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

Answer:

a) sample mean x⁻ = 73

b) The margin of error (M.E)  = 3

c) Sample standard deviation(σ) = 8.9404

Step-by-step 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

$(x^{-} - t_{\frac{\alpha }{2} } \frac{S.D}{\sqrt{n} } , x^{-} +t_{\frac{\alpha }{2} } \frac{S.D}{\sqrt{n} })$

Given sample size n =26

The degrees of freedom ν=n-1 =26-1=25

$t_{\frac{0.10}{2} } = t_{0.05} = 1.711$

$(x^{-} -1.711 \frac{S.D}{\sqrt{n} } , x^{-} +1.711 \frac{S.D}{\sqrt{n} }) = (70 ,76)$

$x^{-} - M.E = 70$  …(i)

$x^{-} + M.E = 76$ …(ii)

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

$2x^{-} = 146$

$x^{-} = \frac{146}{2} = 73$

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

$M.E = t_{\frac{\alpha }{2} } \frac{S.D}{\sqrt{n} }$

we have margin of error = 3

Given sample size 'n' =26

$t_{\frac{\alpha }{2} } = t_{\frac{0.10}{2} } = t_{0.05} =1.711$

$3 = 1.711\frac{S.D}{\sqrt{26} }$

Cross multiplication , we get

$3 \sqrt{26} =1.711 S.D$

$S.D =\frac{3\sqrt{26} }{1.711} = 8.9404$

Standard deviation = 8.9404