Beginning Statistics, 2nd Edition
Beginning Statistics, 2nd Edition
2nd Edition
ISBN: 9781932628678
Author: Carolyn Warren; Kimberly Denley; Emily Atchley
Publisher: Hawkes Learning Systems
Question
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Chapter 11.4, Problem 9E
To determine

(a)

To find:

The null and alternative hypotheses for given scenario.

Expert Solution
Check Mark

Answer to Problem 9E

Solution:

The alternate hypothesis is p2p1<0 and the null hypothesis is p2p10.

Explanation of Solution

Given information:

There is an old wives’ tale that women who eat chocolate during pregnancy are more likely to have happy babies. A pregnancy magazine wants to test this claim, and it gathers 100 randomly selected pregnant women for its study. Half of the women sampled agree to cat chocolate at least once a day, while the other half agree to forego chocolate for the duration of their pregnancies. A year later, the ladies complete a survey regarding the overall happiness of their babies. The results are given in the following table. At the 0.01 level of significance, test the claim of the old wives’ tale.

Number of babies
Happy babies Unhappy Babies
With chocolate 24 26
Without chocolate 22 28

Concept:

The null hypothesis is a statement of no difference, that there is no significant difference between the two phenomena. It is considered to be true until it is nullified by statistical evidence for an alternative hypothesis.

An alternative hypothesis is a contradicting statement to the null hypothesis and states a significant difference between the two phenomena It is accepted when the null hypothesis is false.

Assume, Beginning Statistics, 2nd Edition, Chapter 11.4, Problem 9E , additional homework tip  1be the estimate proportion for the happy babies whose mothers ate chocolate during their pregnancies.

And Beginning Statistics, 2nd Edition, Chapter 11.4, Problem 9E , additional homework tip  2be the estimate proportion for the happy babies whose mothers did not eat chocolate during their pregnancies.

From the given information, the claim is made that the pregnant women gives happy child after having chocolates, that is,

p2<p1p2p1<0

The required alternate hypothesis is p2p1<0.

It is known that the null hypothesis for the given proportion is H0:p2p10. The required null hypothesis is p2p10.

To determine

(b)

To find:

The distribution is to be used for the statistical test and its level of significance.

Expert Solution
Check Mark

Answer to Problem 9E

Solution:

The standard normal variate z-test will be applied for given samples at the level of significance is 0.01 or 1%.

Explanation of Solution

Given information:

There is an old wives’ tale that women who eat chocolate during pregnancy are more likely to have happy babies. A pregnancy magazine wants to test this claim, and it gathers 100 randomly selected pregnant women for its study. Half of the women sampled agree to cat chocolate at least once a day, while the other half agree to forego chocolate for the duration of their pregnancies. A year later, the ladies complete a survey regarding the overall happiness of their babies. The results are given in the following table. At the 0.01 level of significance, test the claim of the old wives’ tale.

Number of babies
Happy babies Unhappy Babies
With chocolate 24 26
Without chocolate 22 28

Since the both sample size is greater than 30, normal distribution will be followed and the standard normal variate z-test will be applied.

According to the null hypothesis, left-tailed test is to be applied.

From the given information, the level of significance is 0.01 or 1%.

To determine

(c)

To find:

The test statistic for a hypothesis test for two population means.

Expert Solution
Check Mark

Answer to Problem 9E

Solution:

The test statistic value for given population is 0.401286.

Explanation of Solution

Given information:

There is an old wives’ tale that women who eat chocolate during pregnancy are more likely to have happy babies. A pregnancy magazine wants to test this claim, and it gathers 100 randomly selected pregnant women for its study. Half of the women sampled agree to cat chocolate at least once a day, while the other half agree to forego chocolate for the duration of their pregnancies. A year later, the ladies complete a survey regarding the overall happiness of their babies. The results are given in the following table. At the 0.01 level of significance, test the claim of the old wives’ tale.

Number of babies
Happy babies Unhappy Babies
With chocolate 24 26
Without chocolate 22 28

Formula used:

The formula used to perform a statistic test for a hypothesis test for two population proportions is,

z=(p^1p^2)(p1p2)p¯(1p¯)(1n1+1n2)

Where, p^1 and p^2 are the estimate proportion for samples 1 and 2 respectively, that be calculated as p^1=x1n1 and p^2=x2n2 respectively. And p1 and p2 are the estimate proportion for samples 1 and 2 respectively.

