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Chapter 11: Problem 10

A recent study compared the time spent together by single- and dual-earner couples. According to the records kept by the wives during the study, the mean amount of time spent together watching television among the single-earner couples was 61 minutes per day, with a standard deviation of 15.5 minutes. For the dual-earner couples, the mean number of minutes spent watching television was 48.4 minutes, with a standard deviation of 18.1 minutes. At the .01 significance level, can we conclude that the single-earner couples on average spend more time watching television together? There were 15 single-earner and 12 dual-earner couples studied.

Short Answer

Expert verified
Fail to reject the null hypothesis, no evidence to conclude single-earner couples spend more time.

Step by step solution

01

State the Hypotheses

We are testing if single-earner couples spend more time watching TV than dual-earner couples. Thus, we set up the null and alternative hypotheses as follows:- Null Hypothesis \(H_0:\) \( \mu_1 = \mu_2 \) (no difference in time spent)- Alternative Hypothesis \(H_a:\) \( \mu_1 > \mu_2 \) (single-earner couples spend more time)
02

Determine the Significance Level

The significance level is given as 0.01. This means we have a 1% risk of concluding that there is a difference when there actually is no difference.
03

Calculate the Test Statistic

We use a two-sample t-test for this comparison. The formula for the t-test statistic is:\[t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]where:- \(\bar{x}_1 = 61\) and \(\bar{x}_2 = 48.4\) are the sample means,- \(s_1 = 15.5\) and \(s_2 = 18.1\) are the standard deviations,- \(n_1 = 15\) and \(n_2 = 12\) are the sample sizes.
04

Substitute Values Into the Formula

Substituting the known values into the formula, we calculate:\[t = \frac{61 - 48.4}{\sqrt{\frac{15.5^2}{15} + \frac{18.1^2}{12}}} = \frac{12.6}{\sqrt{\frac{240.25}{15} + \frac{327.61}{12}}}\]
05

Simplify the Denominator

Calculate the values inside the square root:- \(\frac{240.25}{15} = 16.0167\)- \(\frac{327.61}{12} = 27.3008\)Add these results:\[ 16.0167 + 27.3008 = 43.3175 \]Take the square root:\[ \sqrt{43.3175} \approx 6.58 \]
06

Calculate the t-statistic

Now compute the t-statistic:\[t = \frac{12.6}{6.58} \approx 1.915 \]
07

Determine the Degrees of Freedom

The degrees of freedom for a two-sample t-test can be approximated by the formula:\[ df = \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1 - 1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2 - 1}} \]Substitute the values to approximate the degrees of freedom.
08

Compare the t-statistic to the Critical Value

Using statistical tables or software, find the critical t-value for a right-tailed test at \(\alpha = 0.01\) with the calculated degrees of freedom. The calculated t-value (approximately 1.915) should be compared to the critical t-value.
09

Make the Decision

If the calculated t-statistic is greater than the critical t-value, we reject the null hypothesis. If it is less, we fail to reject the null hypothesis.
10

Conclusion

After comparing the t-statistic to the critical value, the t-statistic is not greater. Therefore, we fail to reject the null hypothesis. We do not have sufficient evidence at the 0.01 significance level to conclude that single-earner couples spend more time watching television together than dual-earner couples.

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Key Concepts

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

t-test
The t-test is a statistical tool used to determine if there is a significant difference between the means of two groups. In this context, it allows us to assess whether single-earner couples spend more TV time together than dual-earner couples. A t-test is particularly useful when sample sizes are small, and the data is approximately normally distributed. It compares the observed difference in means to what we might expect by chance. Using the formula for the t-test statistic, we calculate a value based on group averages, standard deviations, and sample sizes to reach our conclusion on whether any observed difference is significant or not.
null and alternative hypotheses
In hypothesis testing, we begin with statements known as the null and alternative hypotheses. These provide a framework for statistical testing. The null hypothesis (\(H_0\) ) typically suggests no effect or no difference and serves as the statement that we attempt to disprove. For example, \(H_0: \mu_1 = \mu_2\), meaning there is no difference in TV time between single-earner and dual-earner couples. The alternative hypothesis (\(H_a\) ) is what we hope to demonstrate; it's the opposite of the null. For our study, \(H_a: \mu_1 > \mu_2\), suggests single-earner couples watch more TV together than dual-earner couples. The result of the hypothesis test will tell us which of these assumptions we should support based on the data.
significance level
The significance level, denoted by \(\alpha\), represents the risk we are willing to take of making an incorrect conclusion. It is the probability of rejecting the null hypothesis when it is true. A common choice is 0.05, but in our study, we selected a more stringent \(\alpha = 0.01\). This means there is only a 1% chance of concluding a difference when there is none. A smaller alpha level indicates greater confidence in the test results but also requires stronger evidence to reject the null hypothesis. This control ensures that the findings are not due to random variation.
degrees of freedom
Degrees of freedom (df) are a vital concept in hypothesis testing as they account for the sample size and influence the shape of the t-distribution we use for decision making. In a two-sample t-test, the degrees of freedom are calculated to adjust for the sample sizes of both groups and are typically not integers. They can be computed using a rather complex formula: \[ df = \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1 - 1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2 - 1}} \]This calculation is crucial as it determines the critical t-values. The degrees of freedom affect the tail area of the t-distribution, which we rely on to ascertain whether the calculated t-statistic falls within the significant region, ultimately guiding us on whether to reject the null hypothesis.

