The mean number of sick days an employee takes per year is believed to be about 10. Members of a personnel department do not believe this figure. They randomly survey 8 employees. The number of sick days they took for past year are as follows: 11; 5; 14; 3; 10; 9; 8; 9. Let X = the number of sick days they took for the past year. Should the personnel team believe that the mean number is about 10? Conduct a hypothesis test at the 5% leve Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.) O Part (a) O Part (b) O Part (c) O Part (d) O Part (e) O Part (f) O Part (g) O Part (h) Indicate the correct decision ("reject" or "do not reject" the null hypothesis), the reason for it, and write an appropriate conclusion. (1) Alpha (Enter an exact number as an integer, fraction, decimal.) (ii) Decision: O reject the null hypothesis O do not reject the null hypothesis (ii) Reason for decision: O Since a < p-value, we do not reject the null hypothesis. O Since a < p-value, we reject the null hypothesis. O Since a > p-value, we reject the null hypothesis. O Since a > p-value, we do not reject the null hypothesis. (iv) Conclusion: O There is sufficient evidence to conclude that the average number of sick days used per year by an employee is not equal to 10 days. O There is not sufficient evidence to conclude that the average number of sick days used per vear by an emplovee is not egual to 10 days.

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The mean number of sick days an employee takes per year is believed to be about 10. Members of a personnel department do not believe this figure. They randomly survey 8 employees. The number of sick days they took for the past year are as follows: 11; 5; 14; 3; 10; 9; 8; 9. Let $ X $ = the number of sick days they took for the past year. Should the personnel team believe that the mean number is about 10? Conduct a hypothesis test at the 5% level.

**Note**: If you are using a Student’s t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

- Part (a)
- Part (b)
- Part (c)
- Part (d)
- Part (e)
- Part (f)
- Part (g)
- Part (h)

Indicate the correct decision ("reject" or "do not reject" the null hypothesis), the reason for it, and write an appropriate conclusion.

(i) Alpha (Enter an exact number as an integer, fraction, or decimal.)
$ \alpha = \underline{\hspace{2cm}} $

(ii) Decision:
- $ \bigcirc $ reject the null hypothesis
- $ \bigcirc $ do not reject the null hypothesis

(iii) Reason for decision:
- $ \bigcirc $ Since $ \alpha < $ p-value, we do not reject the null hypothesis.
- $ \bigcirc $ Since $ \alpha > $ p-value, we reject the null hypothesis.
- $ \bigcirc $ Since $ \alpha > $ p-value, we reject the null hypothesis.
- $ \bigcirc $ Since $ \alpha > $ p-value, we do not reject the null hypothesis.

(iv) Conclusion:
- $ \bigcirc $ There is sufficient evidence to conclude that the average number of sick days used per year by an employee is not equal to 10 days.
- $ \bigcirc $ There is not sufficient evidence to conclude that the average number of sick days used per year by an employee is not equal to 10 days.

Expert Solution

Step 1

Given data,

11,5,14,3,10,9,8,9.

The mean number of sick days an employee takes per year is believed to be about 10.

Hypotheses:

H₀:µ=10

H₁:µ≠10

Level of significance = 0.05.

Step 2

Mean:

$\begin{array}{rcl} X & = & \frac{\sum x}{n} / n \\ = & \frac{11 + 5 + 14 + 3 + 10 + 9 + 8 + 9}{8} \\ = & \frac{69}{8} \\ = & 8.625 \end{array}$

Standard deviation:

$s = \sqrt{\frac{(x i - x)}{n - 1}} = \sqrt{\frac{{(11 - 8.625)}^{2} + {(5 - 8.625)}^{2} + \dots {(8 - 8.625)}^{2} + {(9 - 8.625)}^{2}}{8 - 1}} = 3.42$

Test statistic:

$\begin{array}{rcl} t & = & \frac{\bar{x} - µ}{\frac{s}{\sqrt{n}}} \\ = & \frac{8.625 - 10}{\frac{3.42}{\sqrt{8}}} \\ = & - 1.137 \end{array}$

Test statistic value is -1.137.

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