The weighted estimated proportion for the population is p¯ and calculated by. n1p¯=x1+x2n1+n2

Also x1 and x2 are the successes that happens in samples 1 and 2 respectively, The sample size of samples 1 and 2 are n1 and n2 respectively.

Calculation:

From the problem, the sample successes are x1=24 and x2=22. The sample sizes are n1=24+26=50 and n2=22+28=50. The null hypothesis for the population is H0:p2p10 at α=1%.

It is already assumed that p1 be the estimate proportion for the happy babies whose mothers ate chocolate during their pregnancies. And p2 be the estimate proportion for the happy babies whose mothers did not eat chocolate during their pregnancies.

The sample success and size for sample 1 is x1=24 and n1=50 respectively. The estimate proportion for samples 1 is,

p^1=x1n1

Substitute 24 for x1 and Beginning Statistics, 2nd Edition, Chapter 11.4, Problem 9E , additional homework tip  3for n1

in above equation.

p^1=2450=0.48---(1)

The sample success and size for sample 2 is x2=22 and n2=50 respectively. The estimate proportion for samples 2 is

p^2=x2n2

Substitute 22 for x2 and 50 for n2 in above equation.

p^2=2250=0.44---(2)

The weighted estimated proportion for the population is given as,

p¯=x1+x2n1+n2

Substitute 24 for Beginning Statistics, 2nd Edition, Chapter 11.4, Problem 9E , additional homework tip  4, 22 for Beginning Statistics, 2nd Edition, Chapter 11.4, Problem 9E , additional homework tip  5,Beginning Statistics, 2nd Edition, Chapter 11.4, Problem 9E , additional homework tip  6 for Beginning Statistics, 2nd Edition, Chapter 11.4, Problem 9E , additional homework tip  7and 50 for Beginning Statistics, 2nd Edition, Chapter 11.4, Problem 9E , additional homework tip  8in above equation.

p¯=24+2250+50=46100=0.46---(3)

The statistical test for two population proportions is given as

z=(p^1p^2)(p1p2)p¯(1p¯)(1n1+1n2)

From equations (1),(2) and (3), substitute 0.46 for p¯, 0.48 for p^1, 0.44 for p^2, Beginning Statistics, 2nd Edition, Chapter 11.4, Problem 9E , additional homework tip  9 for n1 and 50 for n2 in above equation.

z=(0.480.44)(p1p2)(0.46)(10.46)(150+150)

It is given that the null hypothesis for the given proportion is H0:p2p10. Substitute 0 for p2p1.

z=(0.04)(0)(0.2484)(0.04)=0.401286

Thus, the test statistic value is z=0.401286.

To determine

(d)

To find:

The value and conclude the decision at given level of significance.

Expert Solution
Check Mark

Answer to Problem 9E

Solution:

The claim is incorrectly made that the pregnant women gives happy child after having chocolates. There will no difference between babies by eating chocolates.

Explanation of Solution

Given information:

There is an old wives’ tale that women who eat chocolate during pregnancy are more likely to have happy babies. A pregnancy magazine wants to test this claim, and it gathers 100 randomly selected pregnant women for its study. Half of the women sampled agree to cat chocolate at least once a day, while the other half agree to forego chocolate for the duration of their pregnancies. A year later, the ladies complete a survey regarding the overall happiness of their babies. The results are given in the following table. At the 0.01 level of significance, test the claim of the old wives’ tale.

Number of babies
Happy babies Unhappy Babies
With chocolate 24 26
Without chocolate 22 28

Formula Used:

The p-value of given population for calculated z=k is,

p=P(zk)

When p>α, then null hypothesis is accepted at the level of significance α.

It is already computed that zcalc=0.401286 for given samples. Its value is,

p=P(z0.401286)=ϕ(0.401286)=0.3441360.3441

And the level of significance is 0.01, that is α=0.01.

0.01<0.3441α<p

The null hypothesis is accepted as p2p10 and the alternative hypothesis is rejected as p2p1<0.

Thus, the claim is incorrectly made that the pregnant women gives happy child after having chocolates. There will no difference between babies by eating chocolates.

