Inference for Quantitative Data: Means
About 10-18\% of the questions on your AP Statistics exam will cover the topic Inference for Quantitative Data: Means.
The t-distribution
Constructing confidence intervals and carrying out hypothesis tests involving means follows the same steps as for proportions. However, the formulas for the standard errors and test statistics all involve the standard deviation, \(\sigma\), of the population in question. Since this is usually unknown, the best we can do is estimate it using the sample standard deviation, \(s\). When this is done, the resulting distribution is no longer normal. Instead, it follows a distribution, called a \(t\) distribution, that resembles a normal distribution but has more of its probability in its tails.
There is not a single \(t\)-distribution. Rather, there is an infinite family of them that depend on a parameter called the degrees of freedom, abbreviated \(d f\). The value of \(d f\) used in a particular situation usually depends on the sizes of the samples involved. As \(d f\) increases, the \(t\) distribution resembles a normal distribution more closely.
Confidence Intervals for One Mean
The conditions that need to be tested when constructing a confidence interval, as usual, are related to independence and normality of the sampling distribution. Specifically, the sample should be random, and when sampling without replacement, should represent less than \(10 \%\) of the population. In addition, the sample size should be at least 30, or the sample data should be symmetric and free of outliers.
If these conditions are met, the standard error is given by \(S E_{\bar{x}}=\frac{s}{\sqrt{n}}\), where \(s\) is the sample standard deviation. Since \(s\) is being used as an estimate for \(\sigma\), the relevant distribution here is a \(t\)-distribution with \(d f=n-1\), where as usual, \(n\) is the sample size. Therefore, the critical value is a \(t\)-value, \(t^*\). The margin of error, in turn, is \(M E=t * \cdot \frac{s}{\sqrt{n}}\). The confidence interval is then constructed as before. Since the point estimate of \(\mu\) is \(\bar{x}\), the interval is
$
(\bar{x}-M E, \bar{x}+M E)=\left(\bar{x}-t * \cdot \frac{s}{\sqrt{n}}, \bar{x}+t * \cdot \frac{s}{\sqrt{n}}\right) .
$
For example, a sociologist is interested in the number of hours per week that adults in the United States spend on social media. He collects a random sample of 78 adults, and surveys them on their social media habits. He finds a sample mean of 4.7 hours per week, with a standard deviation of 1.4 .
To construct a \(90 \%\) confidence interval for the mean number of hours per week spent on social media by all adults in the United States, we begin by checking the conditions. The sample size is greater than 30 , and the data is from a random sample that is certainly less than \(10 \%\) of the population.
The point estimate is \(\bar{x}=4.7\) and the sample standard deviation is \(s=1.4\). Calculating the standard error, we get \(S E_{\bar{x}}=\frac{1.4}{\sqrt{78}} \approx 0.16\). The critical \(t\)-value with \(n-1=77\) degrees of freedom for which \(90 \%\) of the distribution lies between \(-t\) and \(t\) is 1.665 . Therefore, the margin of error is \(M E=1.665 \cdot 0.16=0.2664\). Finally, the \(90 \%\) confidence interval is \((4.7-0.2664,4.7+0.2664)=(4.4336,4.9664)\).
We are \(90 \%\) confident that the mean number of hours per week spent on social media by adults in the United States is between 4.4336 and 4.9664 .
Confidence Intervals for Two Means
Constructing a confidence interval for the difference between two independent means are like those for a single mean. The sampling should be random, consist of less than \(10 \%\) of each population, and both samples should either be at least 30 or symmetric and free of outliers.
The point estimate is \(\bar{x}_1-\bar{x}_2\) and the standard error is \(S E_{\bar{x}_1-\bar{x}_2}=\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\). The critical value comes from a \(t\)-distribution. The correct value for \(d f\) depends on the sample sizes, but is not usually practical to calculate by hand. It can be calculated using technology or will be provided. As usual, the margin of error is given by \(M E=t^* \cdot S E_{\bar{x}_1-\bar{x}_2}\), and the confidence interval is \(\left(\bar{x}_1-\bar{x}_2-M E, \bar{x}_1-\bar{x}_2+M E\right)\).
