Respuesta :
Answer:
The critical value for this hypothesis test is 6.571.
Step-by-step explanation:
In this case the professor wants to determine whether the average number of minutes that a student needs to complete a statistics exam has a standard deviation that is less than 5.0 minutes.
Then the variance will be, [tex]\sigma^{2}=(5.0)^{2}=25[/tex]
The hypothesis to determine whether the population variance is less than 25.0 minutes or not, is:
H₀: The population variance is not less than 25.0 minutes, i.e. σ² = 25.
Hₐ: The population variance is less than 25.0 minutes, i.e. σ² < 25.
The test statistics is:
[tex]\chi ^{2}_{cal.}=\frac{ns^{2}}{\sigma^{2}}[/tex]
The decision rule is:
If the calculated value of the test statistic is less than the critical value, [tex]\chi^{2}_{n-1}[/tex] then the null hypothesis will be rejected.
Compute the critical value as follows:
[tex]\chi^{2}_{(1-\alpha), (n-1)}=\chi^{2}_{(1-0.05),(15-1)}=\chi^{2}_{0.95, 14}=6.571[/tex]
*Use a chi-square table.
Thus, the critical value for this hypothesis test is 6.571.
Answer:
Critical value for this hypothesis test is 6.571
Step-by-step explanation:
We are given that a random sample of 15 students was selected and the sample standard deviation for the time needed to complete the exam was found to be 4.0 minutes.
A professor would like to test the hypothesis that the average number of minutes that a student needs to complete a statistics exam has a standard deviation that is less than 5.0 minutes.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\sigma[/tex] = 5.0 minutes
Alternate Hypothesis, [tex]H_1[/tex] : [tex]\sigma[/tex] < 5.0 minutes
Now, the test statistics used here for testing population standard deviation is;
T.S. = [tex]\frac{(n-1) s^{2} }{\sigma^{2} }[/tex] ~ [tex]\chi^{2} __n_-_1[/tex]
where, s = sample standard deviation = 4.0 minutes
n = sample size = 15
So, test statistics = [tex]\frac{(15-1) 4^{2} }{5^{2} }[/tex] = 8.96
At, 5% level of significance the chi-square table gives critical value of 6.571 at 14 degree of freedom. Since our test statistics is more than the critical value as 8.96 > 6.571 so we have insufficient evidence to reject null hypothesis.
