A psychologist obtains a random sample of 20 mothers in the first trimester of t

Question

A psychologist obtains a random sample of 20 mothers in the first trimester of their pregnancy. The mothers are asked to play Mozart in the house at least 30 minutes each day until they give birth After 5 years, the child is administered an IQ test. It is known that IQs are normally distributed with a mean of 100. It the IQs of the 20 children in the study result in a sample mean of 104.4 and sample standard deviation of 16, is there evidence that the children have higher IQs? Use the alpha = 0 05 level of significance. Complete parts (a) through (d) Determine the null and alternative hypotheses. Calculate the P-value. (Round to throe decimal places as needed) State the conclusion for the test Choose the correct answer below Reject H_0 because the P-value is less the alpha = 0.05 level of significance. Reject H_0 because the P-value is greater than the alpha = 0 05 level of significance Do not reject H_0 because the P-value is greater than the alpha = 0 05 level of significance. Do not reject H_0 because the P-value is loss than the alpha = 0 05 level of significance State the conclusion in context of the problem. State the conclusion in context of the problem. There sufficient evidence at the alpha = 0.05 level of significance to conclude that mothers who listen to Mozari have children with higher IQs.

Explanation / Answer

Part a

Solution:

The null and alternative hypothesis is given as below:

H0: µ = 100

Ha: µ > 100

Part b

Solution:

P-value = 0.1169

Part c

Solution:

Correct alternative = (C) Do not reject H0 because P-value is greater than the alpha = 0.05 level of significance

Part d

Solution:

There is no sufficient evidence that at 0.05 level of significance that mothers who listen to Mozart have children with higher IQs.

(Use following results for test for more reference.)

t Test for Hypothesis of the Mean

Data

Null Hypothesis                m=

100

Level of Significance

0.05

Sample Size

20

Sample Mean

104.4

Sample Standard Deviation

16

Intermediate Calculations

Standard Error of the Mean

3.5777

Degrees of Freedom

19

t Test Statistic

1.2298

Upper-Tail Test

Upper Critical Value

1.7291

p-Value

0.1169

Do not reject the null hypothesis

t Test for Hypothesis of the Mean

Data

Null Hypothesis                m=

100

Level of Significance

0.05

Sample Size

20

Sample Mean

104.4

Sample Standard Deviation

16

Intermediate Calculations

Standard Error of the Mean

3.5777

Degrees of Freedom

19

t Test Statistic

1.2298

Upper-Tail Test

Upper Critical Value

1.7291

p-Value

0.1169

Do not reject the null hypothesis

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