Question

Senior management of a consulting services firm is concerned about a growing decline in the firm’s weekly number of billable hours. The firm expects each professional employee to spend at least 40 hours per week on work. In an effort to understand this problem better, management would like to estimate the standard deviation of the number of hours their employees spend on work-related activities in a typical week. Rather than reviewing the records of all the firm’s full-time employees, the management randomly selected a sample of size 50 from the available frame. The sample mean and sample standard deviations were 46.4 and 7.2 hours, respectively. Construct a 97% confidence interval for the standard deviation of the number of hours this firm’s employees spend on work-related activities in a typical week

Answer 1

`((49)(7.2)^2)/(72.892) \leq \sigma^2 \leq ((49)(7.2)^2)/(30.036)`

`34.848 \leq \sigma^2 \leq 84.57`

Taking square root in both sides we got the interval desired

`5.903 \leq \sigma \leq 9.196`

Explanation:

Information provided

s=7.2 represent the sample standard deviation

`\bar x=46.4` represent the sample mean

n=50 the sample size

Confidence=97% or 0.97

The confidence interval

We can use the following formula for the confidence interval for the true variance

`((n-1)s^2)/(\chi^2_(\alpha/2)) \leq \sigma^2 \leq ((n-1)s^2)/(\chi^2_(1-\alpha/2))`

The degrees of freedom are given by:

`df=n-1=50-1=49`

The confidence level is 0.97 or 97%, the significance
`\alpha=0.03` and
`\alpha/2 =0.015` and the critical values are:

The excel commands would be: "=CHISQ.INV(0.015,49)" "=CHISQ.INV(0.985,49)". so for this case the critical values are:

`\chi^2_(\alpha/2)=72.892`

`\chi^2_(1- \alpha/2)=30.036`

Replacing the values we have:

`((49)(7.2)^2)/(72.892) \leq \sigma^2 \leq ((49)(7.2)^2)/(30.036)`

`34.848 \leq \sigma^2 \leq 84.57`

Taking square root in both sides we got the interval desired

`5.903 \leq \sigma \leq 9.196`