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Chapter 10: Problem 11

A local college cafeteria has a self-service soft ice cream machine. The cafeteria provides bowls that can hold up to 16 ounces of ice cream. The food service manager is interested in comparing the average amount of ice cream dispensed by male students to the average amount dispensed by female students. A measurement device was placed on the ice cream machine to determine the amounts dispensed. Random samples of 85 male and 78 female students who got ice cream were selected. The sample averages were \(7.23\) and \(6.49\) ounces for the male and female students, respectively. Assume that the population standard deviations are \(1.22\) and \(1.17\) ounces, respectively. a. Let \(\mu_{1}\) and \(\mu_{2}\) be the population means of ice cream amounts dispensed by all male and female students at this college, respectively. What is the point estimate of \(\mu_{1}-\mu_{2} ?\) b. Construct a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). c. Using the \(1 \%\) significance level, can you conclude that the average amount of ice cream dispensed by male college students is larger than the average amount dispensed by female college students? Use both approaches to make this test.

Short Answer

Expert verified
a. The point estimate of \(渭_1 - 渭_2\) is 0.74 ounces. b. A 95% confidence interval for \(渭_1 - 渭_2\) is (0.37, 1.11) ounces. c. Yes, using both the critical value approach and the p-value approach, we conclude at a 1% significance level that the average amount of ice cream dispensed by male college students is larger than that dispensed by female college students.

Step by step solution

01

Calculation of the Point Estimate

The point estimate of \(渭_1 - 渭_2\) is the difference between the sample means \(x虅_1 - x虅_2\). That is, \(7.23 - 6.49 = 0.74\) ounces.
02

Confidence Interval Construction

A 95% confidence interval for \(渭_1 - 渭_2\) is given by \((x虅_1 - x虅_2) 卤 z(蟽_1/鈭歯_1 + 蟽_2/鈭歯_2)\), where z is the value from the standard normal distribution for the desired confidence level (1.96 for 95%), \(蟽_1, 蟽_2\) are the standard deviations, and \(n_1, n_2\) are the sample sizes. Plugging in the given data, it becomes \(0.74 卤 1.96 * 鈭�((1.22^2/85) + (1.17^2/78))\), which calculates to \(0.74 卤 0.37\). Thus, the confidence interval is \((0.37, 1.11)\) ounces.
03

Hypothesis Testing

We perform the z-test for comparing differences in means of independent samples. The null hypothesis is \(H_0: 渭_1 鈮� 渭_2\) (i.e., males do not take more ice cream on average), and the alternative hypothesis is \(H_1: 渭_1 > 渭_2\) (i.e., males take more ice cream on average). The test statistic is given by \(Z = (x虅_1 - x虅_2) / 鈭�((蟽_1^2/n_1) + (蟽_2^2/n_2))\), which calculates to \(2.679\). For 伪=0.01, the critical value from the standard normal table is 2.33. Since our calculated Z value of 2.679 exceeds the critical value, we reject the null hypothesis.
04

p-value Approach

Alternatively, we can use the p-value approach to test the hypothesis. The p-value for the calculated Z score of 2.679 (using standard normal table) is 0.0036. Since this p-value is less than the given significance level of 0.01, we again reject the null hypothesis, thereby providing strong evidence that males on average dispense more ice cream than females.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Confidence Interval
A confidence interval gives us a range of values, which is likely to contain the true difference between the population means, \(\mu_1 - \mu_2\). For the problem at hand, this means finding a range for the average difference in ice cream amounts dispensed by male and female students.
The formula used here is: \[(x虅_1 - x虅_2) \pm z\left(\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}\right)\]
Where:
  • \(x虅_1\) and \(x虅_2\) are the sample means.
  • \(\sigma_1\) and \(\sigma_2\) are the population standard deviations.
  • \(n_1\) and \(n_2\) are the sample sizes.
  • \(z\) is the z-score for the desired confidence level (here 1.96 for 95%).

Plugging in the numbers gives us a confidence interval of \(0.74 \pm 0.37\), resulting in \( (0.37, 1.11)\). This interval suggests that we can be 95% confident that male students dispense between 0.37 and 1.11 ounces more ice cream than female students.
Point Estimate
The point estimate in statistical analysis provides a single value estimate of a population parameter. In our context, it helps estimate the difference in behavior between groups, like the amount of ice cream consumed by male and female students.
For comparing two sample means, the point estimate is the difference between the means, denoted as \(x虅_1 - x虅_2\). For the question at hand, the calculation is straightforward:
\( 7.23 - 6.49 = 0.74\) ounces.
This point estimate indicates that on average, male students dispense 0.74 ounces more ice cream than female students. It's a simple yet effective way to understand the disparity between average values from two different samples.
Z-test
A z-test is a statistical method used to determine if there's a significant difference between two sample means. It compares the sample data against a null hypothesis, often assumed to reflect no effects or difference.
In our exercise, we apply the z-test to see if males dispense more ice cream than females. The steps are:
  • Define the null hypothesis, \(H_0: \mu_1 \leq \mu_2\), meaning males don't dispensed more ice cream on average.
  • The alternative hypothesis is \(H_1: \mu_1 > \mu_2\).
  • Calculate the test statistic: \[ Z = \frac{x虅_1 - x虅_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} = 2.679 \]
  • Compare this calculated z value to the critical z value from the standard normal distribution (2.33 for \(\alpha = 0.01\)).
  • Since our z value of 2.679 is greater, we reject the null hypothesis, supporting that males dispense more ice cream than females.
Sample Means
Sample means help us make inferences about population means. They represent the average of data collected from a sample, a portion of the total population.
In this exercise, \( x虅_1 = 7.23 \) ounces for male students and \( x虅_2 = 6.49 \) ounces for female students. These are summary statistics derived from random samples of male and female students, respectively.
The sample means (\(x虅_1\) and \(x虅_2\)) are central to the analysis:
  • They form the basis of our point estimate.
  • They are used in determining the confidence interval and test statistic.
While they provide an average value for the sample, they help infer the behavior of the entire population. By law of large numbers, as sample size increases, sample means tend to be a more accurate reflection of population means.

