Question

For each of the following tests of hypotheses, determine whether the null hypotheses should be rejected. Use the classical method by finding the lower xL and upper xU limits for two-tail tests (or one of the two for a one-tail test). Additionally, don't forget to use a t-value (rather than a z-value) when n<30.

A) Ho: µ = 6,000 Ha: µ ≠ 6,000; n = 70; x = 6,150; σ = 2,600; α = .01

B) Ho: µ = 50 Ha: µ > 50; n = 24;  x = 55; s = 7; α = .05

C) Ho: µ ≥ 400 Ha: µ < 400; n = 56;  x = 350; s = 62; α =.03

Answer 1

A) For the given hypothesis:

Ho: µ = 6,000

Ha: µ ≠ 6,000;

and given details sample size n = 70, sample mean x = 6,150; population standard deviation σ = 2,600; and at significnce level α = 0.01.

A two-tailed Z-test is conducted since population standard deviation is known and the sample size is greater than 70 hence Z statistic is used to conduct the hypothesis.

Rejection region:

Based on the given significance level the critical scores are calculated using excel formula for normal distribution which is =NORM.S.INV(0.995), thus Zc computed as +/- 2.576

Thus reject Ho if Z >Zc or Z <-Zc.

Test statistic:

Conclusion:

Since Z <Zc hence we fail to reject the null hypothesis.

B) Given that the hypotheses are:

Ho: µ = 50

Ha: µ > 50;

and the given details n = 24;  x = 55; s = 7; α = .05

Based on the hypothesis it will be a right-tailed test. And since the population standard deviation is unknown and the sample size is less than 30 hence we use T-statistic to conduct the hypothesis.

Rejection region:

Based on the given significance level and sample size the critical value of tc is computed using excel formula for t-distribution, the degree of freedom = n-1= 24-1=23.

The formula used is =T.INV(0.95,23), thus tc is computed as 1.714 so reject Ho if t > tc.

Test statistic:

Conclusion:

Since t >tc hence we can reject the null hypothesis and conclude that there is enough evidence to support the claim.

C) Given hypotheses are:

Ho: µ ≥ 400

Ha: µ < 400;

and give details are:

n = 56;  x = 350; s = 62; α =.03

Based on the hypothesis it will be a left-tail test and since the sample population standard deviation is unknown hence we use t-distribution.

Rejection region:

Based on the given significance level and sample size the critical value of tc is computed using excel formula for t-distribution, the degree of freedom = n-1= 56-1=55.

The formula used is =T.INV(0.03,55), thus tc computed as tc = -1.92

So, reject Ho if t <tc

Test statistic:

Conclusion:

Since t >tc hence we cannot reject the null hypothesis and conclude that there is insufficient evidence to support the claim.