Respuesta :
Answer:
(a) Null Hypothesis, [tex]H_0[/tex] : [tex]\sigma[/tex] = 10 beats per minute
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\sigma\neq[/tex] 10 beats per minute
(b) The value of chi-square test statistics is 35.704.
(c) P-value = 0.4360.
(d) We conclude that the pulse rates of men have a standard deviation equal to 10 beats per minute.
Step-by-step explanation:
We are given that a simple random sample of 36 men from a normally distributed population results in a standard deviation of 10.1 beats per minute.
If the range rule of thumb is applied to that normal range, the result is a standard deviation of 10 beats per minute.
Let [tex]\sigma[/tex] = population standard deviation for the pulse rates of men.
(a) So, Null Hypothesis, [tex]H_0[/tex] : [tex]\sigma[/tex] = 10 beats per minute {means that the pulse rates of men have a standard deviation equal to 10 beats per minute}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\sigma\neq[/tex] 10 beats per minute {means that the pulse rates of men have a standard deviation different from 10 beats per minute}
The test statistics that will be used here is One-sample chi-square test for standard deviation;
T.S. = [tex]\frac{(n-1)\times s^{2} }{\sigma^{2} }[/tex] ~ [tex]\chi^{2}__n_-_1[/tex]
where, s = sample standard deviation = 10.1 beats per minute
n = sample of men = 36
So, the test statistics = [tex]\frac{(36-1)\times 10.1^{2} }{10^{2} }[/tex] ~ [tex]\chi^{2}__3_5[/tex]
= 35.70 4
(b) The value of chi-square test statistics is 35.704.
(c) Also, the P-value of the test statistics is given by;
P-value = P([tex]\chi^{2}__3_5[/tex] > 35.704) = 0.4360
(d) Since the P-value of our test statistics is more than the level of significance as 0.4360 > 0.10, so we have insufficient evidence to reject our null hypothesis as the test statistics will not fall in the rejection region.
Therefore, we conclude that the pulse rates of men have a standard deviation equal to 10 beats per minute.
