and \(Y_2=\sum_{i=1}^{n}X_i\) are joint sufficient statistics for \(\theta_1\) and \(\theta_2\). And, the one-to-one functions of \(Y_1\) and \(Y_2\), namely:

\( \bar{X} =\dfrac{Y_2}{n}=\dfrac{1}{n}\sum_{i=1}^{n}X_i \)

and

\( S_2=\dfrac{Y_1-(Y_{2}^{2}/n)}{n-1}=\dfrac{1}{n-1} \left[\sum_{i=1}^{n}X_{i}^{2}-n\bar{X}^2 \right] \)

are also joint sufficient statistics for \(\theta_1\) and \(\theta_2\). Aha! We have just shown that the intuitive estimators of \(\mu\) and \(\sigma^2\) are also sufficient estimators. That is, the data contain no more information than the estimators \(\bar{X}\) and \(S^2\) do about the parameters \(\mu\) and \(\sigma^2\)! That seems like a good thing!

Answer

The probability density function of a normal random variable with mean \(\theta_1\) and variance \(\theta_2\) can be written in exponential form as:

Therefore, the statistics \(Y_1=\sum_{i=1}^{n}X^{2}_{i}\) and \(Y_2=\sum_{i=1}^{n}X_i\) are joint sufficient statistics for \(\theta_1\) and \(\theta_2\).

Answer

In this case, the engineer commits a Type I error if his observed sample mean falls in the rejection region, that is, if it is 172 or greater, when the true (unknown) population mean is indeed 170. Graphically, \(\alpha\), the engineer's probability of committing a Type I error looks like this:

Now, we can calculate the engineer's value of \(\alpha\) by making the transformation from a normal distribution with a mean of 170 and a standard deviation of 10 to that of \(Z\), the standard normal distribution using:

\(Z= \frac{\bar{X}-\mu}{\sigma / \sqrt{n}} \)

Doing so, we get:

So, calculating the engineer's probability of committing a Type I error reduces to making a normal probability calculation. The probability is 0.1587 as illustrated here:

\(\alpha = P(\bar{X} \ge 172 \text { if } \mu = 170) = P(Z \ge 1.00) = 0.1587 \)

A probability of 0.1587 is a bit high. We'll learn in this lesson how the engineer could reduce his probability of committing a Type I error.

Answer

In this case, the engineer commits a Type II error if his observed sample mean does not fall in the rejection region, that is, if it is less than 172, when the true (unknown) population mean is 173. Graphically, \(\beta\), the engineer's probability of committing a Type II error looks like this:

Again, we can calculate the engineer's value of \(\beta\) by making the transformation from a normal distribution with a mean of 173 and a standard deviation of 10 to that of \(Z\), the standard normal distribution. Doing so, we get:

So, calculating the engineer's probability of committing a Type II error again reduces to making a normal probability calculation. The probability is 0.3085 as illustrated here:

\(\beta= P(\bar{X} < 172 \text { if } \mu = 173) = P(Z < -0.50) = 0.3085 \)

A probability of 0.3085 is a bit high. We'll learn in this lesson how the engineer could reduce his probability of committing a Type II error.

Answer

Setting \(\alpha\), the probability of committing a Type I error, to 0.05, implies that we should reject the null hypothesis when the test statistic \(Z\ge 1.645\), or equivalently, when the observed sample mean is 106.58 or greater:

because we transform the test statistic \(Z\) to the sample mean by way of:

\(Z=\dfrac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}}\qquad \Rightarrow \bar{X}=\mu+Z\dfrac{\sigma}{\sqrt{n}} \qquad \bar{X}=100+1.645\left(\dfrac{16}{\sqrt{16}}\right)=106.58\)

Now, that implies that the power, that is, the probability of rejecting the null hypothesis, when \(\mu=108\) is 0.6406 as calculated here (recalling that \(Phi(z)\) is standard notation for the cumulative distribution function of the standard normal random variable):

\( \text{Power}=P(\bar{X}\ge 106.58\text{ when } \mu=108) = P\left(Z\ge \dfrac{106.58-108}{\frac{16}{\sqrt{16}}}\right) \\ = P(Z\ge -0.36)=1-P(Z<-0.36)=1-\Phi(-0.36)=1-0.3594=0.6406 \)

and illustrated here:

In summary, we have determined that we have (only) a 64.06% chance of rejecting the null hypothesis \(H_0:\mu=100\) in favor of the alternative hypothesis \(H_A:\mu>100\) if the true unknown population mean is in reality \(\mu=108\).

