Answer to Question #286166 in Microeconomics for mukti

Question

Consider the following hypothesis test:

H 0 : μ ≤ 13

H a : μ > 13

A sample of 26 provided a sample mean x̄ = 15 and a sample standard deviation s = 5.32.

a) Compute the value of the test statistic.

b) Use the t distribution table to compute a range for the p-value.

c) At α = 0.01, what is your conclusion?

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

Answer 1

$H_0: \mu ≤ 13 \\ H_1: \mu > 13 \\ n= 26 \\ \bar{x} = 15 \\ s = 5.32$

a) Test-statistic

$t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \\ t = \frac{15-13}{5.32/ \sqrt{26}} = 1.92$

b) Degrees of freedom df = 26-1 = 25

we can find p-value using excel function = T.DIST.RT(1.92,25) = 0.0332

c) α=0.01

p-value > α

Fail to reject the null hypothesis.

Conclusion: $\mu ≤ 13$ at 0.01 level of significance.

d) One-tailed test

Reject H₀ if $t ≤ -t_{crit}$

$d.f. = 25 \\ α=0.01 \\ t_{crit} = 2.485 \\ 1.92> -2.485$

Fail to reject the null hypothesis.

Conclusion: $\mu ≤ 13$ at 0.01 level of significance.