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Chapter 9: Q. 77. (page 607)

Pressing pills A drug manufacturer forms tablets by compressing a granular material that contains the active ingredient and various fillers. The hardness of a sample from each batch of tablets produced is measured to control the compression process. The target value for the hardness is μ=11.5The hardness data for a random sample of 20 tablets from one large batch are

Is there convincing evidence at the 5%level that the mean hardness of the tablets in this batch differs from the target value?

Short Answer

Expert verified

No, there is sufficient evidence to show that the mean hardness is different from the targeted value.

Step by step solution

01

Given information

The data set is:

02

The objective is to find whether there is sufficient evidence to show that the mean hardness is different from the targeted value of 11.5at 5%the significance level.

We know,

The test statistic formula is: t=x¯-μosn

The null and alternative hypotheses are as follows:

H0:μ=11.5Ha:μnotequalto11.5

The alternative hypothesis denotes a two-tailed test.

The Minitab output is as follows:

The p-value is 0.449

Here, p-value (0.449)>α(0.05)The null hypothesis does not fail.

There is insufficient evidence at the 5%level of significance to show that the mean hardness of the tablets differs from the target value of11.5.

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Most popular questions from this chapter

A significance test allows you to reject a null hypothesis H0H0in favor of an alternative hypothesisHaaat the 5%significance level. What can you say about significance at the1%level?

a.H0H0can be rejected at the1%significance level.

b. There is insufficient evidence to rejectH0H0at the1%significance level.

c. There is sufficient evidence to accept H0H0at the 1%significance level.

d.HaHacan be rejected at the 1%significance level.

e. The answer can't be determined from the information given.

Interpreting a P-value A student performs a test of H0:μ=100H0:μ=100versus Ha: μ>100Ha:μ>100and gets a P-value of 0.044.The student says, "There is a0.044 probability of getting the sample result I did by chance alone." Explain why the student's explanation is wrong.

Fonts and reading ease Does the use of fancy type fonts slow down the reading of text on a computer screen? Adults can read four paragraphs of a certain text in the common Times New Roman font in an average time of 22 seconds. Researchers asked a random sample of 24 adults to read this text in the ornate font named Gigi. Here are their times (in seconds):

Do these data provide convincing evidence that it takes adults longer than 22 seconds, on average, to read these four paragraphs in Gigi font?

Jump around Student researchers Haley, Jeff, and Nathan saw an article on the Internet claiming that the average vertical jump for teens was 15 inches. They wondered if the average vertical jump of students at their school differed from 15 inches, so they obtained a list of student names and selected a random sample of 20 students. After contacting these students several times, they finally convinced them to allow their vertical jumps to be measured. Here are the data (in inches):

Do these data provide convincing evidence at the α=0.10 level that the average vertical jump of students at this school differs from 15 inches?

After once again losing a football game to the archrival, a college’s alumni association conducted a survey to see if alumni were in favor of firing the coach. An SRS of 100 alumni from the population of all living alumni was taken, and 64 of the alumni in the sample were in favor of firing the coach. Suppose you wish to see if a majority of all living alumni is in favor of firing the coach. The appropriate standardized test statistic is

(a)z=0.64-0.50.64(0.36)100z=0.64-0.50.64(0.36)100

role="math" localid="1654432946823" (b)t=0.64-0.50.64(0.36)100t=0.64-0.50.64(0.36)100

(c)z=0.64-0.50.5(0.5)100z=0.64-0.50.5(0.5)100

(d)z=0.64-0.50.64(0.36)64z=0.64-0.50.64(0.36)64

(e)z=0.5-0.640.5(0.5)100z=0.5-0.640.5(0.5)100
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