Question

(1 point) Working backwards, Part I. A 90% confidence interval for a population mean is (70, 76). The population distribution is approximately normal and the population standard deviation is unknown. This confidence interval is based on a simple random sample of 26 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 4 decimal places.

Answer 1

a) sample mean x⁻ = 73

b) The margin of error (M.E) = 3

c) Sample standard deviation(σ) = 8.9404

Explanation:

Explanation:-

Step(i):-

Given 90% confidence interval for a population mean is (70, 76)

We know that 90% of confidence intervals are determined by

`(x^(-) - t_{(\alpha )/(2) } (S.D)/(√(n) ) , x^(-) +t_{(\alpha )/(2) } (S.D)/(√(n) ))`

Given sample size n =26

The degrees of freedom ν=n-1 =26-1=25

`t_{(0.10)/(2) } = t_(0.05) = 1.711`

`(x^(-) -1.711 (S.D)/(√(n) ) , x^(-) +1.711 (S.D)/(√(n) )) = (70 ,76)`

`x^(-) - M.E = 70` …(i)

`x^(-) + M.E = 76` …(ii)

Adding (i) and (ii) and simplification , we get

`2x^(-) = 146`

`x^(-) = (146)/(2) = 73`

Sample mean = 73

Step(ii):-

Substitute x⁻ = 73 in equation(i)

`x^(-) - M.E = 70`

`73 - M.E = 70`

`73 - 70 = M.E`

Margin of error = 3

Step(iii):-

The margin of error is determined by

`M.E = t_{(\alpha )/(2) } (S.D)/(√(n) )`

we have margin of error = 3

Given sample size 'n' =26

`t_{(\alpha )/(2) } = t_{(0.10)/(2) } = t_(0.05) =1.711`

`3 = 1.711(S.D)/(√(26) )`

Cross multiplication , we get

`3 √(26) =1.711 S.D`

`S.D =(3√(26) )/(1.711) = 8.9404`

Standard deviation = 8.9404