Related questions with answers

As part of a study of corporate employees, the director of human resources for PNC Inc. wants to compare the distance traveled to work by employees at its office in downtown Cincinnati with the distance for those in downtown Pittsburgh. A sample of 35 Cincinnati employees showed they travel a mean of 370 miles per month. A sample of 40 Pittsburgh employees showed they travel a mean of 380 miles per month. The population standard deviation for the Cincinnati and Pittsburgh employees are 30 and 26 miles, respectively. At the .05 significance level, is there a difference in the mean number of miles traveled per month between Cincinnati and Pittsburgh employees?

A recent study examined the amount of time that couples with one income and those with two earn. The single-earner couples watched television together on average 61 minutes each day, with a standard deviation of 15.5 minutes, according to the data kept by the spouses during the study. The average amount of time spent viewing television by the dual-earner couples was 48.4 minutes, with a standard deviation of 18.1 minutes. Can we infer that, on average, couples with one earner watch more television together at the.01 significant level? The study included 12 couples with two incomes and 15 couples with one income.

The null and alternate hypotheses are:

H_0:μ_1=μ_2; H_1:μ_1≠μ_2

A random sample of 15 observations from the first population revealed a sample mean of 350 and a sample standard deviation of 12. A random sample of 17 observations from the second population revealed a sample mean of 342 and a sample standard deviation of 15. At the .10 significance level, is there a difference in the population means?

Question

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 singleearner and 12 dual-earner couples studied.

Solution

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Given:

\begin{align*} n_1&=\text{Sample size}=15 \\ n_2&=\text{Sample size}=12 \\ \overline{x}_1&=\text{Sample mean}=61 \\ \overline{x}_2&=\text{Sample mean}=48.4 \\ s_1&=\text{Sample standard deviation}=15.5 \\ s_2&=\text{Sample standard deviation}=18.1 \\ \alpha&=\text{Significance level}=0.01 \end{align*}

Given claim: The means differ

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis and the alternative hypothesis state the opposite of each other. The null hypothesis needs to include an equality.

H_0:\mu_1=\mu_2

H_1: \mu_1 \neq \mu_2

If the alternative hypothesis $H_1$ contains $<$ , then the test is left-tailed.

If the alternative hypothesis $H_1$ contains $>$ , then the test is right-tailed.

If the alternative hypothesis $H_1$ contains $\neq$ , then the test is two-tailed.

Two-tailed

The rejection region of a two-tailed test with $\alpha=0.01$ contains all t-values below the t-value $-t_0$ that has a probability of $0.01/2=0.005$ to its left and all t-values above the t-value $t_0$ that has a probability of $0.01/2=0.005$ to its right.

P(t<-t_0)=P(t>t_0)=\frac{0.01}{2}=0.005

Determine the critical values from the Student’s T distribution table in the appendix in the row with $df=n_1+n_2-2=15+12-2=25$ and in the column with $\alpha=0.01$ (two-tailed).

t=3.078

The rejection region then contains all values smaller than $-3.078$ and all values greater than $3.078$ , thus we reject $H_0$ when $t<-3.078$ or $t>3.078$ .