Homework Soursethe US census bureau reports that the average number of empl
the US census bureau reports that the average number of employees in a small business is 16.1. suppose a sample of 49 small businesses showed a mean of 15 employees with a standard deviation of 25. test whether the population mean number of employees is different from 16.1 using the confidence interval. (significance level a=0.01)
the US census bureau reports that the average number of employees in a small business is 16.1. suppose a sample of 49 small businesses showed a mean of 15 employees with a standard deviation of 25. test whether the population mean number of employees is different from 16.1 using the confidence interval. (significance level a=0.01)
the US census bureau reports that the average number of employees in a small business is 16.1. suppose a sample of 49 small businesses showed a mean of 15 employees with a standard deviation of 25. test whether the population mean number of employees is different from 16.1 using the confidence interval. (significance level a=0.01)
Solution
Mean = 16.1
n=49
Sample Mean = 15
Sample SD = 25
Confidence level = 100(1) % = 100(10.01) % = 99%
Therefore, z value for 99% = 2.576
Confidence Interval Approach:
Margin of Error = (z*SD)/sqrt(n)
= (2.576 * 25)/sqrt(49)
= 9.2
Confidence Interval:
( Mean - Margin of Error , Mean + Margin of error)
= (15 - 9.2 , 15 + 9.2 )
= ( 5.8 , 24.2 )
True Mean 16.1 lies in this interval.
The confidence interval includes all null hypothesis values for the population mean that would be accepted by an hypothesis test at the 1 % significance level.
