How much does a sleeping bag cost? Let\'s say you want a sleeping bag that shoul

Question

How much does a sleeping bag cost? Let's say you want a sleeping bag that should keep you warm in temperatures from 20°F to 45°F. A random sample of prices ($) for sleeping bags in this temperature range is given below. Assume that the population of x values has an approximately normal distribution. 75 35 55 35 80 80 30 23 100 110 105 95 105 60 110 120 95 90 60 70 (a) Use a calculator with mean and sample standard deviation keys to find the sample mean price x and sample standard deviation s. (Round your answers to two decimal places.) (b) Using the given data as representative of the population of prices of all summer sleeping bags, find a 90% confidence interval for the mean price of all summer sleeping bags. (Round your answers to two decimal places.) ower limit upper limit

Explanation / Answer

SOlution:

code in R:

prices_sleepingbags <- c(75,35,55,35,80,80,30,23,100,110,105,95,105,60,110,120,95,90,60,70)
print(prices_sleepingbags)
mean(prices_sleepingbags)
sd(prices_sleepingbags)
t.test(prices_sleepingbags,conf.level=0.90)

Output:

> prices_sleepingbags <- c(75,35,55,35,80,80,30,23,100,110,105,95,105,60,110,120,95,90,60,70)

> print(prices_sleepingbags)

[1] 75 35 55 35 80 80 30 23 100 110 105 95 105 60

[15] 110 120 95 90 60 70

> mean(prices_sleepingbags)

[1] 76.65

> sd(prices_sleepingbags)

[1] 29.66883

> t.test(prices_sleepingbags,conf.level=0.90)

One Sample t-test

data: prices_sleepingbags

t = 11.554, df = 19, p-value = 4.9e-10

alternative hypothesis: true mean is not equal to 0

90 percent confidence interval:

65.17867 88.12133

sample estimates:

mean of x

76.65

ANSWERS FOR QUES

SOLUTIONA:

sample mean=xbar=76.65

sample sd=s=29.67

SOLUTIONB:

lower limit=65.18

upper limit=88.12

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