The following MINITAB output presents the results of a hypothesis test for a population mean μ.
One-Sample Z : X
Test of mu = 73.5 vs not = 73.5
The assumed standard deviation = 2.3634
Variable N Mean StDev SE Mean 95% 01 Z P
X 145 73.2461 2.3634 0.1963 (72.8614, 73.6308) -1.29 0.196
A. Is this a one-tailed or two tailed test?
B. What is the null hypothesis?
C. What is the P-value?
D. Use the output to compute the P-value for the test of H0 : μ ≥ 73.6 versus H1 : μ < 73.6.

Respuesta :

Answer:

A. This is a two tailed test.

B. H0: μ= 73.5

C. P-value= 0.196

D. The P-Value is 0.460172 for one tailed test .

Step-by-step explanation:

The given data is

Variable   N    Mean        StDev     SE Mean         95%               01 Z      P

X         145      73.2461     2.3634    0.1963 (72.8614, 73.6308) -1.29 0.196

A. Is this a one-tailed or two tailed test?

This is a two tailed test because H0: μ= 73.5 against Ha: μ ≠ 73.5

B. What is the null hypothesis?

H0: μ= 73.5

C. What is the P-value?

P-value= 0.196

D. Use the output to compute the P-value for the test of H0 : μ ≥ 73.6 versus H1 : μ < 73.6.

For z= -1.29  the P-Value is 0.460172 for one tailed test .

The result is not significant at p < 0.05.