7.4 - Central Limit Theorem

7.4 - Central Limit Theorem

As we saw at the beginning of this lesson, many of the sampling distributions that you have constructed and worked with this semester are approximately normally distributed. The reason behind this is one of the most important theorems in statistics.

Central Limit Theorem

The Central Limit Theorem states that if the sample size is sufficiently large then the sampling distribution will be approximately normally distributed for many frequently tested statistics, such as those that we have been working with in this course: one sample mean, one sample proportion, difference in two means, difference in two proportions, the slope of a simple linear regression model, and Pearson's r correlation.

Over the next few lessons we will examine what constitutes a "sufficiently large" sample size. Essentially, it is determined by the point at which the sampling distribution becomes approximately normal.

In practice, when we construct confidence intervals and conduct hypothesis tests we often use the normal distribution (or t distributions which you'll see next week) as opposed to bootstrapping or randomization procedures in situations when the sampling distribution is approximately normal. This method is preferred by many because z scores are on a standard scale (i.e., mean of 0 and standard deviation of 1) which makes interpreting results more straight forward. 

Drag the slider at the bottom of the graph to see normal curve fit on the randomization plot.

7.4.1 - Hypothesis Testing

7.4.1 - Hypothesis Testing

Five Step Hypothesis Testing Procedure

In the remaining lessons, we will use the following five step hypothesis testing procedure. This is slightly different from the five step procedure that we used when conducting randomization tests. 

  1. Check assumptions and write hypotheses. The assumptions will vary depending on the test. In this lesson we'll be confirming that the sampling distribution is approximately normal by visually examining the randomization distribution. In later lessons you'll learn more objective assumptions. The null and alternative hypotheses will always be written in terms of population parameters; the null hypothesis will always contain the equality (i.e., \(=\)).
  2. Calculate the test statistic. Here, we'll be using the formula below for the general form of the test statistic.
  3. Determine the p-value. The p-value is the area under the standard normal distribution that is more extreme than the test statistic in the direction of the alternative hypothesis.
  4. Make a decision. If \(p \leq \alpha\) reject the null hypothesis. If \(p>\alpha\) fail to reject the null hypothesis.
  5. State a "real world" conclusion. Based on your decision in step 4, write a conclusion in terms of the original research question.

General Form of a Test Statistic

When using a standard normal distribution (i.e., z distribution), the test statistic is the standardized value that is the boundary of the p-value. Recall the formula for a z score: \(z=\frac{x-\overline x}{s}\). The formula for a test statistic will be similar. When conducting a hypothesis test the sampling distribution will be centered on the null parameter and the standard deviation is known as the standard error.

General Form of a Test Statistic
\(test\;statistic=\dfrac{sample\;statistic-null\;parameter}{standard\;error}\)

This formula puts our observed sample statistic on a standard scale (e.g., z distribution). A z score tells us where a score lies on a normal distribution in standard deviation units. The test statistic tells us where our sample statistic falls on the sampling distribution in standard error units.

7.4.1.1 - Video Example: Mean Body Temperature

7.4.1.1 - Video Example: Mean Body Temperature

Research question: Is the mean body temperature in the population different from 98.6° Fahrenheit?

Video Walkthrough

7.4.1.2 - Video Example: Correlation Between Printer Price and PPM

7.4.1.2 - Video Example: Correlation Between Printer Price and PPM

Research question: Is there a positive correlation in the population between the price of an ink jet printer and how many pages per minute (ppm) it prints?

Video Walkthrough

7.4.1.3 - Example: Proportion NFL Coin Toss Wins

7.4.1.3 - Example: Proportion NFL Coin Toss Wins

Research question: Is the proportion of NFL overtime coin tosses that are won different from 0.50?

StatKey was used to construct a randomization distribution:

175 200 150 100 0.42 0.44 0.46 0.48 0.50 null = 0.5 0.52 0.54 0.56 0.58 125 75 50 25 0 Randomization Test for a Proportion Original Sample Count 240 428 0.561 186 428 0.435 Sample Size Proportion Count Sample Size Proportion Randomization Sample Randomization Dotplot of Null hypothesis: p = NFL Coin Flip Wins Overtime Generate 1 Sample Generate 10 Samples Generate 100 Samples Generate 1000 Samples Proportion 0.5 Reset Plot Edit Data Left Tail Two - Tail Right Tail

 

Step 1: Check assumptions and write hypotheses

From the given StatKey output, the randomization distribution is approximately normal.

