Question

work time lost due to accidents at a large company, the director of research found that the average work time lost by employees due to accidents was 97 hours per year. she used a random sample of 16 employees. the standard deviation of the sample was 5.9 hours. estimate the population mean for the number of hours lost due to accidents for the company, using a 95% confidence interval. assume the variable is normally distributed. round intermediate answers to at least three decimal places. round your final answers to the nearest whole number. \\( \square < \mu < \square \\)

Explanation:

Step1: Identify parameters

Sample mean $\bar{x}=97$, sample size $n=16$, sample std dev $s=5.9$, confidence level 95% (two-tailed). Degrees of freedom $df=n-1=15$.

Step2: Find t-critical value

For $df=15$ and 95% confidence, $t_{\alpha/2}=2.131$ (from t-table).

Step3: Calculate margin of error

Margin of error $E = t_{\alpha/2} \times \frac{s}{\sqrt{n}} = 2.131 \times \frac{5.9}{\sqrt{16}} = 2.131 \times 1.475 ≈ 3.143$

Step4: Compute confidence interval

Lower bound: $\bar{x} - E = 97 - 3.143 ≈ 93.857 ≈ 94$
Upper bound: $\bar{x} + E = 97 + 3.143 ≈ 100.143 ≈ 100$

Answer:

94 < μ < 100