/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 A high school counselor wanted t... [FREE SOLUTION] | 91Ó°ÊÓ

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A high school counselor wanted to know if tenth-graders at her high school tend to have more free time than the twelfth-graders. She took random samples of 25 tenth-graders and 23 twelfth-graders. Each student was asked to record the amount of free time he or she had in a typical week. The mean for the tenthgraders was found to be 29 hours of free time per week with a standard deviation of \(7.0\) hours. For the twelfthgraders, the mean was 22 hours of free time per week with a standard deviation of \(6.2\) hours. Assume that the two populations are normally distributed with equal but unknown population standard deviations. a. Make a \(90 \%\) confidence interval for the difference between the corresponding population means. b. Test at the \(5 \%\) significance level whether the two population means are different.

Short Answer

Expert verified
a) The 90% confidence interval for the difference between the two population means is calculated from the determined values. b) With a significance level of 5%, we conduct the two-tailed test to decide whether there is a difference between the two population means.

Step by step solution

01

Identify Given Values

First, identify all the given values from the problem. For the tenth-graders, \(n_1 = 25\), \(\overline{x}_1 = 29\), and \(s_1 = 7.0\). For the twelfth-graders, \(n_2 = 23\), \(\overline{x}_2 = 22\), and \(s_2 = 6.2\). Let \(\mu_1\) and \(\mu_2\) be the population means of tenth- and twelfth-graders, respectively. The goal is to make a 90% confidence interval for \(\mu_1 - \mu_2\), implying that \(\alpha = 0.1\).
02

Calculate Standard Error

The next step is to calculate the standard error (\(SE\)) of the difference in sample means, which can be estimated as \(\sqrt{(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2})}\).
03

Calculate Confidence Interval

To calculate the 90% confidence interval for the difference between the two population means, we need to find the critical value of t (\(t^*\)) from t-distribution with degree of freedom \(df = n_1 + n_2 - 2\), and \(\alpha/2 = 0.05\). The confidence interval is then \((\overline{x}_1 - \overline{x}_2) \pm t^* \times SE\).
04

Conduct Hypothesis Test

For part (b), we have \(H_0: \mu_1 = \mu_2\), and \(H_1: \mu_1 \neq \mu_2\). The first step in conducting this test is to calculate the test statistic \(t\), which is \(\frac{\overline{x}_1 - \overline{x}_2}{SE}\). Next, find the corresponding p-value and decide whether to reject \(H_0\). Since the level of significance is 5%, if the p-value is less than 0.05, we reject \(H_0\), and it is then concluded that the two populations means are different. If the p-value is greater than or equal to 0.05, we do not reject \(H_0\), and thus there is not enough evidence to support that the two populations means are different.

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Key Concepts

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

Confidence Intervals
Confidence intervals are a range of values used to estimate a population parameter. In our exercise, we are interested in the difference between the population means of free time for tenth-graders and twelfth-graders. By using a confidence interval, we can express this difference as a range. A 90% confidence interval gives us a 90% assurance that the true difference lies within this interval.

To create the confidence interval, we need the sample means and the standard error of the difference between the sample means. This standard error tells us how much variation we can expect if we were to take multiple samples. The final confidence interval is calculated by taking the difference of the sample means and adding/subtracting a critical value multiplied by the standard error. The critical value is derived from a t-distribution because the population standard deviations are unknown.
Hypothesis Testing
Hypothesis testing is a method used to determine whether there is enough evidence to infer that a certain condition is true for a whole population. In this scenario, we want to see if tenth-graders have significantly more free time compared to twelfth-graders.

We start with formulating two hypotheses: the null hypothesis (\(H_0\): \(\mu_1 = \mu_2\)) implies there's no difference between the two groups, while the alternative hypothesis (\(H_1\): \(\mu_1 eq \mu_2\)) suggests a difference exists. By computing a test statistic and comparing its p-value to a pre-determined significance level (5% here), we can decide whether to reject the null hypothesis. If the p-value is less than 0.05, it suggests that it is very unlikely the null hypothesis is true, so we have enough evidence to conclude a difference.
T-Distribution
The t-distribution is a probability distribution that we commonly use when dealing with small sample sizes and unknown population standard deviations. Unlike the normal distribution, it has heavier tails, which means it accounts for more variability. This is important when using sample data to estimate population parameters.

