/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q23 Testing Hypotheses. In Exercises... [FREE SOLUTION] | 91影视

91影视

Testing Hypotheses. In Exercises 13鈥24, assume that a simple random sample has been selected and test the given claim. Unless specified by your instructor, use either the P-value method or the critical value method for testing hypotheses. Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), or critical value(s), and state the final conclusion that addresses the original claim.

Car Booster Seats The National Highway Traffic Safety Administration conducted crash tests of child booster seats for cars. Listed below are results from those tests, with the measurements given in hic (standard head injury condition units). The safety requirement is that the hic measurement should be less than 1000 hic. Use a 0.01 significance level to test the claim that the sample is from a population with a mean less than 1000 hic. Do the results suggest that all of the child booster seats meet the specified requirement?

774 649 1210 546 431 612

Short Answer

Expert verified

There is not sufficient evidence to accept the claim that the sample is from a population with a mean less than 1000 hic.

The results of the crash test of child booster seats suggest that all child booster seats does not meet the specified requirement.

Step by step solution

01

Given information

The results of the crash test are 774, 649,1210, 546, 431, 612.

The safety requirement is that the hic measurement should be less than 1000 hic.

The significance level is 0.01.

02

Check the requirements

Assume that the population follows the normal distribution and the samples are randomly selected.

The sample size (n) of the crash test of child booster seats is 16.

The t-distribution would be used here.

03

Describe the hypothesis

Null hypothesis, H0is a statement of the claim thatsample is from a population with a mean is equal to 1000 hic.

Alternate hypothesis,H1is a statement of the claim thatsample is from a population with a mean less than 1000 hic.

Let be the true population mean.

Mathematically, it can be expressed as,

H0:=1000H1:<1000

The hypothesis is left-tailed.

04

Calculate the test statistic

Formula for test statistic is given by,

t=x-sn

Where,x is the sample mean and s is the standard deviation of sample.

The sample mean is computed as,

x=xin=774+649+...+6126=703.67

The sample standard deviation is,

s=xi-x2n-1=774-703.672+649-703.672+...+612-703.6726-1=272.72

By substituting these values, test statistics is given by,

t=x-sn=703.67-1000272.736=-2.661

05

Calculate the critical value

The significance level is 0.01.

Sample size (n) is 6.

The degree of freedom is computed as,

df=n-1=6-1=5

In the t-distribution table, find the value corresponding to the row value of degree of freedom 5 and column value of area in one tail 0.01 is 3.365 which is critical value role="math" localid="1649063658875" t0.01; but the given test is left tailed therefore use -3.365 as a critical value.

Thus, the critical valuerole="math" localid="1649063667947" t0.01is -3.365.

The rejection region is t:t<-3.365.

06

Compare test statistic and critical value

Test statistic is -2.661 and the critical value is -3.365.

According to this, we can conclude that the test statistic -2.661 will not fall in the rejection region.

Therefore, we failed to reject the null hypothesis.

07

Conclusion

There is not sufficient evidence to accept the claim that the sample is from a population with a mean less than 1000 hic.

The observation value is 1210 which is too high. Therefore, the results suggest that all child booster seats does not meet the specified requirement.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Early-Onset Dementia. Dementia is the loss of the intellectual and social abilities severe enough to interfere with judgment, behavior, and daily functioning. Alzheimer's disease is the most common type of dementia. In the article "Living with Early Onset Dementia: Exploring the Experience and Developing Evidence Based Guidelines for Practice" (Alzheimer's Care Quarterly, Vol. 5, Issue 2, pp. 111-122), P. Harris and J. Keady explored the experience and struggles of people diagnosed with dementia and their families. A hypothesis test is to be performed to decide whether the mean age at diagnosis of all people with early-onset dementia is less than 55 years old.

Lead in Medicine Listed below are the lead concentrations (in ) measured in different Ayurveda medicines. Ayurveda is a traditional medical system commonly used in India. The lead concentrations listed here are from medicines manufactured in the United States (based on data from 鈥淟ead, Mercury, and Arsenic in US and Indian Manufactured Ayurvedic Medicines Sold via the Internet,鈥 by Saper et al., Journal of the American Medical Association,Vol. 300, No. 8). Use a 0.05 significance level to test the claim that the mean lead concentration for all such medicines is less than 14 g/g.

3.0 6.5 6.0 5.5 20.5 7.5 12.0 20.5 11.5 17.5

Test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then the conclusion about the null hypothesis, as well as the final conclusion that address the original claim. Assume that a simple random sample is selected from a normally distributed population.

Pulse Rates of Women Repeat the preceding exercise using the pulse rates of women listed in Data Set 1 鈥淏ody Data鈥 in Appendix B. For the sample of pulse rates of women, n= 147 and s= 12.5. See the accompanying JMP display that results from using the original list of pulse rates instead of the summary statistics. (Hint:The bottom three rows of the display provide P-values for a two-tailed test, a left-tailed test, and a right-tailed test, respectively.) What do the results indicate about the effectiveness of using the range rule of thumb with the 鈥渘ormal range鈥 from 60 to 100 beats per minute for estimating s in this case?

P-Values. In Exercises 17鈥20, do the following:

a. Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed.

b. Find the P-value. (See Figure 8-3 on page 364.)

c. Using a significance level of = 0.05, should we reject H0or should we fail to reject H0?

The test statistic of z = -1.94 is obtained when testing the claim that p=38 .

This exercise contain graphs portraying the decision criterion for a one-mean 2-test. The curve in each graph is the normal curve for the test statistic under the assumption that the null hypothesis is true. For each exercise, determine the

a. rejection region.

c. critical value(s).

b. nonrejection region.

d. significance level.

e. Construct a graph similar to that in Fig. 9.3 on page 361 that depicts your results from parts (a)-(d).

f. Identify the hypothesis test as two tailed, left tailed or right tailed.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.