Question

Use the pulse rates in beats per minute​ (bpm) of a random sample of adult females listed in the data set available below to test the claim that the mean is less than 74 bpm. Use a 0.01

significance level.

Pulse Rate​ (bpm)

60 87

93 92

95 62

92 93

71 67

35 84

54 68

74 36

72 67

100 61

52 42

58 104

67 102

83 95

73 76

38 42

55 85

38 99

84 57

35 51

93 105

40 84

39 103

89 74

64 59

test statistic?

p-value?

-reject/fail to reject, why?

Answer 1

first, we will find the sample information: $\overline{X}=70.98,S=21.52,n=50$

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: μ-74 Ha: μ < 74 This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used (2) Rejection Region Based on the information provided, the significance level is α-0.01, and the critical value for a left- tailed test is te2.405 The rejection region for this left-tailed test is R = t : t <-2.405 (3)_Test Statistics The t-statistic is computed as follows: x-140 70.98-74 s/vn 21.52/50 ー-0.992 4) Decision about the null hypothesis Since it is observed that tー-0.992 > tc-一2.405, itis then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is 20.163 and since p 0.1630.01, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ is less than 74, at the 0.01 significance level.