Test the indicated dlaim about the variances or standard deviations of two popul

Question

Test the indicated dlaim about the variances or standard deviations of two populations Assume that both samples are independent simple random samples from populations having normal distributions. 52) Two types of flares are tested and their burning times are recorded. The summary statistics are 52) given below. Use a 0.05 significance level to test the claim that the burning times for Brand X flares have the same variance as the burning times for Brand Y flares. Brand Y n 41 Brand X n 35 x-19.4 min x 15.1 min s 1.4 min 0.8 min Test statistic: F-3.0625. p-value:.000824 Reject the null hypothesis. There is sufficient evidence to w arrant rejection of the claim that the burning times for Brand X flares have the same variance as the burning times for Brand Y flares. Test statistic: F3.0625. p-value: 8.24 Reject the null hypothesis. There is sufficient evidence to w arrant rejection of the claim that the burning times for Brand X flares have the same variance as the burning times for Brand Y flares.

Explanation / Answer

Question 52

Part A

Solution:

Here, we have to use F test for population variances.

The null and alternative hypotheses are given as below:

H0: ?12 = ?22

H1: ?12 ? ?22

This is a two tailed test.

The test statistic formula is given as below:

F = S12 / S22

We are given

S1 = 1.4

S2 = 0.8

n1 = 35

df1 = 35 – 1 = 34

n2 = 41

df2 = 41 – 1 = 40

? = 0.05

F = 1.4^2/0.8^2

F = 1.96/ 0.64

F = 3.0625

P-value = 0.0008 (by using Ti-83 calculator)

P-value < ? = 0.05

So, we reject the null hypothesis H0

There is sufficient evidence to conclude that the population variances for brand X and brand Y are not same.

Part B

Here, we have to use F test for population variances.

The null and alternative hypotheses are given as below:

H0: ?12 = ?22

H1: ?12 > ?22

This is a one tailed/upper tailed/right tailed test.

The test statistic formula is given as below:

F = S12 / S22

We are given

S1 = 1.4

S2 = 0.8

n1 = 35

df1 = 35 – 1 = 34

n2 = 41

df2 = 41 – 1 = 40

? = 0.05

F = 1.4^2/0.8^2

F = 1.96/ 0.64

F = 3.0625

P-value = 0.0004 (by using Ti-83 calculator)

P-value < ? = 0.05

So, we reject the null hypothesis H0

There is sufficient evidence to conclude that the population variance of brand X is greater than the population variance of brand Y.

Ti-83 calculator tips:

Press 2ND > DISTR, you will get distribution menu.

Press 9 to select Fcdf( and press ENTER, or you can use arrows for selection.

After selection of Fcdf, put F value, 1000, df1, df2

(We put 1000 as upper bound for F which is needed in Ti-83)

Press ENTER

You will get output for P-value.

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