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Chapter 8: Q 8.83 (page 331)

Prices of New Mobile Homes. Recall that a simple random sample of 36 new mobile homes yielded the prices, in thousands of dollars, shown in Table 8.1on page 315 . We found the mean of those prices to be \(63.28thousand.

a. Use this information and Procedure 8.1on page 322 to find a 95%confidence interval for the mean price of all new mobile homes. Recall that σ=\)7.2thousand.

b. Compare your 95%confidence interval in part (a) to the one found in Example 8.2(c) on page 317 and explain any discrepancy that you observe.

Short Answer

Expert verified

(a) 95%confidence interval for μis (60.928,65.632)

(b) Above example provides 95%confidence interval.

Step by step solution

01

Given Information

It is given thatx¯=563.28,n=36,σ=7.2

02

Determination of 95% Confidence Interval of μ

From table, value of zα2=1.96with 95%confidence interval.

Hence, x¯±za2σn=63.28±1.967.236

=63.28±2.352

=(60.928,65.632)

Hence,95%confidence interval ofμis=(60.928,65.632)

03

Comparison of confidence interval here with that of example 8.2(c)

The confidence level from example is (60.88,65.68)

In above part, confidence interval is 60.88,65.88

Hence, it can be concluded that example provides approximately 95%confidence interval.

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Most popular questions from this chapter

Suppose that you have obtained data by taking a random sample from a population. Before performing a statistical inference, what should you do?

What is meant by saying that a statistical procedure is robust?

"Chips Ahoy! 1,000 Chips Challenge." As reported by B. Warner and J. Rutledge in the paper "Checking the Chips Ahoy! Guarantee" (Chanee, Vol. 12. Issue 1. pp. 10-14), a random sample of forty-two 18-ounce bags of Chips Ahoy! cookies yielded a mean of 1261.6 chips per bag with a standard deviation of 117.6 chips per bug.

a. Determine a 95% confidence interval for the mean number of chips per bag for all 18-ounce bags of Chips Ahoy! cookies, and interpret your result in words.

b. Can you conclude that the average 18-ounce bag of Chips Ahoy! cookies contain at least 1000 chocolate chips? Explain your answer.

One-Sided One-Mean z-Intervals. Presuming that the assumptions for a one-mean z-interval are satisfied, we have the following formulas for (1-α)-level confidence bounds for a population mean \(\mu\) :

- Lower confidence bound: x~-zσ·σ/n

- Upper confidence bound: x¯+zα·σ/n

Interpret the preceding formulas for lower and upper confidence bounds in words.

Suppose that you take 1000simple random samples from a population and that, for each sample, you obtain a 95% confidence interval for an unknown parameter. Approximately how many of those confidence intervals will contain the value of the unknown parameter?

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