Answer

Because we are setting \(\alpha\), the probability of committing a Type I error, to 0.05, we again reject the null hypothesis when the test statistic \(Z\ge 1.645\), or equivalently, when the observed sample mean is 106.58 or greater. That means that the probability of rejecting the null hypothesis, when \(\mu=112\) is 0.9131 as calculated here:

\( \text{Power}=P(\bar{X}\ge 106.58\text{ when }\mu=112)=P\left(Z\ge \frac{106.58-112}{\frac{16}{\sqrt{16}}}\right) \\ = P(Z\ge -1.36)=1-P(Z<-1.36)=1-\Phi(-1.36)=1-0.0869=0.9131 \)

and illustrated here:

In summary, we have determined that we now have a 91.31% chance of rejecting the null hypothesis \(H_0:\mu=100\) in favor of the alternative hypothesis \(H_A:\mu>100\) if the true unknown population mean is in reality \(\mu=112\). Hmm.... it should make sense that the probability of rejecting the null hypothesis is larger for values of the mean, such as 112, that are far away from the assumed mean under the null hypothesis.

Answer

Again, because we are setting \(\alpha\), the probability of committing a Type I error, to 0.05, we reject the null hypothesis when the test statistic \(Z\ge 1.645\), or equivalently, when the observed sample mean is 106.58 or greater. That means that the probability of rejecting the null hypothesis, when \(\mu=116\) is 0.9909 as calculated here:

\(\text{Power}=P(\bar{X}\ge 106.58\text{ when }\mu=116) =P\left(Z\ge \dfrac{106.58-116}{\frac{16}{\sqrt{16}}}\right) = P(Z\ge -2.36)=1-P(Z<-2.36)= 1-\Phi(-2.36)=1-0.0091=0.9909 \)

and illustrated here:

In summary, we have determined that, in this case, we have a 99.09% chance of rejecting the null hypothesis \(H_0:\mu=100\) in favor of the alternative hypothesis \(H_A:\mu>100\) if the true unknown population mean is in reality \(\mu=116\). The probability of rejecting the null hypothesis is the largest yet of those we calculated, because the mean, 116, is the farthest away from the assumed mean under the null hypothesis.

Answer

Setting \(\alpha\), the probability of committing a Type I error, to 0.01, implies that we should reject the null hypothesis when the test statistic \(Z\ge 2.326\), or equivalently, when the observed sample mean is 109.304 or greater:

because:

\(\bar{x} = \mu + z \left( \frac{\sigma}{\sqrt{n}} \right) =100 + 2.326\left( \frac{16}{\sqrt{16}} \right)=109.304 \)

That means that the probability of rejecting the null hypothesis, when \(\mu=108\) is 0.3722 as calculated here:

So, the power when \(\mu=108\) and \(\alpha=0.01\) is smaller (0.3722) than the power when \(\mu=108\) and \(\alpha=0.05\) (0.6406)! Perhaps we can see this graphically:

By the way, we could again alternatively look at the glass as being half-empty. In that case, the probability of a Type II error when \(\mu=108\) and \(\alpha=0.01\) is \(1-0.3722=0.6278\). In this case, the probability of a Type II error is greater than the probability of a Type II error when \(\mu=108\) and \(\alpha=0.05\).

All of this can be seen graphically by plotting the two power functions, one where \(\alpha=0.01\) and the other where \(\alpha=0.05\), simultaneously. Doing so, we get a plot that looks like this:

Answer

Setting \(\alpha\), the probability of committing a Type I error, to 0.05, implies that we should reject the null hypothesis when the test statistic \(Z\ge 1.645\), or equivalently, when the observed sample mean is 103.29 or greater:

because:

\( \bar{x} = \mu + z \left(\dfrac{\sigma}{\sqrt{n}} \right) = 100 +1.645\left(\dfrac{16}{\sqrt{64}} \right) = 103.29\)

Therefore, the power function \K(\mu)\), when \(\mu>100\) is the true value, is:

\( K(\mu) = P(\bar{X} \ge 103.29 | \mu) = P \left(Z \ge \dfrac{103.29 - \mu}{16 / \sqrt{64}} \right) = 1 - \Phi \left(\dfrac{103.29 - \mu}{2} \right)\)

Therefore, the probability of rejecting the null hypothesis at the \(\alpha=0.05\) level when \(\mu=108\) is 0.9907, as calculated here:

\(K(108) = 1 - \Phi \left( \dfrac{103.29-108}{2} \right) = 1- \Phi(-2.355) = 0.9907 \)

And, the probability of rejecting the null hypothesis at the \(\alpha=0.05\) level when \(\mu=112\) is greater than 0.9999, as calculated here:

\( K(112) = 1 - \Phi \left( \dfrac{103.29-112}{2} \right) = 1- \Phi(-4.355) = 0.9999\ldots \)

And, the probability of rejecting the null hypothesis at the \(\alpha=0.05\) level when \(\mu=116\) is greater than 0.999999, as calculated here:

\( K(116) = 1 - \Phi \left( \dfrac{103.29-116}{2} \right) = 1- \Phi(-6.355) = 0.999999... \)

Answer

As is always the case, we need to start by finding a threshold value \(c\), such that if the sample mean is larger than \(c\), we'll reject the null hypothesis:

