Mathematics, 24.05.2021 21:50 tefanyc13

The CEO of a large manufacturing company is curious if there is a difference in productivity level of her warehouse employees based on the region of the country the warehouse is located. She randomly selects 35 employees who work in warehouses on the East Coast (Group 1) and 35 employees who work in warehouses in the Midwest (Group 2) and records the number of parts shipped out from each for a week. She finds that East Coast group ships an average of 1276 parts and knows the population standard deviation to be 347.
The Midwest group ships an average of 1439 parts and knows the population standard deviation to be 298.
Using a 0.01 level of significance, test if there is a difference in productivity level. What is the test statistic?
(Round to 4 decimal places)
z =
please use excel to solve

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Answer from: Kayccee9958

The test statistic is z = -2.11.

Step-by-step explanation:

Before finding the test statistic, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Group 1: Sample of 35, mean of 1276, standard deviation of 347.

This means that:

$\mu_1 = 1276, s_1 = \frac{347}{\sqrt{35}} = 58.6537$

Group 2: Sample of 35, mean of 1439, standard deviation of 298.

This means that:

$\mu_2 = 1439, s_2 = \frac{298}{\sqrt{35}} = 50.3712$

Test if there is a difference in productivity level.

At the null hypothesis, we test that there is no difference, that is, the subtraction is 0. So

$H_0: \mu_1 - \mu_2 = 0$

At the alternate hypothesis, we test that there is difference, that is, the subtraction is different of 0. So

$H_1: \mu_1 - \mu_2 \neq 0$

The test statistic is:

$z = \frac{X - \mu}{s}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis and s is the standard error.

0 is tested at the null hypothesis:

This means that $\mu = 0$

From the two samples:

$X = \mu_1 - \mu_2 = 1276 - 1439 = -163$

$s = \sqrt{s_1^2+s_2^2} = \sqrt{58.6537^2+50.3712^2} = 77.3144$

Test statistic:

$z = \frac{X - \mu}{s}$

$z = \frac{-163 - 0}{77.3144}$

z = -2.11

The test statistic is z = -2.11.

Answer from: Quest

First, think of what you know. you know that the number t the end of the equation will need to be subtracted because it is the money he lost from buying the lawn mower. and your first two weeks have negative profit, so that tells you that c and d are wrong. now you can plug any of the weeks into a and b to see which one works. i'll start with plugging week 1 into a. $y = 60(1) - 90 \\ y = 60 - 90 \\ y = - 30$ this isn't right because it says your answer for week one should be -90.so now we need to try the same thing with b. $y = 60(1) - 150 \\ y = 60 - 150 \\ y = - 90$ this one got you the right answer, (sometimes if you only check with one of the weeks multiple will work so you need to check with more of the weeks, but it worked out fine here)this means that b is the correct answer.let me know if you need more clarification. i hope this : )

Answer from: Quest

the answer should be a. snail