For the study of Jordanian children in Exercise 64, the sample mean hemoglobin level was 11.3 mg/dl and the sample standard deviation was 1.6 mg/dl.

(a) Calculate the test statistic.

(b) Find the P-value using Table B. Then obtain a more precise P-value from your calculator.

Short Answer

Expert verified

a. the test statistic is-3.094

b. the P-value using Table B 0.016

Step by step solution

01

Introduction

A p-value estimates the probability of obtaining the noticed outcomes, it is consistent with expectations that the invalid hypothesis. The lower the p-esteem, the more prominent the statistical significance of the noticed difference

02

Explanation Part (a)

The sample mean is x-=11.3

The sample size is n = 50

Standard deviation is 1.6

Using,

localid="1652954330667" t=x-μsn

localid="1663929363504" =11.3121.650=-3.094

Hence the test statistic islocalid="1663929380848" -3.094

03

Explanation Part (b)

Calculating the degree of freedom,

n-1=50-1=49

The p-value is,

=Ptc>|3.094|=0.0016

Hence the P-value is0.0016

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 Vaia!

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

A random sample of 900students at a very large university was asked which social-networking site they used most often during a typical week. Their responses are shown in the table below.

Assuming that gender and preferred networking site are independent, which of the following is the expected count for female and LinkedIn?

a)18.85b)46.11c)87.00d)91.65e)103.35

A distribution that represents the number of cars parked in a randomly selected residential driveway on any night is given by

xi:0 123 4
pi :0.10.20.350.250.15

Which of the following statements is correct?

(a) This is a legitimate probability distribution because each of the pi-values is between 0and 1.

(b) This is a legitimate probability distribution because xiis exactly 10.

(c) This is a legitimate probability distribution because each of the pi-values is between 0and 1and the xiis exactly 10.

(d) This is not a legitimate probability distribution because xiis not exactly 10.

(e) This is not a legitimate probability distribution because role="math" localid="1650523963322" pi is not exactly 1.

Color words (4.2)Let’s review the design of the study.

(a) Explain why this was an experiment and not an observational study.

(b) Did Mr. Starnes use a completely randomized design or a randomized block design? Why do you think he chose this experimental design?

(c) Explain the purpose of the random assignment in the context of the study. The data from Mr. Starnes’s experiment are shown below. For each subject, the time to perform the two tasks is given to the nearest second.

The P-value for the test in Question 5 is 0.0087. A correct interpretation of this result is that

(a) the probability that there is no linear relationship between an average number of putts per hole and total winnings for these 69players is 0.0087.

(b) the probability that there is no linear relationship between the average number of putts per hole and total winnings for all players on the PGA Tour’s world money list is 0.0087.

(c) if there is no linear relationship between an average number of putts per hole and total winnings for the players in the sample, the probability of getting a random sample of 69players yields a least-squares regression line with a slope of -4139198or less is 0.0087.

(d) if there is no linear relationship between an average number of putts per hole and total winnings for the players on the PGA Tour’s world money list, the probability of getting a random sample of 69 players yields a least-squares regression line with a slope of -4139198or less is 0.0087.

(e) the probability of making a Type II error is0.0087.

The cell that contributes most to the chi-square statistic is

(a) men who developed a rash.

(b) men who did not develop a rash.

(c) women who developed a rash.

(d) women who did not develop a rash.

(e) both (a) and (d).

Question refers to the following situation:

Could mud wrestling be the cause of a rash contracted by University of Washington students? Two physicians at the University of Washington student health center wondered about this when one male and six female students complained of rashes after participating in a mud-wrestling event. Questionnaires were sent to a random sample of students who participated in the event. The results, by gender, are summarized in the following table.

Some Minitab output for the previous table is given below. The output includes the observed counts, the expected counts, and the chi-square statistic.

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.

Sign-up for free