That is, in order for our hypothesis test to be conducted at the \(\alpha=0.05\) level, the following statement must hold (using our typical \(Z\) transformation):

\(c = 40 + 1.645 \left( \dfrac{6}{\sqrt{n}} \right) \) (**)

But, that's not the only condition that \(c\) must meet, because \(c\) also needs to be defined to ensure that our power is 0.90 or, alternatively, that the probability of a Type II error is 0.10. That would happen if there was a 10% chance that our test statistic fell short of \(c\) when \(\mu=45\), as the following drawing illustrates in blue:

This illustration suggests that in order for our hypothesis test to have 0.90 power, the following statement must hold (using our usual \(Z\) transformation):

\(c = 45 - 1.28 \left( \dfrac{6}{\sqrt{n}} \right) \) (**)

Aha! We have two (asterisked (**)) equations and two unknowns! All we need to do is equate the equations, and solve for \(n\). Doing so, we get:

\(40+1.645\left(\frac{6}{\sqrt{n}}\right)=45-1.28\left(\frac{6}{\sqrt{n}}\right)\)
\(\Rightarrow 5=(1.645+1.28)\left(\frac{6}{\sqrt{n}}\right), \qquad \Rightarrow 5=\frac{17.55}{\sqrt{n}}, \qquad n=(3.51)^2=12.3201\approx 13\)

Now that we know we will set \(n=13\), we can solve for our threshold value c:

\( c = 40 + 1.645 \left( \dfrac{6}{\sqrt{13}} \right)=42.737 \)

So, in summary, if the agricultural researcher collects data on \(n=13\) corn plots, and rejects his null hypothesis \(H_0:\mu=40\) if the average crop yield of the 13 plots is greater than 42.737 bushels per acre, he will have a 5% chance of committing a Type I error and a 10% chance of committing a Type II error if the population mean \(\mu\) were actually 45 bushels per acre.

Answer

In this case, because we are interested in performing a hypothesis test about a population proportion \(p\), we use the \(Z\)-statistic:

\(Z = \dfrac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \)

Again, we start by finding a threshold value \(c\), such that if the observed sample proportion is larger than \(c\), we'll reject the null hypothesis:

That is, in order for our hypothesis test to be conducted at the \(\alpha=0.01\) level, the following statement must hold:

\(c = 0.5 + 2.326 \sqrt{ \dfrac{(0.5)(0.5)}{n}} \) (**)

But, again, that's not the only condition that c must meet, because \(c\) also needs to be defined to ensure that our power is 0.80 or, alternatively, that the probability of a Type II error is 0.20. That would happen if there was a 20% chance that our test statistic fell short of \(c\) when \(p=0.55\), as the following drawing illustrates in blue:

This illustration suggests that in order for our hypothesis test to have 0.80 power, the following statement must hold:

\(c = 0.55 - 0.842 \sqrt{ \dfrac{(0.55)(0.45)}{n}} \) (**)

Again, we have two (asterisked (**)) equations and two unknowns! All we need to do is equate the equations, and solve for \(n\). Doing so, we get:

\(0.5+2.326\sqrt{\dfrac{0.5(0.5)}{n}}=0.55-0.842\sqrt{\dfrac{0.55(0.45)}{n}} \\ 2.326\dfrac{\sqrt{0.25}}{\sqrt{n}}+0.842\dfrac{\sqrt{0.2475}}{\sqrt{n}}=0.55-0.5 \\ \dfrac{1}{\sqrt{n}}(1.5818897)=0.05 \qquad \Rightarrow n\approx \left(\dfrac{1.5818897}{0.05}\right)^2 = 1000.95 \approx 1001 \)

Now that we know we will set \(n=1001\), we can solve for our threshold value \(c\):

\(c = 0.5 + 2.326 \sqrt{\dfrac{(0.5)(0.5)}{1001}}= 0.5367 \)

So, in summary, if the pollster collects data on \(n=1001\) voters, and rejects his null hypothesis \(H_0:p=0.5\) if the proportion of sampled voters who favor the political candidate is greater than 0.5367, he will have a 1% chance of committing a Type I error and a 20% chance of committing a Type II error if the population proportion \(p\) were actually 0.55.

Incidentally, we can always check our work! Conducting the survey and subsequent hypothesis test as described above, the probability of committing a Type I error is:

\(\alpha= P(\hat{p} >0.5367 \text { if } p = 0.50) = P(Z > 2.3257) = 0.01 \)

and the probability of committing a Type II error is:

\(\beta = P(\hat{p} <0.5367 \text { if } p = 0.55) = P(Z < -0.846) = 0.199 \)

just as the pollster had desired.

Answer

The p.d.f. of an exponential random variable is:

\(f(x) = \dfrac{1}{\theta}e^{-x/\theta} \)

for \(x ≥ 0\). Under the hypothesis \(H \colon \theta = 3\), the p.d.f. of an exponential random variable is:

\(f(x) = \dfrac{1}{3}e^{-x/3} \)

for \(x ≥ 0\). Because we can uniquely specify the p.d.f. under the hypothesis \(H \colon \theta = 3\), the hypothesis is a simple hypothesis.

