Two suppliers manufacture a plastic gear used in a laser printer. The impact strength of these gears measured in foot-pounds is an important characteristic. A random sample of 10 gears from supplier 1 results in 1 = 290 and s1 = 12, while another random sample of 16 gears from the second supplier results in 2 = 321 and s2 = 22.(a) Is there evidence to support the clain that supplier 2 provides gears with higher mean impact strength? Use α = 0.05, and assume that both populations are normally distributed but the variances are not equal. What is the P-value for this test?(b) Do the data support the claim that the mean impact strength of gears from supplier 2 is at least 25 foot - pounds higher than that of supplier 1? Make the same assumptions as in part (a).(c) Construct a confidence interval estimate for the difference in mean impact strength, and explain how this interval could be used to answer the question posed regarding supplier-to-supplier differences.

Since both populations are normally distributed but the sample sizes are less than 30, we will be using the Student's t Distribution for a comparison of the difference of means.

a)

We compute our t-statistic by the formula

[tex] \bf t=\frac{\bar x_1-\bar x_2}{S_v\sqrt{1/n_1+1/n_2}}[/tex]

where

[tex] \bf \bar x_1[/tex] = 290 is the mean of supplier 1

[tex] \bf \bar x_2[/tex] = 321 is the mean of supplier 2

[tex] \bf n_1[/tex] = 10 is the sample size from supplier 1

[tex] \bf n_2[/tex] = 16 is the sample size from supplier 2

[tex] \bf S_v[/tex] is the pooled standard deviation

[tex] \bf S_v=\sqrt{\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}}[/tex]

[tex] \bf s_1[/tex] = 12 is the standard deviation of supplier 1

[tex] \bf s_1[/tex] = 22 is the standard deviation of supplier 2

We have then

[tex] \bf S_v=\sqrt{\frac{(10-1)12^2+(16-1)22^2}{10+16-2}}=18.8812[/tex]

and out t-statistic is

[tex] \bf t=\frac{290-321}{18.8812\sqrt{1/10+1/16}}=-4.073[/tex]

Our p-value is the area under the t Distribution with 24 degrees of freedom (sample1 -1 + sample2 -1) to the left of -4.073.

We can find this value by using either a table or a spreadsheet.

In Excel use

TDIST(4.073,24,1)

In OpenOffice Calc use

TDIST(4.073;24;1)

We get a p-value = 0.0002

b)

Since 321 is 31 pounds higher than 290 and our p-value is less than the significance level 0.05, the data do support the claim that the mean impact strength of gears from supplier 2 is at least 25 foot - pounds higher than that of supplier 1.

c)

A 95% confidence level interval correspondent to a significance level of 0.05 is given by

[tex] \bf [(\bar x_1-\bar x_2)-t\sqrt{1/n_1+1/n_2},(\bar x_1-\bar x_2)+t\sqrt{1/n_1+1/n_2}]=\\=[-31-4.073*0.4031,-31+4.073*0.4031]=[-32.642,-29.358][/tex]

This means that there is a 95% probability that the mean impact strength of gears from supplier 2 is between 29.358 and 32,642 foot - pounds higher than that of supplier 1