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Listed below are the numbers of words spoken in a day by each member of eight different randomly selected couples. Complete parts (a) and (b) below.

\begin{tabular}{lcccccccc}
\hline
Male & 16,275 & 25,989 & 1,418 & 7,921 & 19,192 & 15,904 & 14,323 & 26,696 \\
\hline
Female & 23,914 & 12,888 & 17,819 & 17,515 & 13,425 & 17,291 & 15,738 & 18,182 \\
\hline
\end{tabular}

a. Use a 0.01 significance level to test the claim that among couples, males speak fewer words in a day than females.

In this example, [tex]$\mu_{d}$[/tex] is the mean value of the differences d for the population of all pairs of data, where each individual difference d is defined as the words spoken by the male minus words spoken by the female. What are the null and alternative hypotheses for the hypothesis test?

[tex]
\begin{array}{l}
H_0: \mu_{d} = 0 \\
H_1: \mu_{d} \ \textless \ 0 \\
\end{array}
[/tex]

(Type integers or decimals. Do not round.)



Answer :

GinnyAnswer
To test the claim that among couples, males speak fewer words in a day than females, we will conduct a paired sample t-test. Here's a detailed step-by-step solution:

### Null and Alternative Hypotheses
First, we need to state the null and alternative hypotheses:

- The null hypothesis [tex]\( H_0 \)[/tex]: The mean difference in the number of words spoken by males and females is zero, i.e., [tex]\( \mu_d = 0 \)[/tex]. This suggests that there is no difference in the number of words spoken between males and females.

- The alternative hypothesis [tex]\( H_1 \)[/tex]: The mean difference in the number of words spoken by males and females is less than zero, i.e., [tex]\( \mu_d < 0 \)[/tex]. This suggests that males speak fewer words than females.

So, the hypotheses are:
[tex]\[ \begin{align*} H_0: \mu_d = 0 \\ H_1: \mu_d < 0 \end{align*} \][/tex]

### Differences Calculation
The differences [tex]\( d \)[/tex] between each pair (words spoken by males - words spoken by females) are:
[tex]\[ \begin{array}{cccccccc} -7639 & 13101 & -16401 & -9594 & 5767 & -1387 & -1415 & 8514 \end{array} \][/tex]

### Calculate the Mean and Standard Deviation of Differences
The mean difference [tex]\( \bar{d} \)[/tex] and the standard deviation [tex]\( s_d \)[/tex] are given:
[tex]\[ \bar{d} = -1131.75 \quad \text{and} \quad s_d = 9931.578593413176 \][/tex]

### Number of Pairs
The number of pairs [tex]\( n \)[/tex] is:
[tex]\[ n = 8 \][/tex]

### Calculate the T-Statistic
The t-statistic is calculated as:
[tex]\[ t = \frac{\bar{d} - \mu_{d_0}}{s_d / \sqrt{n}} \][/tex]
Since the null hypothesis [tex]\( H_0 \)[/tex] states that [tex]\( \mu_d = 0 \)[/tex], we have:
[tex]\[ t = \frac{-1131.75}{9931.578593413176 / \sqrt{8}} \approx -0.3223 \][/tex]

### Calculate the P-Value
The p-value for a one-tailed t-test with [tex]\( n-1 = 7 \)[/tex] degrees of freedom and the t-statistic of [tex]\( -0.3223 \)[/tex] is given as:
[tex]\[ p \approx 0.7566 \][/tex]

### Conclusion
We compare the p-value to our significance level [tex]\( \alpha = 0.01 \)[/tex]:

- If [tex]\( p \leq \alpha \)[/tex], we reject the null hypothesis.
- If [tex]\( p > \alpha \)[/tex], we fail to reject the null hypothesis.

Here, [tex]\( p = 0.7566 \)[/tex] which is much greater than [tex]\( 0.01 \)[/tex].

Therefore, we do not have enough evidence to reject the null hypothesis. We fail to reject the null hypothesis and thus cannot conclude that males speak fewer words in a day than females.

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