Answer

Again, the p.d.f. of an exponential random variable is:

\(f(x) = \dfrac{1}{\theta}e^{-x/\theta} \)

for \(x ≥ 0\). Under the hypothesis \(H \colon \theta > 2\), the p.d.f. of an exponential random variable could be:

\(f(x) = \dfrac{1}{3}e^{-x/3} \)

for \(x ≥ 0\). Or, the p.d.f. could be:

\(f(x) = \dfrac{1}{22}e^{-x/22} \)

for \(x ≥ 0\). The p.d.f. could, in fact, be any of an infinite number of possible exponential probability density functions. Because the p.d.f. is not uniquely specified under the hypothesis \(H \colon \theta > 2\), the hypothesis is a composite hypothesis.

Answer

The p.d.f. of a normal random variable is:

\(f(x)= \dfrac{1}{\sigma\sqrt{2\pi}} exp \left[-\dfrac{(x-\mu)^2}{2\sigma^2} \right] \)

for \(−∞ < x <  ∞, −∞ < \mu < ∞\), and \(\sigma > 0\). Under the hypothesis \(H \colon \mu = 12\), the p.d.f. of a normal random variable is:

\(f(x)= \dfrac{1}{\sigma\sqrt{2\pi}} exp \left[-\dfrac{(x-12)^2}{2\sigma^2} \right] \)

for \(−∞ < x < ∞\) and \(\sigma > 0\). In this case, the mean parameter \( \mu = 12\) is uniquely specified in the p.d.f., but the variance \(\sigma^2\) is not. Therefore, the hypothesis \(H \colon \mu = 12\) is a composite hypothesis.

Answer

Because both the null and alternative hypotheses are simple hypotheses, we can apply the Neyman Pearson Lemma in an attempt to find the most powerful test. The lemma tells us that the ratio of the likelihoods under the null and alternative must be less than some constant k. Again, because we are dealing with just one observation X, the ratio of the likelihoods equals the ratio of the probability density functions, giving us:

\( \dfrac{L(\theta_0)}{L(\theta_\alpha)}= \dfrac{3x^{3-1}}{2x^{2-1}}= \dfrac{3}{2}x \le k \)

That is, the lemma tells us that the form of the rejection region for the most powerful test is:

\( \dfrac{3}{2}x \le k \)

or alternatively, since (2/3)k is just a new constant \(k^*\), the rejection region for the most powerful test is of the form:

\(x < \dfrac{2}{3}k = k^* \)

Now, it's just a matter of finding \(k^*\), and our work is done. We want \(\alpha\) = P(Type I Error) = P(rejecting the null hypothesis when the null hypothesis is true) to equal 0.05. In order for that to happen, the following must hold:

\(\alpha = P( X < k^* \text{ when } \theta = 3) = \int_{0}^{k^*} 3x^2dx = 0.05 \)

Doing the integration, we get:

\( \left[ x^3\right]^{x=k^*}_{x=0} = (k^*)^3 =0.05 \)

And, solving for \(k^*\), we get:

\(k^* =(0.05)^{1/3} = 0.368 \)

That is, the Neyman Pearson Lemma tells us that the rejection region of the most powerful test for testing \(H_{0} \colon \theta = 3 \) against \(H_{A} \colon \theta = 2 \), under the assumed probability distribution, is:

\(x < 0.368 \)

That is, among all of the possible tests for testing \(H_{0} \colon \theta = 3 \) against \(H_{A} \colon \theta = 2 \), based on a single observation X and with a significance level of 0.05, this test has the largest possible value for the power under the alternative hypohthesis, that is, when \(\theta = 2\).

Answer

Because the variance is specified, both the null and alternative hypotheses are simple hypotheses. Therefore, we can apply the Neyman Pearson Lemma in an attempt to find the most powerful test. The lemma tells us that the ratio of the likelihoods under the null and alternative must be less than some constant k:

\( \dfrac{L(10)}{L(15)}= \dfrac{(32\pi)^{-16/2} exp \left[ -(1/32)\sum_{i=1}^{16}(x_i -10)^2 \right]}{(32\pi)^{-16/2} exp \left[ -(1/32)\sum_{i=1}^{16}(x_i -15)^2 \right]} \le k \)

Simplifying, we get:

\(exp \left[ - \left( \dfrac{1}{32} \right) \left( \sum_{i=1}^{16}(x_i -10)^2 - \sum_{i=1}^{16}(x_i -15)^2 \right) \right] \le k \)

And, simplifying yet more, we get:

Now, taking the natural logarithm of both sides of the inequality, collecting like terms, and multiplying through by 32, we get:

\(-10\Sigma x_i +2000 \le 32ln(k)\)

