Question

1A) Twenty laboratory mice were randomly divided into two groups of 10. Each group was fed according to a prescribed diet. At the end of 3 weeks, the weight gained by each animal was recorded. Do the data in the following table justify the conclusion that the mean weight gained on diet B was greater than the mean weight gained on diet A, at the α = 0.05 level of significance? Assume normality. (Use Diet B - Diet A.)

Diet A	6	9	7	13	11	13	8	11	6	14
Diet B	21	21	12	9	21	14	9	16	10	23

(a) Find t. (Give your answer correct to two decimal places.)


(b) Find the p-value. (Give your answer correct to four decimal places.)

1B) A bakery is considering buying one of two gas ovens. The bakery requires that the temperature remain constant during a baking operation. A study was conducted to measure the variance in temperature of the ovens during the baking process. The variance in temperature before the thermostat restarted the flame for the Monarch oven was 2.2 for 23 measurements. The variance for the Kraft oven was 3.2 for 21 measurements. Does this information provide sufficient reason to conclude that there is a difference in the variances for the two ovens? Assume measurements are normally distributed and use a 0.02 level of significance.

(a) Find F. (Give your answer correct to two decimal places.)


(b) Find the p-value. (Give your answer correct to four decimal places.)

Answer 1

1A) Assuming Normality, to test the equality of means, we first need to test the homogeneity of variance using F test:

To test: Vs   

The test statistic is given by:

where and denotes the sample variances of A and B computed using the formula:

, where 9.8 is the sample mean of Diet A

and , where 15.6 is the sample mean of Diet A

Comparing the F statistic with the critical value at 5% significance level for (10-1,10-1) = (9,9) df:

Since F = 3.38 > 3.18, we do not have sufficient evidence to support the null hypothesis.We may reject H₀ at 5% level.Hence, the we cannot assume that the variances are equal.

To test:

Vs   

where is the mean difference between Diet B and Diet A.

The Welch t test statistic is given by:

  

= 2.91

Comparing the test statistic for

= 14 df (approx.)

Since, t = 2.908 > 1.761, we may reject H₀ at 5% level.

(b)

To find the p value of the test, look for the range of values that covers 2.908 in the row corresponding to 14 df:

Hence, 0.005 < p value < 0.01

To find the exact p value, using excel function:

p value = 0.0057

Since the test is significant, we may conclude that the mean weight gained on diet B was greater than the mean weight gained on diet A.

1B)

To test: Vs   

The test statistic is given by:

b) To find the p-value:

We get p-value = 0.1984. Since the p value = 0.198 > 0.02, we have sufficient evidence to support the claim. We fail to reject H₀. We may conclude that this information does not provide sufficient reason to conclude that there is a difference in the variances for the two ovens.