/*! 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 17 A recent study focused on the nu... [FREE SOLUTION] | 91Ó°ÊÓ

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A recent study focused on the number of times men and women who live alone buy take-out dinner in a month. Assume that the distributions follow the normal probability distribution and the population standard deviations are equal. The information is summarized below. $$ \begin{array}{|lcc|} \hline \text { Statistic } & \text { Men } & \text { Women } \\ \hline \text { Sample mean } & 24.51 & 22.69 \\ \text { Sample standard deviation } & 4.48 & 3.86 \\ \text { Sample size } & 35 & 40 \\ \hline \end{array} $$ At the .01 significance level, is there a difference in the mean number of times men and women order take-out dinners in a month? What is the \(p\) -value?

Short Answer

Expert verified
There is no significant difference in means; the p-value is 0.055.

Step by step solution

01

Define the Hypotheses

We start by stating the null and alternative hypotheses. The null hypothesis (H_0) assumes there is no difference in the mean number of take-out dinners ordered by men and women. The alternative hypothesis (H_a) suggests there is a difference.\[H_0: \mu_{Men} = \mu_{Women} \H_a: \mu_{Men} eq \mu_{Women}\]
02

Determine the Test Statistic

We use a two-sample t-test because we are comparing means from two independent samples. The formula for the test statistic is:\[t = \frac{\bar{X}_1 - \bar{X}_2}{S_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}\]where\(\bar{X}_1 = 24.51\), \(\bar{X}_2 = 22.69\), \(n_1 = 35\), \(n_2 = 40\), and \(S_p\) is the pooled standard deviation.
03

Calculate the Pooled Standard Deviation

The pooled standard deviation \(S_p\) is calculated as:\[S_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}} = \sqrt{\frac{(35-1)(4.48)^2 + (40-1)(3.86)^2}{35 + 40 - 2}}\]Calculate to find \(S_p \approx 4.157\).
04

Calculate the Test Statistic

Substitute the values obtained into the test statistic formula:\[t = \frac{24.51 - 22.69}{4.157 \times \sqrt{\frac{1}{35} + \frac{1}{40}}}\]Simplify the expression to get \(t \approx 1.938\).
05

Determine the Critical Value

At a significance level of \(\alpha = 0.01\), and degrees of freedom \(df = n_1 + n_2 - 2 = 73\), we look up the critical value for a two-tailed test from the t-distribution table. The critical value is approximately \(2.64\).
06

Make the Decision

Compare the calculated test statistic \(t \approx 1.938\) with the critical value \(2.64\). Since \(|1.938| < 2.64\), we fail to reject the null hypothesis \(H_0\).
07

Calculate the p-value

Using a t-distribution table or calculator to find the p-value for \(t = 1.938\) and \(df = 73\), we find the p-value is approximately 0.055.
08

