Respuesta :
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
[tex](x^{-} - t_{\frac{\alpha }{2} } \frac{S.D}{\sqrt{n} } , x^{-} +t_{\frac{\alpha }{2} } \frac{S.D}{\sqrt{n} })[/tex]
Given sample size n =26
The degrees of freedom ν=n-1 =26-1=25
[tex]t_{\frac{0.10}{2} } = t_{0.05} = 1.711[/tex]
[tex](x^{-} -1.711 \frac{S.D}{\sqrt{n} } , x^{-} +1.711 \frac{S.D}{\sqrt{n} }) = (70 ,76)[/tex]
[tex]x^{-} - M.E = 70[/tex] …(i)
[tex]x^{-} + M.E = 76[/tex] …(ii)
Adding (i) and (ii) and simplification , we get
[tex]2x^{-} = 146[/tex]
[tex]x^{-} = \frac{146}{2} = 73[/tex]
Sample mean = 73
Step(ii):-
Substitute x⁻ = 73 in equation(i)
[tex]x^{-} - M.E = 70[/tex]
[tex]73 - M.E = 70[/tex]
[tex]73 - 70 = M.E[/tex]
Margin of error = 3
Step(iii):-
The margin of error is determined by
[tex]M.E = t_{\frac{\alpha }{2} } \frac{S.D}{\sqrt{n} }[/tex]
we have margin of error = 3
Given sample size 'n' =26
[tex]t_{\frac{\alpha }{2} } = t_{\frac{0.10}{2} } = t_{0.05} =1.711[/tex]
[tex]3 = 1.711\frac{S.D}{\sqrt{26} }[/tex]
Cross multiplication , we get
[tex]3 \sqrt{26} =1.711 S.D[/tex]
[tex]S.D =\frac{3\sqrt{26} }{1.711} = 8.9404[/tex]
Standard deviation = 8.9404