\(H_0\colon p=0.50\)

\(H_a\colon p \ne 0.50\)

Step 2: Calculate the test statistic

\(test\;statistic=\dfrac{sample\;statistic-null\;parameter}{standard\;error}\)

The sample statistic is the proportion in the original sample, 0.561. The null parameter is 0.50. And, the standard error is 0.024.

\(test\;statistic=\dfrac{0.561-0.50}{0.024}=\dfrac{0.061}{0.024}=2.542\)

Step 3: Determine the p value

The p value will be the area on the z distribution that is more extreme than the test statistic of 2.542, in the direction of the alternative hypothesis. This is a two-tailed test:

Distribution Plot Normal, Mean=0, StDev=1 0.0 0.1 0.2 0.3 0.4 0 X Density -2.54200 0.0055110 0.0055110 2.542

The p value is the area in the left and right tails combined: \(p=0.0055110+0.0055110=0.011022\)

Step 4: Make a decision

The p value (0.011022) is less than the standard 0.05 alpha level, therefore we reject the null hypothesis.

Step 5: State a "real world" conclusion

There is evidence that the proportion of all NFL overtime coin tosses that are won is different from 0.50

 

7.4.1.4 - Example: Proportion of Women Students

7.4.1.4 - Example: Proportion of Women Students

Research question: Are more than 50% of all World Campus STAT 200 students women?

Data were collected from a representative sample of 501 World Campus STAT 200 students. In that sample, 284 students were women and 217 were not women. 

StatKey was used to construct a sampling distribution using randomization methods:

Randomization Dotplot of Null hypothesis: p = Proportion 0.5 Left Tail Two - Tail Right Tail 200 150 0.44 0.46 0.48 0.50 null = 0.5 0.52 0.54 0.56 0.58 100 50 0

Because this randomization distribution is approximately normal, we can find the p value by computing a standardized test statistic and using the z distribution.

Step 1: Check assumptions and write hypotheses

The assumption here is that the sampling distribution is approximately normal. From the given StatKey output, the randomization distribution is approximately normal. 

\(H_0\colon p=0.50\)
\(H_a\colon p>0.50\)

2. Calculate the test statistic

\(test\;statistic=\dfrac{sample\;statistic-hypothesized\;parameter}{standard\;error}\)

The sample statistic is \(\widehat p = 284/501 = 0.567\).

The hypothesized parameter is the value from the hypotheses: \(p_0=0.50\).

The standard error on the randomization distribution above is 0.022.

\(test\;statistic=\dfrac{0.567-0.50}{0.022}=3.045\)

3. Determine the p value

We can find the p value by constructing a standard normal distribution and finding the area under the curve that is more extreme than our observed test statistic of 3.045, in the direction of the alternative hypothesis. In other words, \(P(z>3.045)\):

3.045 0.0011634 Distribution Plot Normal, Mean=0, StDev=1 0.0 0.1 0.2 0.3 0.4 0 X Density

Our p value is 0.0011634

4. Make a decision

Our p value is less than or equal to the standard 0.05 alpha level, therefore we reject the null hypothesis.

5. State a "real world" conclusion

There is evidence that the proportion of all World Campus STAT 200 students who are women is greater than 0.50.

7.4.1.5 - Example: Mean Quiz Score

7.4.1.5 - Example: Mean Quiz Score

Research question: Is the mean quiz score different from 14 in the population?

StatKey was used to construct a randomization distribution:

Randomization Test for a Mean Original Sample Generate 1 Sample Generate 10 Samples Generate 100 Samples Generate 1000 Samples 14 Reset Plot Edit Data Upload File Change Column(s) Show Data Table Custom Dataset Left Tail Two - Tail Right Tail 120 100 80 60 40 20 30 20 10 0 0 13.5 12.0 12.5 13.0 13.5 14.0 14.5 15.0 13.6 13.7 13.8 13.9 14.0 14.1 14.2 14.3 14.4 14.5 null = 14 13.746 n = 41, mean = 13.746 median = 13.98, stdev = 0.098

Step 1: Check assumptions and write hypotheses

From the given StatKey output, the randomization distribution is approximately normal.