And, moving the constant term on the left-side of the inequality to the right-side, and dividing through by −160, we get:

\(\dfrac{1}{16}\Sigma x_i \ge -\frac{1}{160}(32ln(k)-2000) \)

That is, the Neyman Pearson Lemma tells us that the rejection region for the most powerful test for testing \(H_{0} \colon \mu = 10 \) against \(H_{A} \colon \mu = 15 \), under the normal probability model, is of the form:

\(\bar{x} \ge k^* \)

where \(k^*\) is selected so that the size of the critical region is \(\alpha = 0.05\). That's simple enough, as it just involves a normal probabilty calculation! Under the null hypothesis, the sample mean is normally distributed with mean 10 and standard deviation 4/4 = 1. Therefore, the critical value \(k^*\) is deemed to be 11.645:

That is, the Neyman Pearson Lemma tells us that the rejection region for the most powerful test for testing \(H_{0} \colon \mu = 10 \) against \(H_{A} \colon \mu = 15 \), under the normal probability model, is:

\(\bar{x} \ge 11.645 \)

The power of such a test when \(\mu = 15\) is:

\( P(\bar{X} > 11.645 \text{ when } \mu = 15) = P \left( Z > \dfrac{11.645-15}{\sqrt{16} / \sqrt{16} } \right) = P(Z > -3.36) = 0.9996 \)

The power can't get much better than that, and the Neyman Pearson Lemma tells us that we shouldn't expect it to get better! That is, the Lemma tells us that there is no other test out there that will give us greater power for testing \(H_{0} \colon \mu = 10 \) against \(H_{A} \colon \mu = 15 \).

Answer

For each simple alternative in \(H_A , \mu = \mu_a\), say, the ratio of the likelihood functions is:

\( \dfrac{L(10)}{L(\mu_\alpha)}= \dfrac{(32\pi)^{-16/2} exp \left[ -(1/32)\sum_{i=1}^{16}(x_i -10)^2 \right]}{(32\pi)^{-16/2} exp \left[ -(1/32)\sum_{i=1}^{16}(x_i -\mu_\alpha)^2 \right]} \le k \)

Simplifying, we get:

\(exp \left[ - \left(\dfrac{1}{32} \right) \left(\sum_{i=1}^{16}(x_i -10)^2 - \sum_{i=1}^{16}(x_i -\mu_\alpha)^2 \right) \right] \le k \)

And, simplifying yet more, we get:

Taking the natural logarithm of both sides of the inequality, collecting like terms, and multiplying through by 32, we get:

\( -2(\mu_\alpha - 10) \sum x_i +16 (\mu_{\alpha}^{2} - 10^2) \le 32 ln(k) \)

Moving the constant term on the left-side of the inequality to the right-side, and dividing through by \(-16(2(\mu_\alpha - 10)) \), we get:

\( \dfrac{1}{16} \sum x_i \ge - \dfrac{1}{16(2(\mu_\alpha - 10))}(32 ln(k) - 16(\mu_{\alpha}^{2} - 10^2)) = k^* \)

In summary, we have shown that the ratio of the likelihoods is small, that is:

\(\dfrac{L(10)}{L(\mu_\alpha)} \le k \)

if and only if:

\( \bar{x} \ge k^*\)

Therefore, the best critical region of size \(\alpha\) for testing \(H_0: \mu = 10\) against each simple alternative \(H_A \colon \mu = \mu_a\), where \(\mu_a > 10\), is given by:

\( C= \left\{ (x_1, x_1, ... , x_n): \bar{x} \ge k^* \right\} \)

where \(k^*\) is selected such that the probability of committing a Type I error is \(\alpha\), that is:

\( \alpha = P(\bar{X} \ge k^*) \text{ when } \mu = 10 \)

Because the critical region C defines a test that is most powerful against each simple alternative \(\mu_a > 10\), this is a uniformly most powerful test, and C is a uniformly most powerful critical region of size \(\alpha\).

Answer

If the null parameter space is \(\Omega = {\mu: \mu = 3}\), then the alternative parameter space is everything that is in \(\Omega = {\mu: −∞ < \mu < ∞}\) that is not in \(\Omega\). That is, the alternative parameter space is \(\Omega ' = {\mu: \mu ≠ 3}\). And, so the alternative hypothesis is:

\(H_A : \mu \ne 3\)

In this case, we'd be interested in deriving a two-tailed test.

Answer

If the alternative parameter space is (\omega ' = {\mu: \mu > 3}\), then the null parameter space is \(\omega = {\mu: \mu ≤ 3}\). And, so the null hypothesis is:

\(H_0 : \mu \le 3\)

Now, the reality is that some authors do specify the null hypothesis as such, even when they mean \(H_0: \mu = 3\). Ours don't, and so we won't. (That's why I put that "technically" in parentheses up above.) At any rate, in this case, we'd be interested in deriving a one-tailed test.

