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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 51 from the available frame. The sample mean and sample standard deviations were 48.5 and 7.5 hours, respectively. Construct a 99% confidence interval for the standard deviation of the number of hours this firm’s employees spend on work-related activities in a typical week

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JeanaShupp

Answer:  (45.79, 51.21)

Step-by-step explanation:

Given : Significance level : [tex]\alpha: 1-0.99=0.01[/tex]

Sample size : n= 51 , which is a large sample (n>30), so we use z-test.

Critical value: [tex]z_{\alpha/2}=2.576[/tex]

Sample mean : [tex]\overline{x}= 48.5\text{ hours}[/tex]

Standard deviation : [tex]\sigma=7.5\text{ hours}[/tex]

The confidence interval for population means is given by :-

[tex]\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]

i.e. [tex]48.5\pm(2.576)\dfrac{7.5}{\sqrt{51}}[/tex]

i.e.[tex]48.5\pm2.70534112234\\\\\approx48.5\pm2.71\\\\=(48.5-2.71, 48.5+2.71)=(45.79, 51.21)[/tex]

Hence, 99% confidence interval for the standard deviation of the number of hours this firm’s employees spend on work-related activities in a typical week =  (45.79, 51.21)

You can use the formula for confidence interval when population standard deviation is unknown and sample is large(> 30)

The 99% confidence interval for the standard deviation of the number of hours given firm's employees spend on work-related activities in a typical week is given as [45.79, 51.21]

How to find the confidence interval for large samples (sample size > 30)?

Suppose that we have:

  • Sample size n > 30
  • Sample mean = [tex]\overline{x}[/tex]
  • Sample standard deviation = [tex]s[/tex]
  • Population standard deviation = [tex]\sigma[/tex]
  • Level of significance = [tex]\alpha[/tex]

Then the confidence interval is obtained as

  • Case 1: Population standard deviation is known

[tex]\overline{x} \pm Z_{\alpha /2}\dfrac{\sigma}{\sqrt{n}}[/tex]

  • Case 2: Population standard deviation is unknown.

[tex]\overline{x} \pm Z_{\alpha /2}\dfrac{s}{\sqrt{n}}[/tex]

where [tex]Z_{\alpha/2}[/tex] is critical value of the test statistic at level of significance  [tex]\alpha[/tex]

Using the above formulas for the given condition to get the 99% confidence interval needed

We are given that

  • Sample size = n = 51
  • Sample standard deviation = [tex]s = 7.5 \: \rm hours[/tex]
  • Sample mean = [tex]\overline{x} = 48.5 \: \rm hours[/tex]
  • Level of significance = 100 - 99% = 1% = 0.01

Thus, the confidence interval will be calculated as

[tex]\overline{x} \pm Z_{\alpha /2}\dfrac{s}{\sqrt{n}}[/tex]

Z score at 0.01 level of significance is 2.58 (for two tailed)

Using that, we have:

[tex]\overline{x} \pm Z_{\alpha /2}\dfrac{s}{\sqrt{n}} = 48.5 \pm 2.58 \times \dfrac{7.5}{\sqrt{51}} \approx 48.5 \pm 2.71[/tex]

Thus, the confidence interval, in terms of [a,b] can be written as

[48.5 - 2.71, 48.5 + 2.71] = [45.79, 51.21]

Thus,

The 99% confidence interval for the standard deviation of the number of hours given firm's employees spend on work-related activities in a typical week is given as [45.79, 51.21]

Learn more about confidence interval here:

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