Question

. The expected mean of a normal population is 100, and its standard deviation is 12. A sample of 49 measurements gives a sample mean of 96. Using the α = 0.01 level of significance a test is to be made to decide between “population mean is 100” or “population mean is different than 100.” a) State null H0. b) What conclusion can be drawn at the given level of significance α = 0.01. c) What conclusion can be drawn if α = 0.05? d) What is the p-value of the test? e) State the type I and II errors. f) What is probability of type II error when, if mean μ really is 102 and α = 0.05 ?

Answer 1

The hypothesis being tested is:

Null Hypothesis, H0: The population means is equal to 100 i.e µ1 =100

Alternate Hypothesis, Ha: The population means is different than 100 i.e µ1 ≠100

b)

Test Statistic, z = (x - µ)/σ/√n=(96-100)/(12/√49)=-2.35

The level of significance is α=0.01

P value=0.0188

Since p value>α. We fail to reject Null Thypothesis. There is not sufficient evidence to conclude that the population means is different than 100 at 1% level of significance.

c)The level of significance is α=0.05

P value=0.0188

Since p value<α. We reject Null Thypothesis. There is sufficient evidence to conclude that the population means is different than 100 at 5% level of significance.

d) P value=0.0188

e) Type I error

Type 1 error is the rejection of a true null hypothesis I.e The we concluded that the population means is different than 100 which is not true.

Type II error is the non-rejection of a false null hypothesis i.e we conclude that the population means are the same which is not correct.

f) Critical value z0.05=1.96

-1.960   ≤ (x̄ - µo)/σx ≤   1.960                          
92.640   ≤ x̄ ≤   99.360                          
                                  
Type II error = P ( 92.64 ≤ x̄ ≤   99.36)              
       Z1 = (   92.640   -   102   ) /   1.71429   =   -5.460
       Z2 = (   99.360   -   102   ) /   1.71429   =   -1.540
                                  
   so, P(   -5.46≤ Z ≤   -1.54) = P ( Z ≤-1.54) - P ( Z ≤ -5.46)=0.06178-0.000=0.06178