Question

6.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

Solution-A:

Solution-B:

z statistic

z=xbar-mu/sigma/sqrt(n)

=(96-100)/(12/sqrt(49))

z= -2.333333

P Value ine xcel

=NORM.S.DIST(-2.333333,TRUE)

=0.009815337

2*0.009815337=
0.019630675

p=0.019630675

b) What conclusion can be drawn at the given level of significance α = 0.01.

p=0.019630675

p>0.01

Do not reject Ho
Accept Ho

population mean is 100

What conclusion can be drawn if α = 0.05?

p=0.019630675

p<0.05

reject Ho
Accept Ho

population mean is different than 100

d) What is the p-value of the test

p value,p=0.019630675