Question

A city manager wants to know whether the true mean number of temporary secretaries employed per day is 25. If in a random sample of 10 days, the mean number of temporary secretaries employed is 23.5, with a standard deviation s=1.1, is there at the .01 level of significance sufficient evidnece to reject the null hypothesis that M(mule)=25.

1. State Null Hypothesis and Aternative Hypothesis

2. State level of significance.

3. State criterion and test statistic.

4. Calculate.

5. Decision.

Answer 1

1)

ho: mean number of temporary secretaries employed per day is 25. u=25

h1: mean number of temporary secretaries employed per day is NOT 25. u =/= 25

2)

level of significance, alpha = 1%

3)

critical value, t(a/2,n-1)

test statistic, t = (mean-u)/(sd/sqrt(n))
= (23.5-25)/(1.1/sqrt(10))
-4.3122

reject Ho if t > t(a/2,n-1). fail to reject Ho if t < t(a/2,n-1)

4)

mean= 23.50
sd= 1.10
u= 25.00
n= 10.00
alpha= 1%

critical value, t(a/2,n-1)
t(0.01/2,10-1)
3.250

test statistic, t = (mean-u)/(sd/sqrt(n))
= (23.5-25)/(1.1/sqrt(10))
-4.3122

5)
since |t| > t(a/2,n-1), i reject ho and conclud that mean number of temporary secretaries employed per day is NOT 25. u =/= 25