Refer to the data above Do NOT use the “finite population” corrections for stand
ID: 3312015 • Letter: R
Question
Refer to the data above
Do NOT use the “finite population” corrections for standard deviation calculation in this problem. Just assume we’re dealing with an infinite population.
Provide the formal null and alternative hypotheses for a hypothesis test of the question of whether or not the mean administrator salary is really $85,000.
Assume that the underlying population of administrator salaries is normally distributed.
Conduct this hypothesis test based on the 95% significance level. What is the p-value? Do you reject or fail to reject your null hypothesis? Why?
Create a 99% confidence interval for the mean administrator salary. Explain in a sentence, in layman’s terms, what this confidence interval implies.
If I told you that the underlying population of administrator salaries was NOT normally distributed, how would this change your response to “b” and “c”?
If I told you that the underlying population of administrator salaries was NOT normally distributed, BUT, your final calculations (p-value for “b”, confidence interval for “c”) were based on double the sample size (a sample of 40 administrators, rather than 20), would this change your response to “d”? If so, why? If not, why not?
You may ignore the above questions and return to the raw data at the top..
The university system boasts that over 50% of its administrators have master’s degrees. Provide the formal null and alternative hypotheses for a test of this question. Is this a one or two-tailed hypothesis?
Do you have a large enough sample to conduct a hypothesis test? Explain with a calculation why or why not.
If your answer to part “g” is yes, then conduct this hypothesis test based on the 95% significance level. Do you reject or fail to reject your null hypothesis? Why?
Salary ($) Master's Degree (1 = Yes) 100000 1 76000 0 97200 0 90700 1 101800 1 88700 0 84200 0 97600 1 99400 0 84300 0 78700 0 102700 1 103400 0 83800 0 111300 1 94700 0 69200 0 85400 1 61500 0 108800 1Explanation / Answer
Provide the formal null and alternative hypotheses for a hypothesis test of the question of whether or not the mean administrator salary is really $85,000.
Answer : H0 : = $ 85000
Ha : $ 85000
Assume that the underlying population of administrator salaries is normally distributed.
Conduct this hypothesis test based on the 95% significance level. What is the p-value? Do you reject or fail to reject your null hypothesis? Why?
Here sample mean x = $ 90970
Sample standard deviation s = $ 13148.83
standard error of sample mean se0 = s/ sqrt(n) = 13148.83/ sqrt(20) = $ 2940.17
Test statistic
t = (90970 - 85000)/ 2940.17 = 2.03
here dF = 19 and alpha = 0.05
tcritical = 2.093
so here t < tcritical so we shalln't reject the null hypothesis and can conclude that true population salary is $ 85000
Create a 99% confidence interval for the mean administrator salary. Explain in a sentence, in layman’s terms, what this confidence interval implies.
99% confidence interval = x +- tdf,0.01 se0
= 90970 + 2.8609 * 2940.167
= ($82558.47, $99381.52)
Here this confidence means that if randomly taken samples of size 20 repeatedly then there is 95% chance that the sample will lie in the given confidence interval.
If I told you that the underlying population of administrator salaries was NOT normally distributed, how would this change your response to “b” and “c”?
Yes, as here n < 30 which means that we will not use the t distrbution here as test statistics so we will use the non parametric tests. The responses to "b" and " c" will change in this case.
If I told you that the underlying population of administrator salaries was NOT normally distributed, BUT, your final calculations (p-value for “b”, confidence interval for “c”) were based on double the sample size (a sample of 40 administrators, rather than 20), would this change your response to “d”? If so, why? If not, why not?
Here sample size has been increased more than 30. So, we can assume a normal proportion for sample mean. som that would not change response to d.
You may ignore the above questions and return to the raw data at the top..
The university system boasts that over 50% of its administrators have master’s degrees. Provide the formal null and alternative hypotheses for a test of this question. Is this a one or two-tailed hypothesis?
Here Hypothesis are
H0 : p = 0.50
Ha : p > 0.50
Here p is the proportion people who have master degree.
The test is upper one tailed test.
Do you have a large enough sample to conduct a hypothesis test? Explain with a calculation why or why not.
here sample size n = 20
Here sample proportion p= 8/20 = 0.4
so np = 8 > 5 and nq = 12 > 5 so yes sample size is large enough to do hypothesis testing.
If your answer to part “g” is yes, then conduct this hypothesis test based on the 95% significance level. Do you reject or fail to reject your null hypothesis? Why?
Here
Z = (p^- 0.5)/ se0 = (0.4 - 0.5)/ sqrt [0.5 * 0.5/20] = -0.1/ 0.118 = - 0.85
so p- vlue = Pr(Z < -0.85) = 0.1976 > 0.05
so we shalllnot reject the null hypothesis and can conclude that there is university claim that over 50% of its administrators have master’s degrees is false.