Question

Who talks more - men or women? Researchers equipped random samples of 56 male and 56 female students from a large university with a small device that secretly records sound for a random 30 seconds during each 12.5-minute period over two days. Then they counted the number of words spoken by each subject during each recording period and, from this, estimated how many words per day each subject speaks. The female estimates had a mean of 16,177 words per day with a standard deviation of 7520 words per day. For the male estimates, the mean was 16,569 and the standard deviation was 9108.

(a) Do these data provide convincing evidence of a difference in the average number of words spoken in a day by male and female students at this university? Carry out an appropriate test to support your answer.

(b) Interpret the P-value from part (a) in the context of this study.

Short answer

a) No

b) If the population means are equal, then the probability of obtaining a sample with a mean difference of $16177-16569=-392$ or more extreme is higher than$50\backslash \%.$

Step-by-step solution

Part(a) Step 1: Given Information

$$\begin{gathered} {\overline{x}}_1=16177 \\ {\overline{x}}_2=16569 \\ {s}_1=7520 \\ {s}_2=9108 \\ {n}_1=56 \\ {n}_2=56 \end{gathered}$$

Part(a) Step 2: Explanation

Determine the hypothesis:

$H_0:{\mu }_1={\mu }_2$

$H_a:{\mu }_1\ne {\mu }_2$

Determine the test statistic:

localid="1650518880891" role="math" $$\begin{gathered} t=\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} \\ =\frac{16177-16569}{\sqrt{\frac{{7520}^2}{56}+\frac{{9108}^2}{56}}} \\ \approx -0.248 \end{gathered}$$

Determine the degrees of freedom:

localid="1650518751831" $$\begin{gathered} df=\min \left(n_1-1,n_2-1\right) \\ =\min (56-1,56-1) \\ =55>50 \end{gathered}$$

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of Table B containing the t-value in the row$df=50$:

$P>2\times 0.25=0.50$

If the P-value is less than or equal to the significance level, then the null hypothesis is rejected:

$P>0.05\Rightarrow \text{Fail to reject}{H}_0$

There is not sufficient evidence to support the claim of a difference.

Part(b) Step 1: Given Information

$$\begin{gathered} {\overline{x}}_1=16177 \\ {\overline{x}}_2=16569 \\ {s}_1=7520 \\ {s}_2=9108 \\ {n}_1=56 \\ {n}_2=56 \end{gathered}$$

Part(b) Step 2: Explanation

Determine the hypothesis:

$H_0:{\mu }_1={\mu }_2$

$H_a:{\mu }_1\ne {\mu }_2$

Determine the test statistic:

localid="1650518816454" $$\begin{gathered} t=\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} \\ =\frac{16177-16569}{\sqrt{\frac{{7520}^2}{56}+\frac{{9108}^2}{56}}} \\ \approx -0.248 \end{gathered}$$

Determine the degrees of freedom:

localid="1650518837853" $$\begin{gathered} df=\min \left(n_1-1,n_2-1\right) \\ =\min (56-1,56-1) \\ =55>50 \end{gathered}$$

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of Table B containing the t-value in the row $df=20$:

localid="1650518859387" $$\begin{gathered} P>2\times 0.25=0.50 \\ =50\% \end{gathered}$$

This means that the probability of obtaining a sample with a mean difference of $16177-16569=-392$or more extreme is higher than $50\%$, if the population means are equal.