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 2 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. Do these data provide convincing evidence at the α=0.05$\frac{305}{1526}=0.200=20.0\%\alpha =0.05$significance level of a difference in the average number of words spoken in a day by all male and all female students at this university?

Short answer

There is no convincing evidence of a difference in the average number of words spoken in a day by all male and all female students at this university.

Step-by-step solution

Given information

Given that,

$$\begin{gathered} {\overline{x}}_1=16569 \\ {\overline{x}}_2=16177 \\ {n}_1=56 \\ {n}_2=56 \\ {s}_1=9108 \\ {s}_2=7520 \\ \alpha =0.05 \end{gathered}$$

The given claim is that there is a disparity in means.

Calculation

Now we must determine the most appropriate hypotheses for a significance test.

As a result, either the null hypothesis or the alternative hypothesis is the claim. According to the null hypothesis, the population proportions are equal. If the claim is the null hypothesis, the alternative hypothesis is the polar opposite of the null hypothesis.

The appropriate hypotheses for this are:

$$\begin{gathered} H_0:{\mu }_1={\mu }_2 \\ {H}_a:{\mu }_{1}notequalto{\mu }_2 \end{gathered}$$

Where we have,

${\mu }_1$is the true daily average number of words spoken by all male students at this university.

${\mu }_2$is the true average number of words spoken by all female students at this university on a given day.

Locate the following test statistics:

$$\begin{gathered} =\frac{16569-16177-0}{\sqrt{\frac{91082}{56}+\frac{{7520}^2}{56}}} \\ =0.248 \end{gathered}$$

The degree of liberty will now be:

$$\begin{gathered} df=min(n_1-1,n_2-1) \\ =min(56-1,56-1) \\ =55 \end{gathered}$$

Since the student's T distribution table in the appendix does not contain the value of $df=55$so we will take the nearest value $df=50$So the $\mathrm{P}$-value will be:

$P>2(0.25)=0.50$

On the other hand using the calculator command: $2\times tcdf(0.248,1\mathrm{E}99,55)$which results in the $P$-values as $0.80506$
And we know that the null hypothesis is rejected if the P-value is less than or equal to the significance level.

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

Therefore, We conclude that there is no convincing evidence that the average number of words spoken per day by all male and female students at this university differs.