Question

question 7
the human resource director of a large corporation wishes to study absenteeism for their mid - level managers. a study of a random sample of 64 mid - level managers reveals that the mean number of days absent is 6.2 days, with the population standard deviation equal to 7.3 days. construct a 95% confidence interval for the man number of absences for mid - level managers during the year.
point estimate =
alpha =
critical value =+
standard error =
margin of error =+
confidence interval lower limit =
confidence interval upper limit =

Explanation:

Step1: Determine the point estimate

The point estimate for the population mean is the sample mean. Here, the sample mean \( \bar{x}=6.2 \) days.

Step2: Calculate \( \alpha \)

For a 95% confidence interval, \( \alpha = 1 - 0.95=0.05 \)

Step3: Find the critical value

Since the population standard deviation \( \sigma \) is known (\( \sigma = 7.3 \)) and \( n = 64\) (large sample, \( n\geq30 \)), we use the standard normal distribution \( Z \). For \( \alpha=0.05 \), the critical value \( z_{\alpha/2} \). \( \alpha/2=0.025 \), and from the standard - normal table \( z_{0.025}=1.960 \)

Step4: Calculate the standard error

The formula for the standard error \( SE=\frac{\sigma}{\sqrt{n}} \). Substituting \( \sigma = 7.3 \) and \( n = 64\), we get \( SE=\frac{7.3}{\sqrt{64}}=\frac{7.3}{8}=0.9125\approx0.913 \)

Step5: Calculate the margin of error

The margin of error \( E = z_{\alpha/2}\times SE \). Substituting \( z_{\alpha/2}=1.960 \) and \( SE = 0.913 \), we get \( E=1.960\times0.913 = 1.789 \)

Step6: Calculate the confidence interval limits

The lower limit \( LL=\bar{x}-E \). Substituting \( \bar{x}=6.2 \) and \( E = 1.789 \), we get \( LL=6.2-1.789 = 4.411 \)
The upper limit \( UL=\bar{x}+E \). Substituting \( \bar{x}=6.2 \) and \( E = 1.789 \), we get \( UL=6.2 + 1.789=7.989 \)

Answer:

Point estimate \( = 6.2\)
Alpha \( = 0.05\)
Critical Value \(=\pm1.960\)
Standard error \( = 0.913\)
Margin of error \(=\pm1.789\)
Confidence interval lower limit \( = 4.411\)
Confidence interval upper limit \( = 7.989\)