Hypothesis Tests for Means
When conducting a hypothesis test for a mean, the null hypothesis is \(H_0: \mu=\mu_0\), where \(\mu_0\) is the hypothesized value of the mean. The alternative hypothesis is either \(H_a: \mu<\mu_0\), \(H_a: \mu>\mu_0\), or \(H_a: \mu \neq \mu_0\).
The sample should be random, and if collected without replacement, should be less than \(10 \%\) of the population. If the sample size is not at least 30 , the distribution of the sample data should be symmetric.
The test statistic comes from a \(t\)-distribution with \(d f=n-1\), and is \(t=\frac{\bar{x}-\mu_0}{\frac{s}{\sqrt{n}}}\). The \(p\) value is computed from this distribution, and the interpretation remains the same as usual.
When comparing two means, \(\mu_1\) and \(\mu_2\), the null hypothesis is \(H_0: \mu_1=\mu_2\). The possible alternative hypotheses are \(H_a: \mu_1<\mu_2, H_a: \mu_1>\mu_2\), and \(H_a: \mu_1 \neq \mu_2\). Both samples should meet the conditions given for the case of a single mean.
The test statistic is given by \(t=\frac{\bar{x}_1-\bar{x}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\). As mentioned, the degrees of freedom for this \(t\)-distribution are impractical to compute manually, but can be found using technology or will be given. As usual, a \(p\)-value is found using this distribution, and the conclusion then depends on whether \(p \leq \alpha\) or \(p>\alpha\).
Inference with Matched Pairs
When the two means to be compared are matched pairs, they should be treated as a single mean. First, find the difference of each pair, and then follow the inference procedures for single mean confidence intervals and hypothesis tests, using the differences as the sample.
For example, consider the following table, which shows the heart rates, in beats per minute, of seven people both after and before a 100-meter sprint.
Since these are matched pairs, any inference done with them should first begin with finding differences. Here are the seven values obtained by subtracting the “before” heart rate from the “after” heart rate.
To construct a confidence interval for \(\mu_1-\mu_2\), follow the procedures for constructing a confidence interval for a single mean, and use the sample values given by the differences.
Free Response Tip
Be careful when presented with two samples. The fact that two samples have the same size does not mean they are necessarily matched pairs. There must be a meaningful and direct connection between the samples (such as coming from the same person, as in the heart rate example) for them to be matched pairs. In addition, be certain that the order of subtraction is well-defined and applied consistently for every pair.
Suggested Reading
- \(\quad\) Starnes \& Tabor. The Practice of Statistics. \(6^{\text {th }}\) edition. Chapters 810. New York, NY: Macmillan.
- Larson \& Farber. Elementary Statistics: Picturing the World. \(7^{\text {th }}\) edition. Chapter 6-8. New York, NY: Pearson.
- Bock, Velleman, De Veaux, \& Bullard. Stats: Modeling the World. \(5^{\text {th }}\) edition. Chapters 22-24. New York, NY: Pearson.
- Sullivan. Statistics:Informed Decisions Using Data. \(5^{\text {th }}\) edition. Chapters 9-11. New York, NY: Pearson.
- Peck, Short, \& Olsen. Introduction to Statistics and Data Analysis. \(6^{\text {th }}\) edition. Chapters 9-11. Boston, MA: Cengage Learning.
Sample Inference for Quantitative Data: Means Questions
You have measured the daily water intake of a random sample of students that regularly visit the school recreation center. A \(99 \%\) confidence interval for the mean daily water intake (in fluid ounces) for these students is computed to be \((78,90)\). Which of the following statements is a valid interpretation of this interval?
A. If the sampling procedure were repeated many times, then approximately \(99 \%\) of the sample means would be between 78 and 90 .
B. The probability that the sample mean of the students sampled falls between 78 and 90 is equal to 0.99 .
C. If the sampling procedure were repeated many times, then approximately \(99 \%\) of the resulting confidence intervals would contain the mean daily water intake for all st udents who regularly visit the recreation center.
