Question

question 15 (1 point) saved
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 1299 parts and knows the population standard deviation to be 350. the midwest group ships an average of 1456 parts and knows the population standard deviation to be 297.
using a 0.01 level of significance, test if there is a difference in productivity level. what is the p - value? (round to four decimal places) make sure you put the 0 in front of the decimal.
p - value =___

Explanation:

Step1: State the hypotheses

$H_0:\mu_1=\mu_2$ (no difference in productivity), $H_1:\mu_1
eq\mu_2$ (there is a difference in productivity)

Step2: Calculate the test - statistic

The formula for the two - sample z - test statistic when population standard deviations are known is $z=\frac{(\bar{x}_1 - \bar{x}_2)-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}}$. Here, $\bar{x}_1 = 1299$, $\bar{x}_2=1456$, $\mu_1-\mu_2 = 0$ (under $H_0$), $\sigma_1 = 350$, $\sigma_2=297$, $n_1=n_2 = 35$.
$z=\frac{(1299 - 1456)-0}{\sqrt{\frac{350^{2}}{35}+\frac{297^{2}}{35}}}=\frac{- 157}{\sqrt{\frac{122500}{35}+\frac{88209}{35}}}=\frac{-157}{\sqrt{3500 + 2520.2571}}=\frac{-157}{\sqrt{6020.2571}}\approx\frac{-157}{77.5891}\approx - 2.0235$

Step3: Calculate the p - value

Since this is a two - tailed test, the p - value is $2P(Z\lt|z|)$. We know $|z| = 2.0235$. Using the standard normal distribution table or a calculator, $P(Z\lt - 2.0235)=P(Z\gt2.0235)$.
$P(Z\lt - 2.0235)\approx0.0216$. So the p - value is $2\times0.0216 = 0.0432$

Answer:

$0.0432$