Answer

Because we are interested in testing the null hypothesis \(H_0: \mu = 10\) against the alternative hypothesis \(H_A: \mu ≠ 10\) for a normal mean, our total parameter space is:

\(\Omega =\left \{\mu : -\infty < \mu < \infty \right \}\)

and our null parameter space is:

\(\omega = \left \{10\right \}\)

Now, to find the likelihood ratio, as defined above, we first need to find \(L(\hat{\omega})\). Well, when the null hypothesis \(H_0: \mu = 10\) is true, the mean \(\mu\) can take on only one value, namely, \(\mu = 10\). Therefore:

\(L(\hat{\omega}) = L(10)\)

We also need to find \(L(\hat{\Omega})\) in order to define the likelihood ratio. To find it, we must find the value of \(\mu\) that maximizes \(L(\mu)\). Well, we did that back when we studied maximum likelihood as a method of estimation. We showed that \(\hat{\mu} = \bar{x}\) is the maximum likelihood estimate of \(\mu\). Therefore:

\(L(\hat{\Omega}) = L(\bar{x})\)

Now, putting it all together to form the likelihood ratio, we get:

which simplifies to:

Now, let's step aside for a minute and focus just on the summation in the numerator. If we "add 0" in a special way to the quantity in parentheses:

we can show that the summation can be written as:

\(\sum_{i=1}^{n}(x_i - 10)^2 = \sum_{i=1}^{n}(x_i - \bar{x})^2 + n(\bar{x} -10)^2 \)

Therefore, the likelihood ratio becomes:

which greatly simplifies to:

\(\lambda = exp \left [-\dfrac{n}{4}(\bar{x}-10)^2 \right ]\)

Now, the likelihood ratio test tells us to reject the null hypothesis when the likelihood ratio \(\lambda\) is small, that is, when:

\(\lambda = exp\left[-\dfrac{n}{4}(\bar{x}-10)^2 \right] \le k\)

where k is chosen to ensure that, in this case, \(\alpha = 0.05\). Well, by taking the natural log of both sides of the inequality, we can show that \(\lambda ≤ k\) is equivalent to:

\( -\dfrac{n}{4}(\bar{x}-10)^2 \le \text{ln} k \)

which, by multiplying through by −4/n, is equivalent to:

\((\bar{x}-10)^2 \ge -\dfrac{4}{n} \text{ln} k \)

which is equivalent to:

\(\dfrac{|\bar{X}-10|}{\sigma / \sqrt{n}} \ge \dfrac{\sqrt{-(4/n)\text{ln} k}}{\sigma / \sqrt{n}} =k* \)

Aha! We should recognize that quantity on the left-side of the inequality! We know that:

\(Z = \dfrac{\bar{X}-10}{\sigma / \sqrt{n}} \)

follows a standard normal distribution when \(H_0: \mu = 10\). Therefore we can determine the appropriate \(k^*\) by using the standard normal table. We have shown that the likelihood ratio test tells us to reject the null hypothesis \(H_0: \mu = 10\) in favor of the alternative hypothesis \(H_A: \mu ≠ 10\) for all sample means for which the following holds:

\(\dfrac{|\bar{X}-10|}{ \sqrt{2} / \sqrt{n}} \ge z_{0.025} = 1.96 \)

Doing so will ensure that our probability of committing a Type I error is set to \(\alpha = 0.05\), as desired.

Answer

Our unrestricted parameter space is:

\( \Omega = \left\{ (\mu, \sigma^2) : -\infty < \mu < \infty, 0 < \sigma^2 < \infty \right\} \)

Under the null hypothesis, the mean \(\mu\) is the only parameter that is restricted. Therefore, our parameter space under the null hypothesis is:

\( \omega = \left\{(\mu, \sigma^2) : \mu =\mu_0, 0 < \sigma^2 < \infty \right\}\)

Now, first consider the case where the mean and variance are unrestricted. We showed back when we studied maximum likelihood estimation that the maximum likelihood estimates of \(\mu\) and \(\sigma^2\) are, respectively:

\(\hat{\mu} = \bar{x} \text{ and } \hat{\sigma}^2 = \dfrac{1}{n}\sum_{i=1}^{n}(x_i - \bar{x})^2 \)

Therefore, the maximum of the likelihood function for the unrestricted parameter space is:

which simplifies to:

\( L(\hat{\Omega})= \left[\dfrac{ne^{-1}}{2\pi \Sigma (x_i - \bar{x})^2} \right]^{n/2} \)

Now, under the null parameter space, the maximum likelihood estimates of \(\mu\) and \(\sigma^2\) are, respectively:

\( \hat{\mu} = \mu_0 \text{ and } \hat{\sigma}^2 = \dfrac{1}{n}\sum_{i=1}^{n}(x_i - \mu_0)^2 \)

Therefore, the likelihood under the null hypothesis is:

which simplifies to:

