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	12	7	9	12	10	7	9	9	6	5
Diet B	14	23	12	10	10	19	21	20	18	15

(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 3.6 for 21 measurements. The variance for the Kraft oven was 4.8 for 19 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

1 A) For Diet B:

n1 = 10

For Diet A:

n2 = 10

Null and Alternative hypothesis:

Test statistic:

df = 10+10-2 = 18

P-value: Using excel function = T.DIST.RT(4.597,18)

P-value = 0.0001

Conclusion:

As p= 0.0001< 0.05, we reject the null hypothesis, There is enough evidence to conclude that mean weight gained on diet B was greater than the mean weight gained on diet A at 0.05 significance level.

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1B) Variance in temperature before the thermostat restarted the flame, = 3.6

n1= 21

Variance for the Kraft oven, = 4.8

n2 =19

a) Test statistic:

df1 = 21-1 = 20

df2 = 19-1 = 18

b) P value using excel: =F.DIST.RT(1.33, 20, 18)

= 0.2736