Answer to Question #286165 in Microeconomics for mukti

Question #286165

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?


1
Expert's answer
2022-01-10T10:00:16-0500

"H_0: \\mu = 16 \\\\\n\nH_1: \\mu \u2260 16 \\\\\n\nn = 50 \\\\\n\n\\bar{x} = 15.15 \\\\\n\n\\sigma = 3"

a)

Test-statistic

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

b)

Two-tailed test.

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

"= 2 \\times 0.0225 \\\\\n\n= 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 \u2260 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. Conclution: "\\mu \u2260 16" at 0.05 level of significance.


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