\(H_0\colon \mu = 14\)

\(H_a\colon \mu \ne 14\)

Step 2: Calculate the test statistic

\(test\;statistic=\dfrac{sample\;statistic-null\;parameter}{standard\;error}\)

The sample statistic is the mean in the original sample, 13.746 points. The null parameter is 14 points. And, the standard error, 0.142, can be found on the StatKey output.

\(test\;statistic=\dfrac{13.746-14}{0.142}=\dfrac{-0.254}{0.142}=-1.789\)

Step 3: Determine the p value

The p value will be the area on the z distribution that is more extreme than the test statistic of -1.789, in the direction of the alternative hypothesis:

Distribution Plot Normal, Mean=0, StDev=1 0.0 0.1 0.2 0.3 0.4 0 X Density -1.789 0.0368074 0.0368074 1.78900

This was a two-tailed test. The p value is the area in the left and right tails combined: \(p=0.0368074+0.0368074=0.0736148\)

Step 4: Make a decision

The p value (0.0736148) is greater than the standard 0.05 alpha level, therefore we fail to reject the null hypothesis.

Step 5: State a "real world" conclusion

There is not enough evidence to state that the mean quiz score in the population is different from 14 points. 

7.4.1.6 - Example: Difference in Mean Commute Times

7.4.1.6 - Example: Difference in Mean Commute Times

Research question: Do the mean commute times in Atlanta and St. Louis differ in the population? 

StatKey was used to construct a randomization distribution:

21.97 Randomization Test for a Difference in Means Randomization method Original Sample Generate 1 Sample Generate 10 Samples Generate 100 Samples Generate 1000 Samples Reset Plot Edit Data Upload File Change Column(s) Show Data Table Reallocate Groups Commute Time (Atlanta vs. St. Louis) Left Tail Two - Tail Right Tail 100 80 60 40 20 Atlanta St. Louis Atlanta St. Louis 0 -4.0 29.11 25 50 75 100 125 150 175 24.458 26.622 25 50 75 100 125 150 175 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 null = 0 Randomization Dotplot of x 1 - x 2 , Null hypothesis: μ 1 = μ 2 x 1 - x 2 = 7.14, n 1 = 500, n 2 = 500 Randomization Sample x 1 - x 2 = -2.16, n 1 = 500, n 2 = 500 Show Data Table samples = 5000 mean = -0.040 std. error = 1.136

Step 1: Check assumptions and write hypotheses

 From the given StatKey output, the randomization distribution is approximately normal.

\(H_0: \mu_1-\mu_2=0\)

\(H_a: \mu_1 - \mu_2 \ne 0\)

Step 2: Compute the test statistic

\(test\;statistic=\dfrac{sample\;statistic - null \; parameter}{standard \;error}\)

The observed sample statistic is \(\overline x _1 - \overline x _2 = 7.14\). The null parameter is 0. And, the standard error, from the StatKey output, is 1.136.

\(test\;statistic=\dfrac{7.14-0}{1.136}=6.285\)

Step 3: Determine the p value

The p value will be the area on the z distribution that is more extreme than the test statistic of 6.285, in the direction of the alternative hypothesis:

-6.28500 6.285 0.0000000 0.0000000 Distribution Plot Normal, Mean=0, StDev=1 0.0 0.1 0.2 0.3 0.4 0 X Density

This was a two-tailed test. The area in the two tailed combined is 0.000000. Theoretically, the p value cannot be 0 because there is always some chance that a Type I error was committed. This p value would be written as p < 0.001.

Step 4: Make a decision

The p value is smaller than the standard 0.05 alpha level, therefore we reject the null hypothesis. 

Step 5: State a "real world" conclusion

There is evidence that the mean commute times in Atlanta and St. Louis are different in the population. 

