PROBLEM STATEMENT 1 It is known that a certain laboratory task takes the average person 2.5 seconds. A developmental psychologist was interested in whether older people take longer to perform this task. The psychologist tested 30 randomly selected 80-year-olds. Their mean time was 2.7 seconds, with an estimated population standard deviation of 1.4 seconds. What should the psychologist conclude (use the .05 level)? 1. Restate the question as a research hypothesis and a null hypothesis about the populations. (2 points) 2. Determine the characteristics of the comparison distribution: Population mean = (1 point) Population variance = (1 point) Standard deviation of the distribution of sample means = (1 point) Degrees of Freedom = (1 point) 3. Determine the significance cutoff. Use the t table. (1 points) 4. Determine your sample’s score on the comparison distribution. (2 points) 5. Decide whether to reject the null hypothesis. (2 points)

Set Up Hypothesis Null Hypothesis H0: U = 2.5 Alternate, mean response time of older people slower than the response time of people in general H1: U>2.5 Test Statistic Population Mean(U)=2.5 Given That X(Mean)=2.7 Standard Deviation(S.D)=1.4 Number (n)=30 we use Test Statistic (Z) = x-U/(s.d/Sqrt(n)) Zo=2.7-2.5/(1.4/Sqrt(30) Zo =0.7825 | Zo | =0.7825 Critical Value The Value of |Z ?| at LOS 0.05% is 1.64 We got |Zo| =0.7825 & | Z ? | =1.64 Make Decision Hence Value of |Zo | < | Z ? | and Here we Do not Reject Ho P-Value : Right Tail - Ha : ( P > 0.7825 ) = 0.217 Hence Value of P0.05 < 0.217, Here We Do not Reject Ho

accept hypothesis, we don't have evidence that mean response time of older people slower than the response time of people in general

Population 1: Population of times taken by the average person

Population 2: Population of times taken by older people

Explanation:

H0: Null Hypothesis:  2.5 (The older people are not slower (do not take longer) on this task

HA: Alternative Hypothesis:  2.5 (The older people are slower ( take longer) on this task (Claim)

n = 30

 = 2.7

 = 1.4

Take  = 0.05

From Table, critical value of Z = 1.64

Test Statistic is given by:

Since calculated value of Z = 0.782 is less than critical value of Z = 1.64, the difference is not significant. Fail to reject null hypothesis.

By Technology,

p value = 0.217

Since p value = 0.217 is greater than  = 0.05, the difference is not significant. Fail to reject null hypothesis.

Conclusion: The data do not support the claim that the older people are slower ( take longer) on this task.

PROBLEM STATEMENT 2 A study tested the effects of science fiction movies on people’s belief in the supernatural. A scale was designed to measure the degree to which a person believes in the supernatural with high scores indicating high levels of belief. Seven participants completed the scale before and after watching a popular science fiction movie. Participants’ scores are listed below. Assume that the researcher had reason to expect participants to believe less in the supernatural after watching the movie. Belief-in-Supernatural Scores, Before and After Watching Science Fiction Movie. Carry out a t test for dependent means (use the .01 significance level).

Participant Before After A 3 3 B 5 3 C 9 6 D 6 8 E 7 8 F 5 2 G 4 1 1. Restate the question as a research hypothesis and a null hypothesis about the populations. (2 points) 2. Determine the characteristics of the comparison distribution: Population mean = (1 point) The variance of the distribution of means = (2 points) The estimated standard deviation of the population of difference scores = (1 point) The SD of distribution of means = (2 points) Degrees of Freedom = (1 point) 3. Determine the significance cutoff. Use the t table. (1 points) 4. Determine your sample’s score on the comparison distribution. (2 points) 5. Decide whether to reject the null hypothesis. (2 points)

Solution:

A. Paired t-test is used because the data is dependent.

B. Step 1: Null Hypothesis (Ho): d  0

Alternative Hypothesis (Ha): d < 0

Step 2: Level of significance,a = 0.05

Step 3: Test Statistics

t = (-1.143 - 0)/ (2.116/7)

t = -1.43

Step 4: Reject or Fail to reject Ho

Degrees of freedom, df = n - 1 = 7 - 1 = 6

Using t-tables, the critical value is

t (a/2, df) = t (0.025, 6) = -2.447

Step 5: Conclusion

Since test statistics is greater than the critical value, we fail to reject Ho.

Hence, there is no significant difference in scores after watching a popular science fiction movie.

Calculations of  and sd

 = -1.143, sd =  = 26.8571/6 = 2.116

C. 95% confidence interval is given by:-

-1.143  t (0.025, 6)*(2.116/(7))

-1.143  2.447*0.7997

-1.143  1.957

-3.100, 0.814

D. Cohen's d = /sd

Cohen's d = 1.143/2.116

Cohen's d = 0.54 (Medium effect)

Step 1:

H0: Null Hypothesis:  (People believed more than or equal in the supernatural after watching the movie)

HA:Alternative Hypothesis:  (People believed less in the supernatural after watching the movie) (Claim)

Step 2:

From the given data, values of d = Before - After are got as follows:

d = Before - After = 0, 2, 3, - 2, - 1, 3 3

From d values, the following statistics are calculated:

n = Sample Size = 7

 = Mean of d values = 1.143

sd = Standard Deviation of d values

Step 3:

 = 0.01

df = 7 - 1 = 6

From Table, critical value of t = 3.143

Step 4:

The Test Statistic is calculated as follows:

Step 5:

Since calculated value of t = 1.429 isless than critical value of t = 3.143, the difference is not significant. Fai lto reject null hypothesis.

Conclusion:

The data do not support the claim that eople believed less in the supernatural after watching the movie.

(b)

Effrect Sizeis given by: