## Related questions with answers

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

VerifiedGiven:

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

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 $t>3.078$.

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