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Most popular questions from this chapter

Lester Hollar is vice president for human resources for a large manufacturing company. In recent years, he has noticed an increase in absenteeism that he thinks is related to the general health of the employees. Four years ago, in an attempt to improve the situation, he began a fitness program in which employees exercise during their lunch hour. To evaluate the program, he selected a random sample of eight participants and found the number of days each was absent in the six months before the exercise program began and in the six months following the exercise program. Below are the results. At the .05 significance level, can he conclude that the number of absences has declined? Estimate the \(p\) -value. $$ \begin{array}{|lcc|} \hline \text { Employee } & \text { Before } & \text { After } \\ \hline \text { Bauman } & 6 & 5 \\ \text { Briggs } & 6 & 2 \\ \text { Dottellis } & 7 & 1 \\ \text { Lee } & 7 & 3 \\ \text { Perralt } & 4 & 3 \\ \text { Rielly } & 3 & 6 \\ \text { Steinmetz } & 5 & 3 \\ \text { Stoltz } & 6 & 7 \\ \hline \end{array} $$

The president of the American Insurance Institute wants to compare the yearly costs of auto insurance offered by two leading companies. He selects a sample of 15 families, some with only a single insured driver, others with several teenage drivers, and pays each family a stipend to contact the two companies and ask for a price quote. To make the data comparable, certain features, such as the deductible amount and limits of liability, are standardized. The data for the sample of families and their two insurance quotes are reported below. At the .10 significance level, can we conclude that there is a difference in the amounts quoted? $$ \begin{array}{|lcc|} \hline \text { Family } & \begin{array}{c} \text { Midstates } \\ \text { Car Insurance } \end{array} & \begin{array}{c} \text { Gecko } \\ \text { Mutual Insurance } \end{array} \\ \hline \text { Becker } & \$ 2,090 & \$ 1,610 \\ \text { Berry } & 1,683 & 1,247 \\ \text { Cobb } & 1,402 & 2,327 \\ \text { Debuck } & 1,830 & 1,367 \\ \text { DuBrul } & 930 & 1,461 \\ \text { Eckroate } & 697 & 1,789 \\ \text { German } & 1,741 & 1,621 \\ \text { Glasson } & 1,129 & 1,914 \\ \text { King } & 1,018 & 1,956 \\ \text { Kucic } & 1,881 & 1,772 \\ \text { Meredith } & 1,571 & 1,375 \\ \text { Obeid } & 874 & 1,527 \\ \text { Price } & 1,579 & 1,767 \\ \text { Phillips } & 1,577 & 1,636 \\ \text { Tresize } & 860 & 1,188 \\ \hline \end{array} $$

(a) state the decision rule, (b) compute the pooled estimate of the population variance, (c) compute the test statistic, (d) state your decision about the null hypothesis, and (e) estimate the \(p\) -value. The null and alternate hypotheses are: $$ \begin{array}{l} H_{0}: \mu_{1}=\mu_{2} \\ H_{1}: \mu_{1} \neq \mu_{2} \end{array} $$ A random sample of 10 observations from one population revealed a sample mean of 23 and a sample standard deviation of \(4 .\) A random sample of 8 observations from another population revealed a sample mean of 26 and a sample standard deviation of 5 . At the .05 significance level, is there a difference between the population means?

A cell phone company offers two plans to its subscribers. At the time new subscribers sign up, they are asked to provide some demographic information. The mean yearly income for a sample of 40 subscribers to Plan \(A\) is \(\$ 57,000\) with a standard deviation of \(\$ 9,200 .\) For a sample of 30 subscribers to Plan \(B\), the mean income is \(\$ 61,000\) with a standard deviation of \(\$ 7,100 .\) At the .05 significance level, is it reasonable to conclude the mean income of those selecting Plan \(\mathrm{B}\) is larger? What is the \(p\) -value?

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?
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