Since the P-value is greater than or less than the significance level, fail to reject or reject the null hypothesis. There is or is not sufficient evidence to support the claim that presidents tend to be taller than their opponents. Since the confidence interval cotains zero, only positive values, or only zero values | reject or warrent rejection of the null hypothesis. A popular theory is that presidential candidates have an advantage if they are taller than their main opponents. Listed are heights (in centimeters) of randomly selectedpresidents along with the heights of their main opponents. Complete parts (a) and (b) below.Height (cm) of President178 186 172 179 184 175Height (cm) of Main Opponent 170 184 167 167 180 183a. Use the sample data with a 0.05 significance level to test the claim that for the population of heights for presidents and their main opponents, the differences have amean greater than 0 cm.In this example, μg 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 president's heightminus their main opponent's height. What are the null and alternative hypotheses for the hypothesis test?Ho: Hd = 0 cmH₁: Hd> 0 cm(Type integers or decimals. Do not round.)Identify the test statistic.t=C(Round to two decimal places as needed.) 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.

The question is about paired t test Given : To find : a ) Test the given claim b ) Confidence intervalFirstly need to find the sample mean and sample std. deviation for the differences ( d ) No. of obs. ( n ) = 6 The sample mean can be find as d¯ = ∑ dn = 236 = 3.833333 And, The formula of sample std. deviation is given by sd = ∑ ( d - d¯ )2n-1 = 228.833336-1sd = 6.765107 a ) Null and alt. hypo : H0 : μd = 0H1 : μd > 0 Test stat : t = d¯ - μdsdn = 3.833333 - 06.7651076t = 1.387962 Rounding value to 2 dec. place as t = 1.39 p value approach : df = n - 1 = 6 - 1 = 5 Since alt. hypo. has > sign, so it is right tailed test p value can be find by using excel command, =T.DIST.RT ( Test stat, df ) So, p value = 0.111618 Rounding value to 3 dec. place as p value = 0.112 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 the P-value is greater than the signif. level, fail to reject the null hypo.. There is not suffi. evid. to support the claim that presidents tend to be taller than their opponents. b ) The formula of CI is given by Lower limit = d¯ - tcritical * sdnUpper limit = d¯ + tcritical * sdnWhere,tcritical : Critical value at given significance level t critical value : df = n - 1 = 6 - 1 = 5 critical value can be find by using excel command, T.INV.2T ( α, df ) So, tcritical = 2.570582 Plugging all these values in formulae of CI Lower limit = 3.833333 - 2.570582 * 6.7651076 = -3.26621Upper limit = 3.833333 + 2.570582 * 6.7651076 = 10.93288 Rounding value to 1 dec. place as Lower limit = -3.3 Upper limit = 10.9 Thus, the CI can be represented as -3.3 < μd < 10.9 Since the Cl contains zero, warrant rejection of the null hypo.