I would like solutions ONLY TO parts E, F, G and H please. For a random sample o
ID: 3291921 • Letter: I
Question
I would like solutions ONLY TO parts E, F, G and H please.
For a random sample of 16 students, the sample mean time to complete a particular STAT 400 homework assignment was 162 minutes with sample standard deviation of 20 minutes. Assume that the time to complete this assignment is normally distributed. a) Construct a 95% (two-sided) confidence interval for the overall average time to complete this assignment. b) A certain STAT 400 instructor claims that the overall average time to complete this assignment is at most 150 minutes. Perform the appropriate test at a 5% significance level. State the null and alternative hypotheses for this test in terms of the relevant parameter, report the value of the test statistic, the critical value(s), and state your decision. c) Using the t distribution table only, what is the p-value of the test in part (b)? (You may give a range.) Use a computer to find the p-value of the test in part (b). e) Construct a 90% (two-sided) confidence interval for the overall standard deviation of the time needed to complete this assignment. f) Test H_0: sigma lessthanorequalto 17 minutes vs. H_1: sigma > 17 minutes at a 10% significance level. Report the value of the test statistic, the critical value(s), and state your decision. g) Using the chi-square distribution table only, what is the p-value of the test in part (f)? (You may give a range.) h) Use a computer to find the p-value of the test in part (f).Explanation / Answer
Question e)
s = 20
n = 16
Confidence level = 90%
Formula:
sqrt ((n-1)*s^2/ X^2R ) and sqrt ((n-1)*s^2 / X^2L)
X^2R = right side chi-square critical value. We get it from chi-square table as 24.996 at 0.05 level of significance for 15 degrees of freedom.
X^2L = right side chi-square critical value. We get it from chi-square table as 7.261at 0.05 level of significance for 15 degrees of freedom.
sqrt (((16-1)*20^2)/ 24.996) and sqrt (((16-1)*20^2)/ 7.261)
The 90% confidence level is 15.493 and 28.746
Question f)
H0: s 17
H1: s > 17
Test statistics:
Chi-square = (n-1)*(s^2/sigma^2) = (16-1)*(20^2/17^2) = 20.761
Test Statistics = 20.761
We find the critical value for 90% level at 15 degrees of freedom as 8.547.
The test statistics is greater than (20.761>5.547); we fail to reject the null hypothesis.
The population standard deviation is not greater than 17 minutes.
Question g)
From Chi-square table we get the p-value will lie in between 0.1 and 0.9
Question h)
Excel function:
=CHIDIST(20.761,15)
=0.1446
p-value is 0.1446