Answered: Eastland and Westland are two…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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Question

Eastland and Westland are two neighboring countries. A random sample of 101 families from Eastland has a sample mean income of $52 thousand and a sample standard deviation $10.3 thousand.

a. Find and interpret a 95% confidence interval for the population mean income in Eastland?

An independent sample of 31 families from Westland has sample mean income of $48 thousand and a sample standard deviation of 11.6 thousand.

b. Test the null hypothesis that the population mean income in Eastland equals the population mean income in Westland against a two-sided alternative. Include in your answer the null and alternative hypotheses. Show calculations and state the conclusion.

c. Find a 90% confidence interval for the difference between the population mean income in Eastland and the population mean income in Westland

d. Clearly describe what happens to the interval estimate in (c) when the sample size increases.

e. It is claimed that mean income in Eastland equals the population mean income in Westland. Using your results in (c). do you reject this claim? Clearly explain why. Is your answer consistent with your conclusion in (b)? Explain.

Expert Solution

Step 1

Comments : As per the company norms we have to solved first three parts only.

Solution:

Given information:

$n_{1} = 101 S a m p l e s i z e o f E a s t l a n d n_{2} = 31 S a m p l e s i z e o f W e s t l a n d x_{1} = $ 52 S a m p l e m e a n o f E a s t l a n d x_{2} = $ 48 S a m p l e m e a n o f W e s t l a n d s_{1} = $ 10.3 S a m p l e s \tan d a r d d e v i a t i o n s o f E a s t l a n d s_{2} = $ 11.6 S a m p l e s \tan d a r d d e v i a t i o n s o f W e s t l a n d α = 0.05 L e v e l o f s i g n i f i c a n c e$

To find pooled variance

${s_{p}}^{2} = \frac{(n_{1} - 1) \times {s_{1}}^{2} + (n_{2} - 1) \times {s_{2}}^{2}}{n_{1} + n_{2} - 2} {s_{p}}^{2} = \frac{(101 - 1) \times 10 . 3^{2} + (31 - 1) \times 11 . 6^{2}}{101 + 31 - 2} {s_{p}}^{2} = \frac{100 \times 106.09 + 30 \times 134.56}{132 - 2} {s_{p}}^{2} = \frac{10609 + 4036.8}{130} {s_{p}}^{2} = \frac{14645.8}{130} {s_{p}}^{2} = 112.66$

Step 2

a) The 95% confidence interval for the population mean income in Eastland is

$x_{1} \pm t_{α / 2, n - 1} \times \frac{s_{1}}{\sqrt{n_{1}}} A t α = 0.05 t_{α / 2, n - 1} = t_{0.025, 100} = 1.984 F r o m e x c e l (52 \pm 1.984 \times \frac{10.3}{\sqrt{101}}) (52 \pm 1.984 \times \frac{10.3}{10.049875}) (52 \pm 1.984 \times 1.0248883) (52 \pm 2.0333785) (49.96662, 54.033378) (49.9666, 54.0334)$

The 95% confidence interval for the population mean income in Eastland is 49.9666 to 54.0334

Interpretation of confidence interval:

We can be 95% confident that the true population mean income in Eastland is between 49.9666 to 54.0334

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