Interpret the Result

At the 0.01 significance level, the p-value of 0.055 is greater than 0.01, indicating insufficient evidence to claim a difference in means.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a method used to decide if a hypothesis about a population parameter should be accepted or rejected. In our scenario, it's applied to see if there are differences in the average number of take-out dinners ordered by men and women living alone. We begin by establishing two contradictory hypotheses: the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_a\)). - The null hypothesis (\(H_0: \mu_{Men} = \mu_{Women}\)) assumes no difference exists.- The alternative hypothesis (\(H_a: \mu_{Men} eq \mu_{Women}\)) suggests that there is a difference between the two means. In this method, results are analyzed to determine if there is enough statistical evidence to reject the null hypothesis in favor of the alternative. When applying hypothesis testing, it is important to decide in advance what level of uncertainty you will accept within the test, known as the significance level.
Significance Level
The significance level, denoted as \(\alpha\), is the probability of rejecting the null hypothesis when it is actually true. Essentially, it represents the risk of making a Type I error, which occurs when you mistakenly claim a significant effect. Commonly chosen significance levels include 0.05, 0.01, and 0.10. In our example, a significance level of 0.01 is used. This means there is a 1% risk of concluding a difference between men's and women's take-out habits if no true difference exists. The chosen significance level affects the critical value from the statistical distribution being used, influencing whether you can convincingly reject the null hypothesis. - A lower \(\alpha\), like 0.01, reflects more stringent testing criteria, reducing the chance of a false positive but increasing the chance of a Type II error (failing to detect a difference when one actually exists).- Each increase in the level, such as setting \(\alpha = 0.05\), makes it easier to find significance but also raises the risk of a Type I error.
Pooled Standard Deviation
The pooled standard deviation (\(S_p\)) is a method used to estimate a standard deviation that combines the variability of two or more groups. When you assume equal population variances, it merges these into a single value, offering a more efficient estimator for the populations' shared standard deviation. Calculating the pooled standard deviation involves the formula:\[S_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}}\]This factor becomes crucial in the two-sample t-test, affecting the test statistic's calculation. - Here, - - \(n_1\) and \(n_2\) represent the sample sizes. - \(s_1\) and \(s_2\) are the sample standard deviations.After substituting the values from our exercise, the pooled standard deviation is found to be approximately \(4.157\). This value smooths out differences in variability between the men's and women's samples and is used to compute the t-test value.
p-value Calculation
The p-value is a critical statistic in hypothesis testing that helps determine the significance of your test results. It gives the probability of obtaining test results at least as extreme as the observed results under the null hypothesis. In essence, the p-value lets us understand how "likely" the data observed would appear by random chance if \(H_0\) were true. - A lower p-value (< \(\alpha\)) typically indicates stronger evidence to reject \(H_0\), suggesting significant differences exist.- In our example, the p-value of approximately 0.055 is computed for the test statistic \(t = 1.938\) with 73 degrees of freedom.By comparing the p-value to the significance level, a decision is made:- If the p-value is less than or equal to \(\alpha\), reject the null hypothesis.- Here, the p-value (0.055) is higher than \(\alpha = 0.01\), indicating we fail to reject the null hypothesis. Thus, there isn’t enough evidence at the 1% level to conclude that men and women differ in their frequency of ordering take-out.

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Most popular questions from this chapter

Fry Brothers Heating and Air Conditioning Inc. employs Larry Clark and George Murnen to make service calls to repair furnaces and air-conditioning units in homes. Tom Fry, the owner, would like to know whether there is a difference in the mean number of service calls they make per day. A random sample of 40 days last year showed that Larry Clark made an average of 4.77 calls per day. For a sample of 50 days, George Murnen made an average of 5.02 calls per day. Assume the population standard deviation is 1.05 calls per day for Larry Clark and 1.23 calls per day for George Murnen. At the .05 significance level, is there a difference in the mean number of calls per day between the two employees? What is the \(p\) -value?

Gibbs Baby Food Company wishes to compare the weight gain of infants using its brand versus its competitor's. A sample of 40 babies using the Gibbs products revealed a mean weight gain of 7.6 pounds in the first three months after birth. For the Gibbs brand, the population standard deviation of the sample is 2.3 pounds. A sample of 55 babies using the competitor's brand revealed a mean increase in weight of 8.1 pounds. The population standard deviation is 2.9 pounds. At the .05 significance level, can we conclude that babies using the Gibbs brand gained less weight? Compute the \(p\) -value and interpret it.

Listed below are the 25 players on the opening-day roster of the 2016 New York Yankees Major League Baseball team, their salaries, and fielding positions. Sort the players into two groups, all pitchers (relief and starting) and position players (all others). Assume equal population standard deviations for the pitchers and the position players. Test the hypothesis that mean salaries of pitchers and position players are equal, using the . 01 significance level.

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A sample of 40 observations is selected from one population with a population standard deviation of \(5 .\) The sample mean is \(102 .\) A sample of 50 observations is selected from a second population with a population standard deviation of \(6 .\) The sample mean is 99. Conduct the following test of hypothesis using the .04 significance level. $$ \begin{array}{l} H_{0}: \mu_{1}=\mu_{2} \\ H_{1}: \mu_{1} \neq \mu_{2} \end{array} $$ a. Is this a one-tailed or a two-tailed test? b. State the decision rule. c. Compute the value of the test statistic. d. What is your decision regarding \(H_{0} ?\) e. What is the \(p\) -value?

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