Question

Consider the following hypothesis test: H 0: 50 H a: > 50 A sample of 50 is used and the population standard deviation is 6. Use the critical value approach to state your conclusion for each of the following sample results. Use = .05. a. With = 52.5, what is the value of the test statistic (to 2 decimals)? Can it be concluded that the population mean is greater than 50? Select b. With = 51, what is the value of the test statistic (to 2 decimals)? Can it be concluded that the population mean is greater than 50? Select c. With = 51.8, what is the value of the test statistic (to 2 decimals)? Can it be concluded that the population mean is greater than 50? Select

Answer 1

Part a:

The calculate value of Z = 2.946 lies in the critical region so the null hypothesis is rejected and alternate hypothesis is accepted that population mean is greater than 50.

Part b:

The calculate value of Z = 1.17851 does not lie in the critical region so the null hypothesis is accepted that population mean is not greater than 50.

Part c:

The calculate value of Z = 2.1213 lies in the critical region so the null hypothesis is rejected and concluded that population mean is greater than 50.

Explanation:

Data as given

H 0: μ ≤ 50 against the claim H a: μ> 50 one tailed test

Z (0.05)= ± 1.645

Sample size n= 50

Using the central limit theorem

Population Standard Deviation= σ=s=6

X= 52.5 , 51 and 51.8

Part a:

Z= x-μ/ s/ √n

Z= 52.5- 50 / 6/ √50

Z= 2.5/0.84853

z= 2.946

The calculate value of Z = 2.946 lies in the critical region so the null hypothesis is rejected and alternate hypothesis is accepted that population mean is greater than 50.

Part b:

Z= x-μ/ s/ √n

Z= 51- 50 / 6/ √50

Z= 1/0.84853

z= 1.17851

The calculate value of Z = 1.17851 does not lie in the critical region so the null hypothesis is accepted that population mean is not greater than 50.

Part c:

Z= x-μ/ s/ √n

Z= 51.8- 50 / 6/ √50

Z= 1.8/0.84853

z= 2.1213

The calculate value of Z = 2.1213 lies in the critical region so the null hypothesis is rejected and concluded that population mean is greater than 50.