Question

A random sample of 85 supervisors revealed that they worked an average of 6.5 years before being promoted. The population standard deviation was 1.7 years. Using the 0.95 degree of confidence, what is the confidence interval for the population mean? Select one: a. 6.99 and 7.99 b. 4.15 and 7.15 c. 6.14 and 6.86 d. 6.49 and 7.49

Answer 1

Answer: c. 6.14 and 6.86

Explanation:

The confidence interval for population mean is given by :-

`\overline{x}\pm z^*((\sigma)/(√(n)))`

, where n= sample size

`\overline{x}` = Sample mean

`\sigma` = Population standard deviation

z*= Critical z-value.

As per given , we have

n= 85

`\overline{x}=6.5`

`\sigma=1.7`

For 0.95 degree of confidence interval , the critical z-value is 1.96. (By z-table).

So , the required confidence interval will be:

`6.5\pm (1.96)((1.7)/(√(85)))`

`6.5\pm (1.96)((1.7)/( 9.21954445729))`

`6.5\pm (0.3614)`

`=(6.5-0.3614,\ 6.5+0.3614)=(6.1386,\ 6.8614)\approx(6.14,\ 6.86)`

Hence, the confidence interval for the population mean is c. 6.14 and 6.8 .