A psychologist theorized that people can hear better when they have just eaten a

Question

A psychologist theorized that people can hear better when they have just eaten a large meal. Six individuals were randomly assigned to eat either a large meal or a small meal. After eating the meal, their hearing was tested. The hearing ability scores (high numbers indicate greater ability) are given m the following table. Using the .05 level, do the results support the psychologist's theory? (a) Use the steps of hypothesis testing, (b) sketch the distributions involved, and (c) explain your answers to someone who has never had a course in statistics. Twenty students randomly assigned to an experimental group receive an instructional program; 30 in a control group do not. After 6 months, both groups are tested on their knowledge. The experimental group has a mean of 38 on the test (with an estimated population standard deviation of 3); the control group has a mean of 35 (with an estimated population standard deviation of 5). Using the .05 level, what should the experimenter conclude? (a) Use the steps of hypothesis testing, (b) sketch the distributions involved, and (c) explain your answer to someone who is familiar with the t test for a single sample but not with the t test for independent means. A study of the effects of color on easing anxiety compared anxiety test scores of participants who completed the test printed on either soft yellow paper or on ' harsh green paper. The scores for five participants who completed the test printed on the yellow paper were 17, 19, 28, 21, and 18. The scores for four participants who completed the test on the green paper were 20, 26, 17, and 24. Using the .05 level, one-tailed (predicting lower anxiety scores for the yellow paper), what should the researcher conclude? (a) Use the steps of hypothesis testing, (b) sketch the distributions involved, and (c) explain your answers to someone who is familiar with the t test for a single sample but not with the t test for independent means. Figure the estimated effect size for problems (a) 16, (b) 17, and (c) 18. (d) Explain your answer to part (a) to a person who understands the t test for independent means but is unfamiliar with effect size. Figure the approximate power of a t test for independent means for each of the following planned studies:

Explanation / Answer

Question 17.a.

Solution:

Here, we have to use the two sample t test for the population mean. The null and alternative hypotheses for this test are given as below:

H0: µ1 = µ2

Ha: µ1 > µ2

This is a one tailed test. This is a upper tailed or right tailed test.

The level of significance for this test is given as alpha = 0.05.

The test statistic formula for this test is given as below:

Test statistic = t = (X1bar – X2bar) / sqrt[(S1^2/N1)+(S2^2/N2)]

From the given data we have

N1 = 3

N2 = 3

X1bar = 24

X2bar = 21

S1 = 1.7321

S2 = 2.0000

Degrees of freedom = 3 + 3 – 2 = 4

Critical value = 2.1318

Test statistic = (24 – 21) / sqrt[(1.7321^2/3)+(2^2/3)]

Test statistic = 3/sqrt[2.333390137]

Test statistic = 3/1.527543825

Test statistic = 1.963937107

P-value = 0.06

Alpha value = 0.05

P-value > Alpha value

So, we do not reject the null hypothesis that there is no any significant difference in the average hearing score when taking big meal and small meal.

Question 19.a.

Here, we have to use the two sample t test for the population mean. The null and alternative hypotheses for this test are given as below:

H0: µ1 = µ2

Ha: µ1 < µ2

This is a one tailed test. This is a lower tailed or left tailed test.

The level of significance for this test is given as alpha = 0.05.

The test statistic formula for this test is given as below:

Test statistic = t = (X1bar – X2bar) / sqrt[(S1^2/N1)+(S2^2/N2)]

From the given data we have

N1 = 4

N2 = 4

X1bar = 21.25

X2bar = 21.75

S1 = 4.7871

S2 = 4.0311

Df = 4+4-2 = 6

Test statistic = t = (21.25 – 21.75) / sqrt[(4.7871^2/4)+(4.0311^2/4)]

Test statistic = t = -0.50/sqrt[9.791523405]

Test statistic = t = -0.50/3.129141001

Test statistic = t = -0.159788261

P-value = 0.4397

Alpha value = 0.05

P-value > alpha value

So, we do not reject the null hypothesis that there is no any significant difference in the average anxiety test scores by using soft yellow paper and harsh green paper.

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