D. About \(99 \%\) of the sample of students has a daily water intake between 78 and 90 fluid ounces.
E. About \(99 \%\) of the students who regularly visit the recreation center have a daily water intake between 78 and 90 fluid ounces.
▶️Answer/Explanation
Explanation:
The correct answer is \(\mathbf{C}\). There are various ways to interpret “confidence,” and the one presented in this choice is one of them. Choice A is incorrect because you cannot draw such a conclusion about the means of other samples. The interpretation of a confidence interval concerns how many such intervals we would expect to contain the true mean daily water intake. Choice \(\mathrm{B}\) is incorrect because the sample mean is definitely in this interval; in fact, it is the midpoint since the margin of error is added and subtracted from it to generate the interval. So, this probability is 1.00 . Choice \(D\) is incorrect because without having the actual raw data, we cannot determine the number of students for which this is true. There could be a couple of extreme outliers in the data set that could throw off the percentage of those with daily water intake amounts in the interval. Choice \(\mathrm{E}\) is incorrect because you cannot use a single confidence interval to make such a conclusion about the whole population.
A rock gym manager used a random sample of 300 rock climbers to obtain a \(95 \%\) confidence interval for the mean time (in minutes) it takes to complete a difficult climbing route. This interval was \((12.2,14.3)\). If he had used a \(90 \%\) confidence interval instead, the confidence interval would have been
A. wider and would have involved a smaller risk of being incorrect.
B. narrower and would have involved a smaller risk of being incorrect.
C. wider, but it cannot be determined whether the risk of being incorrect would be greater or smaller.
D. wider and would have involved a larger risk of being incorrect.
E. narrower and would have involved a larger risk of being incorrect.
▶️Answer/Explanation
Explanation:
The correct answer is E. As the confidence interval decreases, we are less “confident” that the interval we produce actually contains the true mean. The only way for this to occur is to shrink the interval (or make the target smaller, so to speak). In so doing, there is a greater chance that the interval does not contain the true mean, so the risk of being incorrect is larger.
You have sampled 30 amateur tennis players in New York to determine the mean first-serve ball speed. A \(95 \%\) confidence interval for the mean first-serve ball speed is 85 to 94 miles per hour. Which of the following statements gives a valid interpretation of this interval?
A. If this sampling procedure were repeated several times, \(95 \%\) of the resulting confidence intervals would contain the true mean first-serve ball speed of amateur tennis players in New York.
B. If this sampling procedure were repeated several times, \(95 \%\) of the sample means would be between 85 and 94 .
C. If 100 samples were taken and a \(95 \%\) confidence interval was computed, 5 of the intervals would be included within the confidence interval \((85,94)\).
D. \(95 \%\) of the population of all New York amateur tennis players have a first-serve ball speed between 85 and 94 miles per hour.
E. \(95 \%\) of the 30 amateur tennis players sampled have a mean first-serve ball speed between 85 and 94 miles per hour.
▶️Answer/Explanation
Explanation:
The correct answer is A. There are various ways to interpret “confidence,” one of which is described here as the result of repeated sampling. Choice B is incorrect because 85 and 94 were the endpoints of a specific confidence interval that was linked to a particular sample, and hence a particular sample mean. Different samples will generate different sample means, and the one that was used to produce the given confidence interval could have contained extreme values not characteristic of the overall population. Choice \(\mathrm{C}\) is incorrect because there is no reason to believe that a confidence interval formed from a single sample of 30 players should have this strong of a relationship to confidence intervals formed using different samples. The 100 confidence intervals formed may very well overlap in various ways, but they will not, in general, be included within any given confidence interval. Choice \(D\) is incorrect because different samples will generate different sample means, and the one that was used to produce the given confidence interval could have contained extreme values not characteristic of the overall population. So, there is no way to tell if \(95 \%\) of all such tennis players have a first-serve ball speed in this interval. Choice \(E\) is incorrect because there is no way to determine this from the information provided. The tennis players could have put forth much more of an effort to produce their fastest serves when the data was being collected. So, their individual average first-serve ball speeds may very well be much lower.