Question

Please complete the following problems. Show as much work as you can, and complete the problems as neatly as possible. The x in [x] after each problem denotes the point value.

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

Answer:-

Given That:-

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?

	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

s = 36.053

	Tallest Buildings	Tallest Buildings2
	452	204304
	442	195364
	415	17225
	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 is

s = 77.521

Null and Alternative Hypothesis:

The following null and alternative hypothesis need to be tested.

The corresponding 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.

Rejection Region

based on the information provided, the significnace level is 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 = 2131, for and df = 15

Test Statitstics

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

t = -8.464

Decision about the null hypothesis

since it is observed that |t| = 8.464 > t_c = 2131, 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.

Conclusion

It is concluded that the null hypothesis H0 is rejected. Therefore, there is enough evidence to claim that population mean is deifferent than , at the 0.05 significance level.

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