In: Statistics and Probability
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 5% level of significance. Give answer to at least 4 decimal places.
Hours |
---|
46 |
45 |
55 |
54 |
48 |
68 |
51 |
60 |
46 |
46 |
52 |
52 |
What are the correct hypotheses?
H0: Select an answer p σ² μ s² x̄ p̂ s σ ?
> ≤ ≥ ≠ = < hours
H1: Select an answer p σ² p̂ σ μ s² x̂ s ?
> ≤ = ≥ < ≠ hours
Based on the hypotheses, find the following:
Test Statistic=
p-value=
The correct decision is to Select an answer Accept the null hypothesis Reject the null hypothesis Accept the alternative hypotheis Fail to reject the null hypothesis .
The correct summary would be: Select an answer There is enough evidence to reject the claim There is not enough evidence to support the claim There is not enough evidence to reject the claim There is 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.
: Mean work week;
Claim : employees at start-up companies work more on average then most working adults in the US
H0:
= 47 hours
H1:
> 47 hours
Right tailed test:
Hypothesized mean : = 47
xi : lengths in hours of work week for ith engineering friend
Sample size : n= 12
Sample mean :Sample mean hours of work
Sample standard deviation :s
xi:Hours | ||
46 | -5.9167 | 35.0069 |
45 | -6.9167 | 47.8403 |
55 | 3.0833 | 9.5069 |
54 | 2.0833 | 4.3403 |
48 | -3.9167 | 15.3403 |
68 | 16.0833 | 258.6736 |
51 | -0.9167 | 0.8403 |
60 | 8.0833 | 65.3403 |
46 | -5.9167 | 35.0069 |
46 | -5.9167 | 35.0069 |
52 | 0.0833 | 0.0069 |
52 | 0.0833 | 0.0069 |
=623 | =506.9167 | |
=623/12=51.9167 |
Based on the hypotheses, find the following:
Test Statistic= 2.5089
For right ailed test:
Degrees of freedom = n-1 =12-1 =11
For 11 degrees of freedom,
P(t>2.5089) =0.0145
p-value = 0.0145
As P-Value i.e. is less than Level of significance i.e
(P-value:0.0145 < 0.05:Level of significance); Reject Null
Hypothesis
The correct decision is to : Reject the null
hypothesis
The correct summary would be:
There is 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.