4. A study was performed to determine whether men and women have the sae varianc

Question

4. A study was performed to determine whether men and women have the sae variance in sleeping time. Two samples of 25 men and 21 women are selected randomly. The two sample standard deviations of sleeping time were s0.914 and women 1.093. Are there any evidonce to support that men and women have different variance in sleeping time? Assume both the sleeping tie of men and women are ormally distributed and independent of each other (a) Write down Ho and Hi (b) Perform the hypothesis test using = 0.05 (c) Construct the 95% two-sided confidence interval on the ratio of the variances. 2en/o2omen Also explain how this confidence interval confirms your conclusion n part b) 5. Suppose we are interested in the burning rate of a solid propellant used to power aircrew escape systems. Specifically, we are interested in deciding whether or not the mean burning rate is 50 cm/s. We plan to take sample of n = 10 specimens. Further, we know that the standard deviation of the burning rates of all the specimens is = 2.5 and that the burning rates are normally distributed (a) State the appropriate nu and alternative hypotheses. (b) Prior to collecting the data, we have determined a decision rule for this test. We wil reject Ho if our sample a X is less than 48.5 or 51.5. Find the signifi cance level, a, of the test (c) Rather than a sample of size 10, suppose we take a sample of size n 16. Using the same decision criteria, find the significance level of the test (d) Consider Again the original scenario in which we are taking a sample of size n = 10 However, suppose that the true average burning rate is 52 cm/s. Using the samoe decision rule as in (b), find the probability of a Type II Error, () Find the probability of correctly rejecting the nu hypothesis when the true av erage burning rate is 52 cm/s.

Explanation / Answer

Q1. We have to check that variance of men is same as the variance of women .

a. Consider Var(m) is varianc of men and Var(w) is variance of women .

So, Ho : Var(m) = Var(w)

   H1 : Var(m) not equal to Var(w)

We have given n1 = 25 , s1 = 0.914 , n2 = 21 , s2 = 1.093

b. We have to test hypothesis at significance level 0.05

Using TI84 we can do this as :

Press STAT --- > Select TESTS ---- > Scroll down and select E : 2 SampFTest hit ENTER

Here we have to plug values as :

Inout :    select Stats

Sx1 = 0.914

n1 = 25

Sx2 = 1.093

n2 = 21

sigma1 not equal to sigma 2

Select calculate and hit ENTER

It gives you output as :

F = 0.69928

p = 0.40018

Decision rule :    If p value < significance level , then we reject the null hypothesis

                        If p value > significance level , then we fail to reject the null hypothesis

Here p = 0.40018 > 0.05 , so fail to rejct the null hypothesis .

That is variances are equal .

c. Formula for the confidence interval for the ration sigma12 / sigma22 :

S12 / S22    <     sigma12 / sigma22   <   S12 / S22                                                                                            

--------------                                           ----------------

F(1-alpha/2)                                          Falpha/2

From F table , F(1-alpha/2) = 2.4076 and Falpha/2 = 1 / 2.4076 = 0.4154

S12 / S22 = 0.9142 / 1.0932 = 0.699282

So, Interval is 0.699282 / 2.4076 < sigma12 / sigma22 < 0.699282 / 0.4154

                         0.29045 < sigma12 / sigma22 < 1.6836

Ration of S12 / S22 lies between interval , so we fail to reject h null hypothesis .

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