Question

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.
male 15,194 26,233 1414 7870 19,132 15,436 13,656 25,708
female 24,048 13,776 18,783 18,346 13,017 16,829 15,625 19,343

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, \\( \mu _ { d } \\) 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?
\\( h _ { 0 } : \mu _ { d } = 0 \\) word(s)
\\( h _ { 1 } : \mu _ { d } < 0 \\) word(s)
(type integers or decimals. do not round.)
identify the test statistic.
\\( t = - 0.53 \\) (round to two decimal places as needed.)
identify the p - value.
\\( p \text { - value } = 0.304 \\) (round to three decimal places as needed.)
what is the conclusion based on the hypothesis test?
since the p - value is greater than the significance level, fail to reject the null hypothesis. there is not sufficient evidence to support the claim that males speak fewer words in a day than females.
b. construct the confidence interval that could be used for the hypothesis test described in part (a). what feature of the confidence interval leads to the same conclusion reached in part (a)?
the confidence interval is \\( \square \text { word } ( s ) < \mu _ { d } < \square \text { word } ( s ) \\).
(round to the nearest integer as needed.)

Explanation:

Step1: Determine the formula for the confidence interval

For a paired - t test, the confidence interval for \(\mu_d\) is given by \(\bar{d}-t_{\alpha/2}\frac{s_d}{\sqrt{n}}<\mu_d <\bar{d}+t_{\alpha/2}\frac{s_d}{\sqrt{n}}\). Here, \(n = 8\) (the number of pairs), \(\alpha=0.01\) (significance level), and since it is a one - tailed test in part (a), for the confidence interval (which is two - tailed in the context of hypothesis testing equivalence), \(\alpha/2=0.005\). The degrees of freedom \(df=n - 1=8 - 1 = 7\). From the t - distribution table, \(t_{0.005,7}=3.499\).

Step2: Calculate \(\bar{d}\) and \(s_d\)

First, calculate the differences \(d\):
\(d_1=15194 - 24048=-8854\), \(d_2=26233 - 13776 = 12457\), \(d_3=1414-18783=-17369\), \(d_4=7870 - 18346=-10476\), \(d_5=19132-13017 = 6115\), \(d_6=15436-16829=-1393\), \(d_7=13656 - 15625=-1969\), \(d_8=25708-19343 = 6365\)
\(\bar{d}=\frac{\sum_{i = 1}^{n}d_i}{n}=\frac{-8854 + 12457-17369-10476 + 6115-1393-1969+6365}{8}=\frac{-15124}{8}=-1890.5\)
\(s_d=\sqrt{\frac{\sum_{i = 1}^{n}(d_i-\bar{d})^2}{n - 1}}\)
\(\sum_{i = 1}^{n}(d_i-\bar{d})^2=(-8854 + 1890.5)^2+(12457 + 1890.5)^2+(-17369+1890.5)^2+(-10476 + 1890.5)^2+(6115 + 1890.5)^2+(-1393+1890.5)^2+(-1969+1890.5)^2+(6365 + 1890.5)^2\)
\(\sum_{i = 1}^{n}(d_i-\bar{d})^2=( - 6963.5)^2+(14347.5)^2+( - 15478.5)^2+( - 8585.5)^2+(8005.5)^2+(497.5)^2+( - 78.5)^2+(8255.5)^2\)
\(\sum_{i = 1}^{n}(d_i-\bar{d})^2=48492392.25+205845956.25+239584702.25+73700890.25+64088030.25+247506.25+6162.25+68153780.25\)
\(\sum_{i = 1}^{n}(d_i-\bar{d})^2=699902420\)
\(s_d=\sqrt{\frac{699902420}{7}}\approx9999.3\)

Step3: Calculate the margin of error \(E\)

\(E=t_{\alpha/2}\frac{s_d}{\sqrt{n}}=3.499\times\frac{9999.3}{\sqrt{8}}\approx3.499\times3535.7\approx12371.4\)

Step4: Calculate the confidence interval

\(\bar{d}-E=-1890.5-12371.4=-14261.9\approx - 14262\)
\(\bar{d}+E=-1890.5 + 12371.4=10480.9\approx10481\)

Answer:

The confidence interval is \(-14262\) word(s) \(<\mu_d<10481\) word(s). The feature is that the confidence interval contains \(0\). Since \(0\) is in the confidence interval for \(\mu_d\), we fail to reject the null hypothesis \(H_0:\mu_d = 0\) (which is equivalent to the conclusion in part (a) where we failed to reject the null hypothesis based on the P - value).