Research Question: Are more than 80% of American's right handed?

In a sample of 100 Americans, 87 were right handed.

\(np_0 = 100(0.80)=80\)

\(n(1-p_0) = 100 (1-0.80) = 20\)

Both \(np_0\) and \(n(1-p_0)\) are at least 10 so we can use the normal approximation method. 

This is a right-tailed test because we want to know if the proportion is greater than 0.80.

\(H_{0}\colon p=0.80\)
\(H_{a}\colon p>0.80\)

\(\widehat{p}=\dfrac{87}{100}=0.87\), \(p_{0}=0.80\), \(n=100\)

\(z= \dfrac{\widehat{p}- p_0 }{\sqrt{\frac{p_0 (1- p_0)}{n}}}= \dfrac{0.87-0.80}{\sqrt{\frac{0.80 (1-0.80)}{100}}}=1.75\)

Our \(z\) test statistic is 1.75.

This is a right-tailed test so we need to find the area to the right of the test statistic, \(z=1.75\), on the z distribution.

Using Minitab , we find the probability \(P(z\geq1.75)=0.0400592\) which may be rounded to \(p\; value=0.0401\).

\(p\leq .05\), therefore our decision is to reject the null hypothesis

Yes, there is statistical evidence to state that more than 80% of all Americans are right handed.

Research Question: Is the percentage of Creamery customers who prefer chocolate ice cream over vanilla less than 80%?

In a sample of 50 customers 60% preferred chocolate over vanilla.

\(np_0 = 50(0.80) = 40\)

\(n(1-p_0)=50(1-0.80) = 10\)

Both \(np_0\) and \(n(1-p_0)\) are at least 10. We can use the normal approximation method.

This is a left-tailed test because we want to know if the proportion is less than 0.80.

\(H_{0}\colon p=0.80\)
\(H_{a}\colon p<0.80\)

\(\widehat{p}=0.60\), \(p_{0}=0.80\), \(n=50\)

\(z= \dfrac{\widehat{p}- p_0 }{\sqrt{\frac{p_0 (1- p_0)}{n}}}= \dfrac{0.60-0.80}{\sqrt{\frac{0.80 (1-0.80)}{50}}}=-3.536\)

Our \(z\) test statistic is -3.536.

This is a left-tailed test so we need to find the area to the right of our test statistic, \(z=-3.536\).

From the Minitab output above, the p-value is 0.0002031

\(p \leq.05\), therefore our decision is to reject the null hypothesis.

Yes, there is evidence that the percentage of all Creamery customers who prefer chocolate ice cream over vanilla is less than 80%.

If you have data in a Minitab worksheet, then you have what we call "raw data."  This is in contrast to "summarized data" which you'll see on the next page.

In order to use the normal approximation method both \(np_0 \geq 10\) and \(n(1-p_0) \geq 10\). Before we can conduct our hypothesis test we must check this assumption to determine if the normal approximation method or exact method should be used. This must be checked manually. Minitab will not check assumptions for you.

In the example below, we want to know if there is evidence that the proportion of students who are male is different from 0.50.

\(n=226\) and \(p_0=0.50\)

\(np_0 = 226(0.50)=113\) and \(n(1-p_0) = 226(1-0.50)=113\)

Both \(np_0 \geq 10\) and \(n(1-p_0) \geq 10\) so we can use the normal approximation method. 

We could summarize these results using the five-step hypothesis testing procedure:

\(np_0 = 226(0.50)=113\) and \(n(1-p_0) = 226(1-0.50)=113\) therefore the normal approximation method will be used.

 \(H_0\colon p = 0.50\)

 \(H_a\colon p \ne 0.50\)

From the Minitab output, \(z\) = -1.86

From the Minitab output, \(p\) = 0.0625

\(p > \alpha\), fail to reject the null hypothesis

There is NOT enough evidence that the proportion of all students in the population who are male is different from 0.50.

We could summarize these results using the five-step hypothesis testing procedure:

\(np_0=129(0.40)=51.6\) and \(n(1-p_0)=129(1-0.40)=77.4\) both values are at least 10 so we can use the normal approximation method.

\(H_0\colon p = 0.40\)

\(H_a\colon p < 0.40\)

From output, \(z\) = -2.62

From output, \(p\) = 0.004

\(p \leq \alpha\), reject the null hypothesis

There is evidence that the proportion of women in the population who think they are overweight is less than 40%.