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Chapter 8: Q 8.130. (page 343)

Digital Viewing Times. According to marketing, the average time spent per day with digital media in 2010was 3hours and 14minutes. For last year, a random sample of 20American adults spent the following number of hours per day with digital media.

Find and interpret a 90%confidence interval for last year's mean time spent per day with digital media by American adults. (Note: x¯=5.16hrand s=2.30hr)

Short Answer

Expert verified

The 90% confidence interval for American adults' mean daily time spent with digital media last year is (4.271,6.049)

Step by step solution

01

Given information

1.45.05.84.15.6
2.16.52.38.09.5
7.35.99.44.32.9
2.85.14.54.16.6
02

Concept

The formula used: The confidence intervalx¯±ta2sn

03

Calculation

Calculate the 90%confidence interval for American adults' average daily time spent with digital media last year.

Consider x¯=5.16,n=20, and s=2.30

The needed value of ta2for 90%confidence with 19(=20-1)degrees of freedom is 1.729, according to "Table IV Values of ta

Thus, the confidence interval is,

x¯±ta2sn=5.16±1.7292.3020=5.16±1.729(0.514)=5.16±0.889=(4.271,6.049)

As a result, the 90% confidence interval for American adults' mean daily time spent with digital media last year is (4.271,6.049)

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Most popular questions from this chapter

For a fixed confidence level, show that (approximately) quadrupling the sample size is necessary to halve the margin of error. (Hint: Use Formula 8.2.)

A simple random sample of size 100 is taken from a population with an unknown standard deviation. A normal probability plot of the data displays significant curvature but no outliers. Can you reasonably apply the t-interval procedure? Explain your answer.

Explain the effect on the margin of error and hence the effect on the accuracy of estimating a population mean by a sample mean.

Increasing the sample size while keeping the same confidence level.

Digital Viewing Times. Refer to Exercise 8.130

a. Find and interpret a 90%lower confidence bound for last year's mean time spent per day with digital media by American adults.

b. Compare your one-sided confidence interval in part (a) to the (two-sided) confidence interval found in Exercise 8.130.

"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.
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