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 | Female ||----|-------|---------|| 1 | 16,289 | 24,069 || 2 | 26,617 | 13,795 || 3 | 1,457 | 17,768 || 4 | 8,067 | 17,931 || 5 | 18,417 | 13,444 || 6 | 16,054 | 17,307 || 7 | 14,264 | 16,308 || 8 | 24,918 | 18,738 |---**a.** Use a 0.05 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.) **Hypothesis Testing and Confidence Interval Construction****a. Identifying the Test Statistic:**- **t =** (Round to two decimal places as needed.)**Identifying the P-value:**- **P-value =** (Round to three decimal places as needed.)**Conclusion Based on the Hypothesis Test:**- Since the P-value is [ ] the significance level, [ ] the null hypothesis. There [ ] sufficient evidence to support the claim that males speak fewer words in a day than females.**b. Confidence Interval Construction:**- Construct the confidence interval that could be used for the hypothesis test described above: **The confidence interval is** [ ] word(s) < μ₁ < [ ] word(s). (Round to one decimal place as needed.)**Feature of the Confidence Interval Leading to the Same Conclusion:**- What feature of the confidence interval leads to the same conclusion reached in part (a)? - Since the confidence interval contains [ ] the null hypothesis.

The question is about paired t test Given : To find : a ) Test the claim that male speaks fewer than female b ) CIFirstly need to find sample mean and sample std. deviation of the differences No. of pairs of male and female ( n ) = 8 The formula to find sample mean is given by d¯ = ∑ dn = -132778 = -1659.63 And, The formula to find sample std. deviation is given by sd = d - d¯ 2n-1 = 6349155348-1sd = 9523.771 a ) Null and alt. hypo : H0 : μd = 0H1 : μd < 0 Test stat : The formula is given by t = d¯ - μdsdn = -1659.63 - 09523.7718t = -0.49289 Rounding value to 2 dec. place as t = -0.49 p value approach : df = n - 1 = 8 - 1 = 7 Since alt. hypo. has < sign, so it is left tailed test p value can be find by using excel command, =T.DIST ( Test stat, df , TRUE ) So, p value = 0.319558 Rounding value to 3 dec. place as p value = 0.320 Decision rule : If p value < level of signif. ( α ) then reject H0 otherwise fail to reject H0 Given : Level of signif. ( α ) = 0.05 Here, p value > α Hence, fail to reject H0 Conclusion : Since p value is greater than signif. level, fail to reject null hypo. There is not suffi. evid. to support the claim that male speak fewer words in a day than female b ) The formula of CI is given by Lower limit = d¯ - tcritical * sdnUpper limit = d¯ + tcritical * sdnWhere,tcritical : Critical value at given confidnece level t critical value : df = n - 1 = 8 - 1 = 7 critical value can be find by uisng excel command, =T.INV.2T ( α, df ) So, tcritical = 2.364624 Plugging all these values in formulae of CI Lower limit = -1659.63 - 2.364624 * 9523.7718 = -9621.7Upper limit = -1659.63 - 2.364624 * 9523.7718 = 6302.446 Rounding value to 1 dec. place as Lower limit = -9621.7 Upper limit = 6302.4 Thus, -9621.7 < μd < 6302.4 Since CI contains the 0, fail to reject the null hypo.