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Chapter 11 Solutions

Beginning Statistics, 2nd Edition

Ch. 11.1 - Prob. 11ECh. 11.1 - Prob. 12ECh. 11.1 - Prob. 13ECh. 11.1 - Prob. 14ECh. 11.1 - Prob. 15ECh. 11.1 - Prob. 16ECh. 11.1 - Prob. 17ECh. 11.1 - Prob. 18ECh. 11.1 - Prob. 19ECh. 11.1 - Prob. 20ECh. 11.1 - Prob. 21ECh. 11.2 - Prob. 1ECh. 11.2 - Prob. 2ECh. 11.2 - Prob. 3ECh. 11.2 - Prob. 4ECh. 11.2 - Prob. 5ECh. 11.2 - Prob. 6ECh. 11.2 - Prob. 7ECh. 11.2 - Prob. 8ECh. 11.2 - Prob. 9ECh. 11.2 - Prob. 10ECh. 11.2 - Prob. 11ECh. 11.2 - Prob. 12ECh. 11.2 - Prob. 13ECh. 11.2 - Prob. 14ECh. 11.2 - Prob. 15ECh. 11.2 - Prob. 16ECh. 11.2 - Prob. 17ECh. 11.3 - Prob. 1ECh. 11.3 - Prob. 2ECh. 11.3 - Prob. 3ECh. 11.3 - Prob. 4ECh. 11.3 - Prob. 5ECh. 11.3 - Prob. 6ECh. 11.3 - Prob. 7ECh. 11.3 - Prob. 8ECh. 11.3 - Prob. 9ECh. 11.3 - Prob. 10ECh. 11.4 - Prob. 1ECh. 11.4 - Prob. 2ECh. 11.4 - Prob. 3ECh. 11.4 - Prob. 4ECh. 11.4 - Prob. 5ECh. 11.4 - Prob. 6ECh. 11.4 - Prob. 7ECh. 11.4 - Prob. 8ECh. 11.4 - Prob. 9ECh. 11.4 - Prob. 10ECh. 11.4 - Prob. 11ECh. 11.5 - Prob. 1ECh. 11.5 - Prob. 2ECh. 11.5 - Prob. 3ECh. 11.5 - Prob. 4ECh. 11.5 - Prob. 5ECh. 11.5 - Prob. 6ECh. 11.5 - Prob. 7ECh. 11.5 - Prob. 8ECh. 11.5 - Prob. 9ECh. 11.5 - Prob. 10ECh. 11.5 - Prob. 11ECh. 11.5 - Prob. 12ECh. 11.5 - Prob. 13ECh. 11.5 - Prob. 14ECh. 11.5 - Prob. 15ECh. 11.5 - Prob. 16ECh. 11.5 - Prob. 17ECh. 11.5 - Prob. 18ECh. 11.5 - Prob. 19ECh. 11.5 - Prob. 20ECh. 11.5 - Prob. 21ECh. 11.5 - Prob. 22ECh. 11.5 - Prob. 23ECh. 11.5 - Prob. 24ECh. 11.5 - Prob. 25ECh. 11.5 - Prob. 26ECh. 11.6 - Prob. 1ECh. 11.6 - Prob. 2ECh. 11.6 - Prob. 3ECh. 11.6 - Prob. 4ECh. 11.6 - Prob. 5ECh. 11.6 - Prob. 6ECh. 11.6 - Prob. 7ECh. 11.6 - Prob. 8ECh. 11.6 - Prob. 9ECh. 11.6 - Prob. 10ECh. 11.6 - Prob. 11ECh. 11.6 - Prob. 12ECh. 11.6 - Prob. 13ECh. 11.6 - Prob. 14ECh. 11.6 - Prob. 15ECh. 11.6 - Prob. 16ECh. 11.CR - Prob. 1CRCh. 11.CR - Prob. 2CRCh. 11.CR - Prob. 3CRCh. 11.CR - Prob. 4CRCh. 11.CR - Prob. 5CRCh. 11.CR - Prob. 6CRCh. 11.CR - Prob. 7CRCh. 11.CR - Prob. 8CRCh. 11.CR - Prob. 9CRCh. 11.CR - Prob. 10CR
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