7.4.2 - Confidence Intervals

7.4.2 - Confidence Intervals

Standard Normal Distribution Method

The normal distribution can also be used to construct confidence intervals. You used this method when you first learned to construct confidence intervals using the standard error method. Recall the formula you used:

95% Confidence Interval
\(sample\;statistic \pm 2 (standard\;error)\)

The 2 in this formula comes from the normal distribution. According to the 95% Rule, approximately 95% of a normal distribution falls within 2 standard deviations of the mean.

The normal curve showing the empirical rule.
µ−2 σ µ−1 σ µ+1 σ µ−3 σ µ+3 σ µ µ+2 σ 68% 95% 99.7%

Using the normal distribution, we can conduct a confidence interval for any level using the following general formula:

General Form of a Confidence Interval
sample statistic \(\pm\) \(z^*\) (standard error)
\(z^*\) is the multiplier

The \(z^*\) multiplier can be found by constructing a z distribution in Minitab.

 

z* Multiplier for a 90% Confidence Interval

What z* multiplier should be used to construct a 90% confidence interval?

For a 90% confidence interval, we would find the z scores that separate the middle 90% of the z distribution from the outer 10% of the z distribution:

Minitab output: Normal distribution showing the values that separate the outer 10% from the inner 90%
0.05 1.64485 -1.64485 0 0.05 0.0 0.1 0.2 0.3 0.4 Density X DistributionPlot Normal,Mean,StDev=1

For a 90% confidence interval, the \(z^*\) multiplier will be 1.64485.

Note: Refer back to 7.3.3 for directions on using Minitab to find multipliers.

7.4.2.1 - Video Example: 98% CI for Mean Atlanta Commute Time

7.4.2.1 - Video Example: 98% CI for Mean Atlanta Commute Time

Construct a 98% confidence interval to estimate the mean commute time in the population of all Atlanta residents.

This example uses a dataset is built in to StatKey: Confidence Interval for a Mean, Median, Std. The dataset is titled 'Atlanta Commute.'

Video Walkthrough

7.4.2.2 - Video Example: 90% CI for the Correlation between Height and Weight

7.4.2.2 - Video Example: 90% CI for the Correlation between Height and Weight

Construct a 90% confidence interval to estimate the correlation between height and weight in the population of all adult men.

Video Walkthrough

7.4.2.3 - Example: 99% CI for Proportion of Women Students

7.4.2.3 - Example: 99% CI for Proportion of Women Students

Scenario: Data were collected from a representative sample of 501 World Campus STAT 200 students. In that sample, 284 students were women and 217 were not women. Construct a 99% confidence interval to estimate the proportion of all World Campus students who are women. 

StatKey was used to construct a sampling distribution using bootstrapping methods:

Left Tail Two - Tail Right Tail Confidence Interval for a Proportion Generate 1 Sample Generate 10 Samples Generate 100 Samples Generate 1000 Samples Edit Data Custom Data Proportion Bootstrap Dotplot of Count 284 501 0.567 Sample Size Proportion Original Sample samples = 5000 mean = 0.567 std. error = 0.022 Reset Plot 125 100 75 200 175 150 50 25 0 0.50 0.52 0.54 0.56 0.58 0.60 0.62 0.64 0.567

Because this distribution is approximately normal, we can approximate the sampling distribution using the z distribution. We will use the standard error, 0.022, from this distribution.

The original sample statistic was \(\widehat p =\frac{284}{501}=0.567\). 

We can find the \(z^*\) multiplier by constructing a z distribution to find the values that separate the middle 99% from the outer 1%:

-2.57583 2.57583 0.005 0.005 Distribution Plot Normal, Mean=0, StDev=1 0.0 0.1 0.2 0.3 0.4 0 X Density

The \(z^*\) multiplier is 2.57583

Recall the general form of a confidence interval: sample statistic \(\pm\) \(z^*\) (standard error) where \(z^*\) is the multiplier. So in this case we have...

\(0.567 \pm 2.57583 (0.022)\)

\(0.567 \pm 0.057\)

\([0.510, 0.624]\)

I am 99% confident that the proportion of all World Campus students who are women is between 0.510 and 0.624

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