Question

We want to create a 95% confidence interval to estimate the average yearly income of doctors who earned their degrees from public institutions. We sampled 40 doctors to obtain a mean of $145,580 and standard deviation of $13,200. 1) Do we use a t-value or a z-score? 2) Create a 95% confidence interval

Answer 1

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.

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

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

$(145,580-1.96(\frac{13200}{\sqrt{40}) } ,145580+1.96\frac{13200}{\sqrt{40} } )$

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)