Answer to Question #286310 in Microeconomics for MD Tausif zaman

Question

2) Consider the following hypothesis test:

H 0 : μ = 16

H a : μ ≠ 16

A sample of 50 provided a sample mean of 15.15. The population standard deviation is 3.

a) Compute the value of the test statistic.

b) What is the p-value?

c) Write the rejection rule using the p-value. Using α = 0.05, what is your conclusion?

d) Write the rejection rule using the critical value. What is your conclusion?

Answer 1

$H_0: \mu = 16 \\ H_1: \mu ≠ 16 \\ n = 50 \\ \bar{x} = 15.15 \\ \sigma = 3$

a)

Test-statistic

$Z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \\ Z = \frac{15.15 -16}{3 / \sqrt{50}} = -2.003$

b)

Two-tailed test.

P-value = 2P(Z< -2.003)

$= 2 \times 0.0225 \\ = 0.045$

The EXCEL formula to find the p-value for a two-tailed z-test is “=2*(NORM.S.DIST(-2.003, TRUE))”.

=0.045

c) α = 0.05

If the p-value is less than 0.05, we reject the null hypothesis.

0.045<0.05

Reject H₀. Conclusion: $\mu\not = 16$ at 0.05 level of significance.

d) α = 0.05

Reject H₀ if Z ≤ -1.96 or Z ≥ 1.96

-2.003 < -1.96

Reject H₀.

Conclusion: $\mu ≠ 16$ at 0.05 level of significance.