Question

NEED SOLVED ASAP!!

The performance is videotaped and analyzed for the total of times the performer(s) have to wait for the audience laughter to subside. Given below are the times in seconds for 15 different shows from the most recent show. Test the claim that the mean wait time is greater than 63.2 seconds (this information is from a performance held in 2013). Based on the results, did this year’s performance seem to do better than the one in 2013? Use a 0.05 level of significance (question adapted from chapter 7, Mario F. Triola, William Goodman, Richard Law and Gerry Labute (2011) Elementary Statistics, 3rd Canadian edition, Pearson, Canada). Be sure to state the null and alternate hypotheses and the conclusions you arrive at.

86

45

44

78

52

79

86

66

61

57

98

44

61

99

87

Answer 1

null hypothesis: Ho:             μ		=	63.2
Alternate Hypothesis: Ha:μ		>	63.2
0.05 level with right tailed test and n-1= 14 df, critical t=			1.76
Decision rule :                   reject Ho if test statistic t>1.761
	population mean μ=	63.2
	sample mean         x=	69.533
	sample size             n=	15
	sample std deviation s=	19.302
	std error sx=s/√n=	4.9837
	test stat t='(x-μ)*√n/s=	1.271

since test statistic does not falls in rejection region we fail to reject null hypothesis
we do not have have sufficient evidence to conclude that the mean wait time is greater than 63.2 seconds