Question

the work week for adults in the us that work full time is normally distributed with a mean of 47 hours. a newly hired engineer at a start - up company believes that employees at start - up companies work more on average then most working adults in the us. she asks 12 engineering friends at start - ups for the lengths in hours of their work week. their responses are shown in the table below. test the claim using a 10% level of significance. give answer to at least 4 decimal places. hours 50 41 52 50 49 69 46 47 45 49 50 53 what are the correct hypotheses? h0: select an answer 1 0 hours h1: select an answer 1 0 hours based on the hypotheses, find the following: test statistic= p - value= the correct decision is to select an answer the correct summary would be: select an answer that the mean number of hours of all employees at start - up companies work more than the us mean of 47 hours.

Explanation:

Step1: State the hypotheses

The null hypothesis $H_0$ is that the mean work - week of start - up employees is equal to the US mean. The alternative hypothesis $H_1$ is that the mean work - week of start - up employees is greater than the US mean. So, $H_0:\mu = 47$ and $H_1:\mu>47$.

Step2: Calculate the sample mean $\bar{x}$

The data set is $x = \{50,41,52,50,49,69,46,47,45,49,50,53\}$. The sample size $n = 12$.
$\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}=\frac{50 + 41+52+50+49+69+46+47+45+49+50+53}{12}=\frac{591}{12}=49.25$

Step3: Calculate the sample standard deviation $s$

First, calculate the sum of squared differences from the mean:
$\sum_{i = 1}^{n}(x_i-\bar{x})^2=(50 - 49.25)^2+(41 - 49.25)^2+(52 - 49.25)^2+(50 - 49.25)^2+(49 - 49.25)^2+(69 - 49.25)^2+(46 - 49.25)^2+(47 - 49.25)^2+(45 - 49.25)^2+(49 - 49.25)^2+(50 - 49.25)^2+(53 - 49.25)^2$
$=0.5625 + 68.0625+7.5625 + 0.5625+0.0625+390.0625+10.5625+5.0625+18.0625+0.0625+0.5625+14.0625 = 495.375$
The sample standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i - \bar{x})^2}{n - 1}}=\sqrt{\frac{495.375}{11}}\approx6.7177$

Step4: Calculate the test statistic $t$

The test statistic for a one - sample $t$ - test is $t=\frac{\bar{x}-\mu_0}{s/\sqrt{n}}$, where $\mu_0 = 47$, $\bar{x}=49.25$, $s\approx6.7177$, and $n = 12$.
$t=\frac{49.25 - 47}{6.7177/\sqrt{12}}\approx1.1394$

Step5: Calculate the p - value

The degrees of freedom is $df=n - 1=12 - 1 = 11$. Since this is a right - tailed test, the p - value is $P(T>1.1394)$ with $df = 11$. Using a $t$ - distribution table or software, the p - value is approximately $0.1379$.

Step6: Make a decision

Since the level of significance $\alpha = 0.10$ and the p - value ($0.1379$)>$\alpha$, we fail to reject the null hypothesis.

Answer:

$H_0:\mu = 47$
$H_1:\mu>47$
Test Statistic: $1.1394$
p - value: $0.1379$
The correct decision is to fail to reject $H_0$.
The correct summary would be: There is not enough evidence to support the claim that the mean number of hours of all employees at start - up companies work more than the US mean of 47 hours.