Respuesta :
Answer:
1) use z-score
2) The 95% of confidence intervals
(141,489.24 , 149,670.75)
Step-by-step explanation:
(i) we will use z-score
The 95% confidence interval for the mean of the population corresponding to the given sample.
[tex](x^{-} -z_{\alpha } \frac{S.D}{\sqrt{n} } ,x^{-} +z_{\alpha }\frac{S.D}{\sqrt{n} } )[/tex]
Given data the sample size n=40
mean of the sample x⁻ = $145,580
Standard deviation (σ) = $13,200
Level of significance z-score =1.96
ii) The 95% of confidence intervals
[tex](145,580-1.96(\frac{13200}{\sqrt{40}) } ,145580+1.96\frac{13200}{\sqrt{40} } )[/tex]
on calculation, we get
(145,580-4090.75,145,580+4090.75)
(141,489.24 , 149,670.75)
Conclusion:-
The 95% of confidence intervals
(141,489.24 , 149,670.75)
