A random sample of 85 group leaders, supervisors, and similar personnel revealed that on the average a person spent 6.5 years on the job before being promoted. The standard deviation of the sample was 1.7 years. Using the 0.95 degree of confidence, what is the confidence interval within which the population mean lies?
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A. B. C. D. E.E
Interval estimate can be found from x_bar +/-z*s/(n^0.5). Here we have n = 85, x_bar = 6.5 and z = 1.96 (for 95%) and s = 1.7. Therefore 6.5+/-1.96*1.7/9.22 and we get 6.14 and 6.86.
To calculate the confidence interval for the population mean, we can use the formula:
Confidence Interval = Sample Mean ± (Critical Value) * (Standard Deviation / √Sample Size)
First, let's calculate the critical value using the 0.95 confidence level. Since the sample size is 85, the degrees of freedom for the t-distribution is 84 (n - 1). We can consult the t-distribution table or use statistical software to find the critical value.
The critical value for a 95% confidence level and 84 degrees of freedom is approximately 1.992.
Now, let's substitute the given values into the formula:
Confidence Interval = 6.5 ± (1.992) * (1.7 / √85)
Calculating the standard deviation divided by the square root of the sample size:
Confidence Interval = 6.5 ± 1.992 * (1.7 / √85) = 6.5 ± 1.992 * 0.1846 = 6.5 ± 0.3673
Finally, we can calculate the confidence interval:
Confidence Interval = (6.5 - 0.3673, 6.5 + 0.3673) = (6.1327, 6.8673)
Therefore, the confidence interval within which the population mean lies is approximately 6.13 to 6.87 years.
None of the provided answers (A, B, C, D, E) matches the correct confidence interval, so the correct answer would be C, "None of these answers."