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Chapter 9: Q. 68 (page 588)

Anemia For the study of Jordanian children in Exercise 64, the sample mean hemoglobin level was 11.3mg/dl and the sample standard deviation was 1.6mg/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.t=-3.0936

b.p-value=1.638; using calculator,p-value 鈮� 0.998352

Step by step solution

01

Given information

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

02

Explanation (part a)

Hypothesised mean渭鈧赌=12

Sample mean虫虅=11.3

Sample standard deviations=1.6

Sample sizen=50

t-score

t=x炉鈭�渭0s鈰�n

On plugging all the values, we gett=-3.0936

03

Explanation (part b)

Using Table B, the p-value is found to be p-value=1.638

Using calculator, the p-value can be find by,

t-score: the test statistic follows the t-distribution with 49degrees of freedom.

p-value鈮�0.998352

Your result is not statistically significant: there is not enough evidence to reject the null hypothesis.

This decision is made at significance level伪=0.05.

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Most popular questions from this chapter

The French naturalist Count Buffon (1707-1788) tossed a coin 4040 times. He got 2048 heads. That's a bit more than one-half. Is this evidence that Count Buffon's coin was not balanced? To find out, Luisa decides to perform a significance test. Unfortunately, she made a few errors along the way. Your job is to spot the mistakes and correct them.

H0:渭>0.5Ha:x炉=0.5

- Independent 4040(0.5)=2020 and 4040(1-0.5)=2020 are both at least 10 .

- Normal There are at least 40,400 coins in the world.

t=0.5鈭�0.5070.5(0.5)4040=鈭�0.89;P-value=1鈭�0.1867=0.8133

Reject H0because the P-value is so large and conclude that the coin is fair.

A change is made that should improve student satisfaction with the parking situation at a local high school. Right now, 37% of students approve of the parking that鈥檚 provided. The null hypothesis H0:p>0.37is tested against the alternativeHa:p=0.37.

You read that a statistical test at the 伪=0.01 level has a probability of 0.14 of making a Type II error when a specific alternative is true. What is the power of the test against this alternative?

You are thinking of conducting a one-sample t-test about a population mean M using a 0.05 significance level. You suspect that the distribution of the population is not Normal and may be moderately skewed. Which of the following statements is correct?

(a) You should not carry out the test because the population does not have a Normal distribution.

(b) You can safely carry out the test if your sample size is large and there are no outliers.

(c) You can safely carry out the test if there are no outliers, regardless of the sample size.

(d) You can carry out the test only if the population standard deviation is known.

(e) The t procedures are robust鈥攜ou can u

A random sample of 100likely voters in a small city produced 59voters in favor of Candidate A. The observed value of the test statistic for testing the null hypothesis H0:p=0.5versus the alternative hypothesis Ha:p=0.5is

(a)z=0.59-0.50.59(0.41)100

(b). z=0.59-0.50.5(0.5)100

(c). z=0.5-0.590.59(0.41)100

(d). z=0.5-0.590.5(0.5)100

(e).t=0.59-0.50.5(0.5)100
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