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
48
47
59
50
49
63
47
45
47
46
52
51

Explanation:

Step1: Calculate sample mean

Let \(x_1,x_2,\cdots,x_{12}\) be the data - points. The sample mean \(\bar{x}=\frac{\sum_{i = 1}^{12}x_i}{n}\), where \(n = 12\).
\(\sum_{i=1}^{12}x_i=48 + 47+59+50+49+63+47+45+47+46+52+51 = 594\)
\(\bar{x}=\frac{594}{12}=49.5\)

Step2: Calculate sample standard - deviation

The sample standard - deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}\)
\((x_1-\bar{x})^2=(48 - 49.5)^2=(-1.5)^2 = 2.25\)
\((x_2-\bar{x})^2=(47 - 49.5)^2=(-2.5)^2 = 6.25\)
\(\cdots\)
\(\sum_{i = 1}^{12}(x_i-\bar{x})^2=2.25+6.25 + 81+0.25+0.25+182.25+6.25+20.25+6.25+12.25+6.25+2.25=336\)
\(s=\sqrt{\frac{336}{11}}\approx5.5091\)

Step3: State the hypotheses

The null hypothesis \(H_0:\mu\leq47\) and the alternative hypothesis \(H_1:\mu>47\)

Step4: Calculate the test - statistic

The test - statistic for a one - sample t - test is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\), where \(\mu = 47\), \(\bar{x}=49.5\), \(s\approx5.5091\), and \(n = 12\)
\(t=\frac{49.5 - 47}{5.5091/\sqrt{12}}\approx1.5777\)

Step5: Determine the critical value

The degrees of freedom \(df=n - 1=12-1 = 11\). For a one - tailed test with \(\alpha = 0.01\), the critical value \(t_{\alpha,df}=t_{0.01,11}= 2.7181\)

Step6: Make a decision

Since \(t = 1.5777

Answer:

We fail to reject the null hypothesis.