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Most popular questions from this chapter

We wish to estimate the difference between the mean scores on a standardized test of students taught by Instructors \(\mathrm{A}\) and \(\mathrm{B}\). The scores of all students taught by Instructor A have a normal distribution with a standard deviation of 15, and the scores of all students taught by Instructor B have a normal distribution with a standard deviation of 10 . To estimate the difference between the two means, you decide that the same number of students from each instructor's class should be observed. a. Assuming that the sample size is the same for each instructor's class, how large a sample should be taken from each class to estimate the difference between the mean scores of the two populations to within 5 points with \(90 \%\) confidence? b. Suppose that samples of the size computed in part a will be selected in order to test for the difference between the two population mean scores using a \(.05\) level of significance. How large does the difference between the two sample means have to be for you to conclude that the two population means are different? c. Explain why a paired-samples design would be inappropriate for comparing the scores of Instructor A versus Instructor \(\mathrm{B}\).

A random sample of nine students was selected to test for the effectiveness of a special course designed to improve memory. The following table gives the scores in a memory test given to these students before and after this course. $$ \begin{array}{l|lllllllll} \hline \text { Before } & 43 & 57 & 48 & 65 & 81 & 49 & 38 & 69 & 58 \\ \hline \text { After } & 49 & 56 & 55 & 77 & 89 & 57 & 36 & 64 & 69 \\ \hline \end{array} $$ a. Construct a \(95 \%\) confidence interval for the mean \(\mu_{d}\) of the population paired differences, where a paired difference is defined as the difference between the memory test scores of a student before and after attending this course. b. Test at the \(1 \%\) significance level whether this course makes any statistically significant improvement in the memory of all students.

Conduct the following tests of hypotheses, assuming that the populations of paired differences are normally distributed a. \(H_{0}: \mu_{d}=0, \quad H_{1}: \mu_{d} \neq 0, \quad n=26, \quad \bar{d}=9.6, \quad s_{d}=3.9, \quad \alpha=.05\) b. \(H_{0}: \mu_{d}=0, \quad H_{1}: \mu_{d}>0, \quad n=15, \quad \bar{d}=8.8, \quad s_{d}=4.7, \quad \alpha=.01\) c. \(H_{0}: \mu_{d}=0, \quad H_{1}: \mu_{d}<0, \quad n=20, \quad \bar{d}=-7.4, \quad s_{d}=2.3, \quad \alpha=.10\)

A state that requires periodic emission tests of cars operates two emissions test stations, \(\mathrm{A}\) and \(\mathrm{B}\), in one of its towns. Car owners have complained of lack of uniformity of procedures at the two stations, resulting in different failure rates. A sample of 400 cars at Station A showed that 53 of those failed the test; a sample of 470 cars at Station B found that 51 of those failed the test. a. What is the point estimate of the difference between the two population proportions? b. Construct a \(95 \%\) confidence interval for the difference between the two population proportions. c. Testing at the \(5 \%\) significance level, can you conclude that the two population proportions are different? Use both the critical-value and the \(p\) -value approaches.

In parts of the eastern United States, whitetail deer are a major nuisance to farmers and homeowners, frequently damaging crops, gardens, and landscaping. A consumer organization arranges a test of two of the leading deer repellents \(\mathrm{A}\) and \(\mathrm{B}\) on the market. Fifty-six unfenced gardens in areas having high concentrations of deer are used for the test. Twenty-nine gardens are chosen at random to receive repellent \(\mathrm{A}\), and the other 27 receive repellent \(\mathrm{B} .\) For each of the 56 gardens, the time elapsed between application of the repellent and the appearance in the garden of the first deer is recorded. For repellent \(\mathrm{A}\), the mean time is 101 hours. For repellent \(\mathrm{B}\), the mean time is 92 hours. Assume that the two populations of elapsed times have normal distributions with population standard deviations of 15 and 10 hours, respectively. a. Let \(\mu_{1}\) and \(\mu_{2}\) be the population means of elapsed times for the two repellents, respectively. Find the point estimate of \(\mu_{1}-\mu_{2}\). b. Find a \(97 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). c. Test at the \(2 \%\) significance level whether the mean elapsed times for repellents \(\mathrm{A}\) and \(\mathrm{B}\) are different. Use both approaches, the critical-value and \(p\) -value, to perform this test.
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