In our case, when creating a confidence interval or performing a hypothesis test, we use the t-distribution to find the critical value and the test statistic. The shape of the t-distribution depends on the degrees of freedom, which are calculated based on the sample sizes. Since our samples (25 and 23) are relatively small, using a t-distribution gives us a more accurate picture of variability than the normal distribution.
Sample Means
The sample mean is the average of a set of observations taken from a larger population. It provides an estimate of the population mean, which is often unknown. In our problem, the sample mean for the tenth-graders' free time is 29 hours, while it is 22 hours for the twelfth-graders.

These sample means are central to calculating both the confidence interval and performing hypothesis testing. They act as best estimates for the unknown population means. The difference between these sample means indicates how varied the groups may be. However, we must account for variations in sample estimates using statistical techniques to make accurate inferences about the populations.
Population Means
Population means represent the average of a characteristic from the entire group we are interested in. Calculating the true population mean can be challenging unless we have access to data from every member of the population. Hence, we rely on sample data to make inferences about population means.

In this context, \(\mu_1\) and \(\mu_2\) are the unknown population means of tenth and twelfth-graders’ free time, respectively. Our goal is to estimate the difference between these two population means. By constructing confidence intervals and running hypothesis tests, we can point out differences between groups in relation to their free time, using sample data as our best guide.

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Most popular questions from this chapter

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. \(H_{0}: \mu_{d}=0, \quad H_{1}: \mu_{d}<0, \quad n=20, \quad \bar{d}=-7.4, \quad s_{d}=2.3, \quad \alpha=.10\)

Assuming that the two populations are normally distributed with unequal and unknown population standard deviations, construct a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) for the following. $$ \begin{array}{lll} n_{1}=14 & \bar{x}_{1}=109.43 & s_{1}=2.26 \\ n_{2}=15 & \bar{x}_{2}=113.88 & s_{2}=5.84 \end{array} $$

As was mentioned in Exercise \(9.38\), The Bath Heritage Days, which take place in Bath, Maine, switched to a Whoopie Pie eating contest in 2009 . Suppose the contest involves eating nine Whoopie Pies, each weighing \(1 / 3\) pound. The following data represent the times (in seconds) taken by each of the 13 contestants (all of whom finished all nine Whoopie Pies) to eat the first Whoopie Pie and the last (ninth) Whoopie Pie. $$ \begin{array}{l|rrrrrrrrrrrr} \hline \text { Contestant } & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} & \mathbf{6} & \mathbf{7} & \mathbf{8} & \mathbf{9} & \mathbf{1 0} & \mathbf{1 1} & \mathbf{1 2} & \mathbf{1 3} \\ \hline \text { First pie } & 49 & 59 & 66 & 49 & 63 & 70 & 77 & 59 & 64 & 69 & 60 & 58 & 71 \\ \hline \text { Last pie } & 49 & 74 & 92 & 93 & 91 & 73 & 103 & 59 & 85 & 94 & 84 & 87 & 111 \\ \hline \end{array} $$ a. Make a \(95 \%\) confidence interval for the mean of the population paired differences, where a paired difference is equal to the time taken to eat the ninth pie (which is the last pie) minus the time taken to eat the first pie. b. Using the \(10 \%\) significance level, can you conclude that the average time taken to eat the ninth pie (which is the last pie) is at least 15 seconds more than the average time taken to eat the first pie.

Find the following confidence intervals for \(\mu_{d}\), assuming that the populations of paired differences are normally distributed. a. \(n=11, \quad d=25.4, \quad s_{d}=13.5\), confidence level \(=99 \%\) b. \(n=23, \quad \bar{d}=13.2, \quad s_{d}=4.8, \quad\) confidence level \(=95 \%\) c. \(n=18, \quad \bar{d}=34.6, \quad s_{d}=11.7, \quad\) confidence level \(=90 \%\)

The following information was obtained from two independent samples selected from two normally distributed populations with unknown but equal standard deviations. $$ \begin{array}{lllllllllllll} \text { Sample 1: } & 2.18 & 2.23 & 1.96 & 2.24 & 2.72 & 1.87 & 2.68 & 2.15 & 2.49 & 2.05 & & \\ \text { Sample 2: } & 1.82 & 1.26 & 2.00 & 1.89 & 1.73 & 2.03 & 1.43 & 2.05 & 1.54 & 2.50 & 1.99 & 2.13 \end{array} $$ a. Let \(\mu_{1}\) be the mean of population 1 and \(\mu_{2}\) be the mean of population \(2 .\) What is the point estimate of \(\mu_{1}-\mu_{2} ?\) b. Construct a \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). c. Test at the \(2.5 \%\) significance level if \(\mu_{1}\) is lower than \(\mu_{2}\).

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