3. Sick-leave records obtained from a random sample of 200 social workers showed a mean number of days sick leave equal to 25.6 last year. If the sample standard deviation is 5.1, construct a 90% confidence interval for the mean number of days sick leave for all social workers last year.
Formula with your data included:
CI90 = 25.6 + or - 1.645(5.1/√200)
Calculate for your interval.
To construct a confidence interval for the mean number of days sick leave for all social workers last year, we can use the formula:
Confidence Interval = Sample Mean ± (Critical Value) * (Standard Deviation / √Sample Size)
1. First, determine the critical value corresponding to a 90% confidence level. For a two-tailed test at a 90% confidence level, the critical value is found using a t-distribution with (n - 1) degrees of freedom. Since we have a sample size of 200, the degrees of freedom is (200 - 1) = 199.
Looking up the critical value in a t-table or using a statistical calculator, the critical value for a 90% confidence level with 199 degrees of freedom is approximately 1.651.
2. Next, calculate the standard error of the mean, which is the standard deviation divided by the square root of the sample size:
Standard Error = Sample Standard Deviation / √Sample Size
= 5.1 / √200
≈ 0.361
3. Now, plug the values into the confidence interval formula:
Confidence Interval = 25.6 ± (1.651) * (0.361)
Calculating the upper and lower bounds of the confidence interval:
Upper Bound = 25.6 + (1.651) * (0.361)
≈ 25.6 + 0.596
≈ 26.196
Lower Bound = 25.6 - (1.651) * (0.361)
≈ 25.6 - 0.596
≈ 25.004
Therefore, the 90% confidence interval for the mean number of days sick leave for all social workers last year is approximately 25.004 to 26.196.