H0: =, >, <, or ≠ _______________ year (s). H1: =, >, <, or ≠ _________________ year (s). Since the P-value is less than or equal to, or more than or equal to the significance level, reject or do not reject the null hypothesis. There is or is not sufficient evidence to support the claim that actress are generally younger when they won the award than actors. Since the confidence interval contains only zero, positive number values, or only negative number values, reject or warrent rejection of the null hypothesis. The following data lists the ages of a random selection of actresses when they won an award in the category of Best Actress, along with the ages of actors when theywon in the category of Best Actor. The ages are matched according to the year that the awards were presented. Complete parts (a) and (b) below.Actress (years) 28 31 30 29 36 28 27 40 33 3528 33 48 35 39 40Actor (years)59 413835a. Use the sample data with a 0.05 significance level to test the claim that for the population of ages of Best Actresses and Best Actors, the differences have a meanless than 0 (indicating that the Best Actresses are generally younger than Best Actors).In this example, 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 actress's age minusthe actor's age. What are the null and alternative hypotheses for the hypothesis test?Ho: Hayear(s)H₁: Hayear(s)(Type integers or decimals. Do not round.) Identify the test statistic.t= (Round to two decimal places as needed.)Identify the P-value.P-value =(Round to three decimal places as needed.)What is the conclusion based on the hypothesis test?Since the P-value isthe significance level,actresses are generally younger when they won the award than actors.the null hypothesis. Thereb. Construct the confidence intervall that could be used for the hypothesis test described in part (a). What feature of the confidence interval leads to the sameconclusion reached in part (a)?The confidence interval is year(s)<<(Round to one decimal place as needed.)What feature of the confidence interval leads to the same conclusion reached in part (a)?Since the confidence interval containsyear(s).▼sufficient evidence to support the claim thatthe null hypothesis.

Given data: Actress Actor 28 59 31 41 30 38 29 35 36 28 28 33 27 48 40 35 33 39 35 40 (a) The relevant hypotheses are: Null Hypothesis H0:μd=0 Alternative Hypothesis H1:μd<0 This is a left-tailed test. The test statistic is given by the formula: t=d¯-μdsdn Calculations: Actress Actor difference (d) 28 59 -31 31 41 -10 30 38 -8 29 35 -6 36 28 8 28 33 -5 27 48 -21 40 35 5 33 39 -6 35 40 -5 Mean of the differences is: d¯=-31-10-8-6+8-5-21+5-6-510=-7.9 Standard deviation of the differences is: sd=-31+7.92+-10+7.92+-8+7.92+......+-5+7.9210-1=11.31813 Substitute the values in the formula: t=d¯-μdsdn=-7.9-011.3181310≈-2.21 Thus, the values of the test statistic is -2.21 Degrees of freedom: df=n-1=10-1=9 Use Excel function "=T.DIST(-2.21,9,TRUE)", the p-value obtained is 0.027 Since the p-value is less than or equal to the significance level, reject the null hypothesis. There is sufficient evidence to support the claim that actress are generally younger when they won the award than actors. (b) Since this is a paired test, the confidence level is: 1-2α=1-2*0.05=0.90 The 90% confidence interval for the mean difference is computed using the formula: C.I=d¯±t*sdn where t* is the t-critical value at df=9 and 90% confidence level. Using the t-table, the t-critical value is: 1.833 Substitute the values in the formula: C.I=d¯±t*sdn=-7.9±1.83311.3181310=-7.9±6.560503068≈-14.5, -1.3 [rounded to one decimal place] Thus, the confidence interval is -14.5 years<μd<-1.3 years Since the confidence interval contains only negative number values, reject the null hypothesis.