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Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test. At a significance level of a = 0.05, what is the correct conclusion?

a. There is enough evidence to conclude that the mean number of hours is more than 4.75

b. There is enough evidence to conclude that the mean number of hours is more than 4.5

c. There is not enough evidence to conclude that the mean number of hours is more than 4.5

d. There is not enough evidence to conclude that the mean number of hours is more than 4.75

Short Answer

Expert verified

c. There is not enough evidence to conclude that the mean number of hours is more than 4.5

Step by step solution

01

Find H0 and Ha: We want to test if an Organization Teenagers take more than 4.5 hours per week, on average, on the phone.

H0:μ=4.5;Ha:μ>4.5

02

Compare α and the p-value:Indicate the correct decision (“reject” or “do not reject” the null hypothesis), the reason for it, and write an appropriate conclusion, using complete sentences.

alphadecisionreason for decision
0.05Do not reject H0
p-value<0.05

Conclusion: There is not enough evidence to conclude that the mean number of hours is more than 4.5

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Most popular questions from this chapter

The student academic group on a college campus claims that freshman students study at least 2.5 hours per day, on average. One Introduction to Statistics class was skeptical. The class took a random sample of 30 freshman students and found a mean study time of 137 minutes with a standard deviation of 45 minutes. At α=0.01 level, is the student academic group’s claim correct?

The US Department of Energy reported that 51.7% of homes were heated by natural gas. A random sample of 221 homes in Kentucky found that 115 were heated by natural gas. Does the evidence support the claim for Kentucky at the α = 0.05 level in Kentucky? Are the results applicable across the country? Why?

H0:μ1,Ha:μ>1

Assume the p-value is 0.1243. What type of test is this? Draw the picture of the p-value.

La Leche League International reports that the mean age of weaning a child from breastfeeding is age four to five worldwide. In America, most nursing mothers wean their children much earlier. Suppose a random survey is conducted of21 U.S. mothers who recently weaned their children. The mean weaning age was nine monthsrole="math" localid="1649394170256" (3/4year) with a standard deviation of 4 months. Conduct a hypothesis test to determine if the mean weaning age in the U.S. is less than four years old.

"Dalmatian Darnation," by Kathy Sparling

A greedy dog breeder named Spreckles

Bred puppies with numerous freckles

The Dalmatians he sought

Possessed spot upon spot

The more spots, he thought, the more shekels.

His competitors did not agree

That freckles would increase the fee.

They said, “Spots are quite nice

But they don't affect price;

One should breed for improved pedigree.”

The breeders decided to prove

This strategy was a wrong move.

Breeding only for spots

Would wreak havoc, they thought.

His theory they want to disprove.

They proposed a contest to Spreckles

Comparing dog prices to freckles.

In records they looked up

One hundred one pups:

Dalmatians that fetched the most shekels.

They asked Mr. Spreckles to name

An average spot count he'd claim

To bring in big bucks.

Said Spreckles, “Well, shucks,

It's for one hundred one that I aim.”

Said an amateur statistician

Who wanted to help with this mission.

“Twenty-one for the sample

Standard deviation's ample:

They examined one hundred and one

Dalmatians that fetched a good sum.

They counted each spot,

Mark, freckle and dot

And tallied up every one.

Instead of one hundred one spots

They averaged ninety six dots

Can they muzzle Spreckles’

Obsession with freckles

Based on all the dog data they've got?

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