\( L(\hat{\omega})= \left[\dfrac{ne^{-1}}{2\pi \Sigma (x_i - \mu_0)^2} \right]^{n/2} \)

And now taking the ratio of the two likelihoods, we get:

which reduces to:

\( \lambda = \left[ \dfrac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{\sum_{i=1}^{n}(x_i - \mu_0)^2} \right] ^{n/2}\)

Focusing only on the denominator for a minute, let's do that trick again of "adding 0" in just the right away. Adding 0 to the quantity in the parentheses, we get:

which simplifies to:

\( \sum_{i=1}^{n}(x_i - \mu_0)^2 = \sum_{i=1}^{n}(x_i - \bar{x})^2 +n(\bar{x} - \mu_0)^2 \)

Then, our likelihood ratio \(\lambda\) becomes:

\( \lambda = \left[ \dfrac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{\sum_{i=1}^{n}(x_i - \mu_0)^2} \right] ^{n/2} = \left[ \dfrac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{ \sum_{i=1}^{n}(x_i - \bar{x})^2 +n(\bar{x} - \mu_0)^2} \right] ^{n/2} \)

which, upon dividing through numerator and denominator by \( \sum_{i=1}^{n}(x_i - \bar{x})^2 \) simplifies to:

Therefore, the likelihood ratio test's critical region, which is given by the inequality \(\lambda ≤ k\), is equivalent to:

which with some minor algebraic manipulation can be shown to be equivalent to:

So, in a nutshell, we've shown that the likelihood ratio test tells us that for this situation we should reject the null hypothesis \(H_0: \mu= \mu_0\) in favor of the alternative hypothesis \(H_A: \mu ≠ \mu_0\) if:

\( \dfrac{(\bar{x}-\mu_0)^2 }{s^2 / n} \ge k^{*} \)

Well, okay, so I started out this page claiming that the t-test for a mean \(\mu\) is the likelihood ratio test. Is it? Well, the above critical region is equivalent to rejecting the null hypothesis if:

\( \dfrac{|\bar{x}-\mu_0| }{s / \sqrt{n}} \ge k^{**} \)

Does that look familiar? We previously learned that if \(X_1, X_2, \dots, X_n\) are normally distributed with mean \(\mu\) and variance \(\sigma^2\), then:

\( T = \dfrac{\bar{X}-\mu}{S / \sqrt{n}} \)

follows a T distribution with n − 1 degrees of freedom. So, this tells us that we should use the T distribution to choose \(k^{**}\). That is, set:

\(k^{**} = t_{\alpha /2, n-1}\)

and we have our size \(\alpha\) t-test that ensures the probability of committing a Type I error is \(\alpha\).

Answer

The research question involves the study of one group, namely college students. The research question involves a binary response... either a student does or does not work during the semester. The research question is an "is it this?" question, and therefore involves conducting a hypothesis test. If p is the (unknown) proportion of college students who work during the semester, then we are specifically interested in testing the null hypothesis \(H_0: p = 0.50\) against the alternative hypotheses \(H_A: p > 0.50\). We can enter the resulting data into Minitab and then ask Minitab to conduct a Z-test for one proportion for us.

Incidentally, this discussion of whether or not we should conduct a hypothesis test or calculate a confidence interval is a bit like splitting hairs. That's because, as you might recall, a confidence interval can always be used to answer an "is it this?" question, too. For example, in this case, we could calculate a confidence interval for p, and then if the confidence interval only contains values greater than 0.50, then we can reject the null hypothesis \(H_0: p = 0.50\) in favor of the alternative hypotheses \(H_A: p > 0.50\). In practice, most statisticians do both, that is, conduct and report the results of both the hypothesis test and the confidence interval.

Answer

The research question involves the study of one group, namely college students. The research question involves a binary response... either a student does or does not have an E in his or her last name. The research question is a "what is it?" question, and therefore involves calculating a confidence interval. If p is the (unknown) proportion of college students who have an E in their last name, then we are specifically interested in estimating p. We can enter the resulting data into Minitab and then ask Minitab to calculate a Z-interval for one proportion for us.

Answer

The research question involves the study of two groups, namely elderly males and elderly females. The research question involves a binary response... either a person does or does not snore. The research question is an "is it this?" question, and therefore involves conducting a hypothesis test. In this case, the question involves determining whether or not the difference in the two proportions \(p__1\) and \(p__2\) is 0. That is, if p₁ is the (unknown) proportion of elderly males who snore, and \(p__2\) is the (unknown) proportion of elderly females who snore, then we are specifically interested in testing the null hypothesis \(H_0: p_1−p_2 = 0\) against the alternative hypotheses \(H_A: p_1−p_2 ≠ 0\). We can enter the resulting data into Minitab and then ask Minitab to conduct either a chi-square test or a Z-test for two proportions for us.

