Question

 

For each problem, perform the following steps. Assume that all variables are normally or approximately normally distributed.

1. State the hypothesis and identify the claim.

2. Find the critical value(s).

3. Compute the test value.

4. Make the decision.

5. Summarize the results.

5. The heights (in feet) for a random sample of world famous cathedrals are listed below. In addition, the heights for a sample of the tallest buildings in the world are listed. Is there sufficient evidence at α = 0.05 to conclude that there is a difference in the variances in height between the two groups? [4]

Cathedrals	72	114	157	56	83	108	90	151
Tallest buildings	452	442	415	391	355	344	310	302	209

Answer 1

	Cathedrals	Cathedrals2
	72	5184
	114	12996
	157	24649
	56	3136
	83	6889
	108	11664
	90	8100
	151	22801
Sum =	831	95419

The sample mean is computed as follows:

Also, the sample variance is

Therefore, the sample standard deviation s is

	Tallest buildings	Tallest buildings2
	452	204304
	442	195364
	415	172225
	391	152881
	355	126025
	344	118336
	310	96100
	302	91204
	209	43681
Sum =	3220	1200120

The sample mean is computed as follows:

Also, the sample variance is

Therefore, the sample standard deviation s is

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ1​ = μ2​

Ha: μ1​ ≠ μ2​

This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df = 15 In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

Hence, it is found that the critical value for this two-tailed test is t_c = 2.131, for α=0.05 and df = 15

(3) Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

t = −8.464

(4) Decision about the null hypothesis

Since it is observed that |t| = 8.464 > t_c = 2.131, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p = 0, and since p = 0 < 0.05 , it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean μ1​ is different than μ2​, at the 0.05 significance level.