A sample of 32 observations is selected from a normal population. The sample mea

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

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