Question

Working backwards, Part I. A 90% confidence interval for a population mean is (83, 89). The population distribution is approximately normal and the population standard deviation is unknown. This confidence interval is based on a simple random sample of 25 observations. Calculate the sample mean, the margin of error, and the sample standard deviation. Use the t distribution in any calculations. Round non-integer results to 2 decimal places.

Sample mean =

Margin of error =

Sample standard deviation =

Answer 1

Solution :

Given that,

Lower confidence interval = 83

Upper confidence interval = 89

​  = (Lower confidence interval + Upper confidence interval ) / 2

= (83 + 89) / 2

= 172 / 2 = 86

= 86

Sample mean = 86

Margin of error = E = Upper confidence interval - ​  = 89 - 86 = 3

Margin of error = 3

n = 25

Degrees of freedom = df = n - 1 = 25 - 1 = 24

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,24 = 1.711

Margin of error = E = t_/2,df * (s /n)

s = E * n /  t_/2,df

= 3 * 25 / 1.711

= 15 / 1.711

= 8.77

Sample standard deviation = 8.77

=