Answer

The research question involves the study of four groups, namely freshmen, sophomores, juniors, and seniors. The research question involves a categorical response... either a person classifies him- or herself as having never smoked, as having smoked a few times, as a regular smoker, or as being completely addicted. The research question involves assessing the independence of the two variables, smoking and semester standing. In summarizing the data, we determine the proportion of freshmen falling into each category of smokers, the proportion of sophomores falling into each category of smokers, the proportion of juniors falling into each category of smokers, and the proportion of seniors falling into each category of smokers. We can enter the resulting data into Minitab and then ask Minitab to conduct a chi-square test for us.

Answer

The research question involves the study of one group, namely college students. The research question involves a continuous response... the length of the pointer finger of a randomly selected college student. The research question is a "what is it?" question, and therefore involves calculating a confidence interval. If \(\mu\) is the (unknown) mean length of the pointer finger of college students, then we are specifically interested in estimating \(\mu\). We can enter the resulting data into Minitab and then ask Minitab to calculate a t-interval for one mean for us.

Answer

The research question involves the study of one group, namely graduating college seniors. The research question involves a continuous response... the score on the Stanford-Binet IQ test. The research question is a "is it this?" question, and therefore involves conducting a hypothesis test. If \(\mu\) is the (unknown) mean IQ score of graduating college seniors, then we are specifically interested in testing the null hypothesis \(H_0: \mu = 115\) against the alternative hypothesis \(H_A: \mu > 115\). We can enter the resulting data into Minitab and then ask Minitab to conduct a t-test for one mean for us.

Answer

The research question involves the study of one group, namely American households. The research question involves a continuous response... annual income (in dollars). The research question is a "is it this?" question, and therefore involves conducting a hypothesis test. At this point, because the response is continuous we could conduct a hypothesis test about the mean or the median. However, because it is well known that the distribution of American incomes is highly skewed, the median is a better measure of the "center" of the income distribution. Therefore, our analysis should probably concern the median. That said, if m is the (unknown) median annual income of American households, then we are specifically interested in testing the null hypothesis \(H_0: m = 40,000\) against the alternative hypothesis \(H_A: m > 40,000\). We can enter the resulting data into Minitab and then ask Minitab to conduct either a sign test or a signed rank test for one median for us.

Answer

The research question involves the study of one group, namely people. Oops, actually if you think about it, the question involves the study of two groups, people, before exercise and people after exercise. Although the research question doesn't specifically suggest this, we should all know by now that it would be a good idea to collect the data in a paired way, that is, to measure the same people before and after exercise. Doing otherwise would introduce needless variability into the data. By measuring the same people twice, of course, removes the independence of the groups, and hence we should be thinking paired, paired, paired.

Because the research question involves a continuous response... the pulse rate, we should be thinking mean, mean, mean or median, median, median. So, we have a paired, paired, paired, mean, mean, mean or a paired, paired, paired, median, median, median. (I've been clearly writing too long today.) At any rate, the research question is clearly a "is it this?" question. It is? Clearly? Well, if \(\mu_D = \mu_{After} − \mu_{Before}\), is the (unknown) mean difference in the pulse rates, then we are specifically interested in testing the null hypothesis \(H_0: \mu_D = 0\) against the alternative hypothesis \(H_A: \mu_D > 0\). We can enter the resulting data into Minitab and then ask Minitab to conduct either a paired t-test for one mean or, alternatively, a sign test or signed-rank test for the median difference. Of course, if we went a step further, we could also ask Minitab to calculate a confidence interval for us, so that we can quantify how different the pulse rates are before and after exercise.

Answer

The research question involves the study of two independent groups, namely that of adult males and females. The research question involves a continuous response... resting pulse rates. The research question is a "is it this?" question, and therefore involves conducting a hypothesis test. If \(\mu_M\) is the (unknown) mean pulse rate of adult males, and \(\mu_F\) is the (unknown) mean pulse rate of adult females, then we are specifically interested in testing the null hypothesis:

\(H_0: \mu_M − \mu_F = 0\)

against the alternative hypothesis:

\(H_A: \mu_M − \mu_F ≠ 0\)

We can enter the resulting data into Minitab and then ask Minitab to conduct a two-sample t-test for one mean for us. Of course, we should check, as always, for the normality of the data and the equality of the population variances.

Answer

Because we are only interested in learning whether a linear relationship exists between husbands' and wives' heights, and not the nature of the relationship, we would want to conduct a correlation analysis. We can use Minitab's Stat >> Basic Statistics >> Correlation... command to test the null hypothesis:

\(H_0 : \rho = 0\)

against the alternative hypothesis:

\(H_A : \rho \ne 0\)

Answer

If x denotes the number of times a randomly selected college student goes out to party in one month, and y = the student's grade point average, then we'd be interested in estimating the slope and intercept parameters in the linear regression equation:

\(\mu_y=\alpha + \beta x\)

Of course, that's assuming that the relationship is indeed a linear relationship, but that could be verified when doing the analysis. We could use Minitab's Stat >> Regression >> Regression... command to help complete the analysis.