Mathematics, 10.11.2019 03:31 peoplearecoming

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 () 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.

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Answer from: TamB01

a) Yes, there is. The p-value for this test is 0.0002

b) Yes, it does

c) [-32.642,-29.358]

Step-by-step explanation:

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.

We compute our t-statistic by the formula

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

where

$\bf \bar x_1$ = 290 is the mean of supplier 1

$\bf \bar x_2$ = 321 is the mean of supplier 2

$\bf n_1$ = 10 is the sample size from supplier 1

$\bf n_2$ = 16 is the sample size from supplier 2

$\bf S_v$ is the pooled standard deviation

$\bf S_v=\sqrt{\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}}$

$\bf s_1$ = 12 is the standard deviation of supplier 1

$\bf s_1$ = 22 is the standard deviation of supplier 2

We have then

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

and out t-statistic is

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

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

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.

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

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

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

Answer from: Quest

the correct answer is a) they believe they have a "just cause."

islamic terrorists justify their violent actions because they believe they have a "just cause."

every act of terrorism of islamic sects is done “in the name of allah”. that justifies their actions. islam people who commit those terrorist attacks are not afraid to die. that is why the attacks have horrible consequences. they know they are going to die and they are proud to do it because it is in the name of allah, so it is justified and is for a “good cause.”

since the very beginning of the conflict with israel, islam is fighting a “sacred war”, respecting the words stated in the qur’an. they fight against the domination of the western countries and to defend their occupied territories.

Answer from: Quest

3+x

step-by-step explanation: