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 1% level of significance. give answer to at least 4 decimal places. hours 49 48 52 55 45 63 51 45 47 50 50 55 what are the correct hypotheses? h0: select an answer? hours h1: select an answer? 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: Define hypotheses

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

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

$\bar{x}=\frac{49 + 48+52+55+45+63+51+45+47+50+50+55}{12}=\frac{600}{12}=50$.

Step3: Calculate sample standard deviation $s$

First, calculate the squared differences from the mean for each data - point:
$(49 - 50)^2=1$, $(48 - 50)^2 = 4$, $(52 - 50)^2=4$, $(55 - 50)^2 = 25$, $(45 - 50)^2=25$, $(63 - 50)^2 = 169$, $(51 - 50)^2=1$, $(45 - 50)^2=25$, $(47 - 50)^2=9$, $(50 - 50)^2=0$, $(50 - 50)^2=0$, $(55 - 50)^2=25$.
The sum of squared differences $\sum(x_i-\bar{x})^2=1 + 4+4+25+25+169+1+25+9+0+0+25 = 288$.
The sample standard deviation $s=\sqrt{\frac{\sum(x_i - \bar{x})^2}{n - 1}}=\sqrt{\frac{288}{11}}\approx5.1138$.

Step4: Calculate the test statistic $t$

The test statistic for a one - sample $t$ - test is $t=\frac{\bar{x}-\mu_0}{\frac{s}{\sqrt{n}}}$, where $\bar{x}$ is the sample mean, $\mu_0$ is the hypothesized population mean, $s$ is the sample standard deviation, and $n$ is the sample size.
Here, $\bar{x}=50$, $\mu_0 = 47$, $s\approx5.1138$, and $n = 12$.
$t=\frac{50 - 47}{\frac{5.1138}{\sqrt{12}}}=\frac{3}{\frac{5.1138}{3.4641}}\approx2.0267$.

Step5: Calculate the p - value

The degrees of freedom is $df=n - 1=12-1 = 11$.
Using a $t$ - distribution table or a calculator, for a one - tailed test with $t = 2.0267$ and $df = 11$, the p - value is $P(t_{11}>2.0267)\approx0.0332$.

Step6: Make a decision

Since the p - value ($0.0332$) is greater than the significance level $\alpha = 0.01$, we fail to reject the null hypothesis.

Answer:

$H_0:\mu = 47$
$H_1:\mu>47$
Test Statistic $\approx2.0267$
p - value $\approx0.0332$
The correct decision is to fail to reject $H_0$.
The correct summary would be: There is not enough evidence at the 1% level of significance that the mean number of hours of all employees at start - up companies work more than the US mean of 47 hours.