A sample of 32 observations is selected from a normal population. The sample mea
ID: 3264343 • Letter: A
Question
A sample of 32 observations is selected from a normal population. The sample mean is 66, and the population standard deviation is 5. Conduct the following test of hypothesis using the 0.10 significance level. H0: = 68 H1: 68 rev: 10_12_2016_QC_CS-65155 12. value: 1.00 points Required information
a. Is this a one- or two-tailed test?
b. What is the decision rule?
Reject H0 if z < -1.645 or z > 1.645
Reject H0 if -1.645 < z < 1.645
c. What is the value of the test statistic? (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)
Value of the test statistic
d. What is your decision regarding H0?
Reject H0
Fail to reject H0
e. What is the p-value? (Round z value to 2 decimal places and final answer to 4 decimal places.)
p-value
Explanation / Answer
given n = 32
mean = 68
standard deviation =5
significance level = 0.10 means 90% ci
H0 : µ = 68
Ha : µ not equal= 68 (Null Hypothesis)
(Alternative Hypothesis, also called H1 ) This is a two tailed test,
a) this is a two tailed test
b)Reject H0 if z < -1.645 or z > 1.645
c) tstat = (68-66)/5/srt32 = 2 /(0.8838)= 2.2628
e) p (2.262) = 0.9881.from standard z table
as its two tailed test there fore 2*0.9881=1.9762
so as p= greater than 0.10
do not reject H0
d) fail to reject