SESSION 2

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QUESTION 3E

Hypothesis:

SPSS Steps:

Result:

Since we are using K-S Test, we only need to look at Kolmogorov-Smirnov's Statistic and Sig

This statistic is D-statistics in the textbook. And Sig. is the significance for the test(aka p-value)

For Maternal Age:

D-Statistic = .149 p-value = 0.2 > 0.05 => Fail to Reject Ho => the test is insignificant.

=> We assume normality for maternal age.

For Birthweights:

D-Statistic = .141 p-value = 0.2 > 0.05 => Fail to Reject Ho => the test is insignificant.

=> We assume normality for birthweights.

QUESTION 3F

Keywords: Regresion analysis, bootstrap

SPSS Steps:

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Result:

Birthweights = $\beta_{o} + \beta_{1}*Maternal\_Age$

$\beta_{0} = -1163.45$

95% confidence of $\beta_{0}$ is (-2561.917,322.417)

$\beta_{1} = 245.15$

95% confidence of $\beta_{1}$ is (162.012,327.252)

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SESSION 1

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QUESTION 2:

Check Variable View & Data View (Data Editor Window)

Part 1: draw scatter plot

Open OutPut Windows, then double click on the graph to activate

Part 2: add regression line

• Result:

Part 3: Relationship between mothers'age and birthweight

Based on the scatterplot, there is a linear relationship between the age of teenage mother and birth weights of their children. On average, the younger the teenage mother is, the smaller the weight of her child.

Part 4:

Result:

QUESTION 3:

Variable View:

Data View:

Part 1:

Result:

Part 2:

• Result:

Part 3:

Part 3A. Create Error Bar Chart comparing scores when married to a goat or a dog for life satisfaction:

• Result:

Part 3B. Create Error Bar Chart comparing scores when married to a goat or a dog for animal likings:

• Result:

Part 4:

Even though the different of the mean love of animals between goat and dog are not statistically significant (Overlappping), the difference of life satisfaction between goat and dog are statistically significant (No Overlapping).

In other words, even though there is not much difference between how people like goat and how people like dog, they live more happily with a dog.

QUESTION 4:

I. Data Viewer:

• Variable View:

• Data View:

II. Generate Differences(or Success Score) From Raw Data:

Differences Data are the difference between pre and post data.

Diffence = Pre - Post

Step 1: (Go to Transform > Compute Variables)

Step 2:

a)

b) The arrows show you the location that you need to fill in as the following photo:

c) Click OK.

Step 3:

You can see in the data view. We now have another column name differences(or Success Score)

III. Distribution of the mean of Success Score (Or Distribution of Differences):

III.A Distribution of Success Score for any Intervention method:

• Step 1:

• Step 2:

• Result:

III.B Distribution of Success Score for Hypnosis Intervention Method:

III.B1) Select Hypnosis Cases;

• Step 1: Create Select Case for Hypnosis: (Data > Select Cases)

• Step 2:

• Step 3:

• Step 4:

• Step 5: Click Continue > Ok

• Result: Now you can see the filter column, which select hypnosis case:

III.B2) Draw Histogram of Success Score for Hypnosis Intervention Method

• Step 1:

• Step 2:

• Step 3: Click OK

• Result:

• Conclusion: Based on the above histogram, we realize that most of the data is in between -2 and 2 and very close to zero. So we can say that the there is not enough evidence to conclude the hypnosis intervention can help people smoke less.

III.C Distribution of Success Score for Nicotine Patch Intervention Method:

III.C1) Select Nicotine Cases;

• Step 1: Right Click > Clear: To delete the old filter

• Step 2: Data > Select Cases

• Result: Now you can see the filter column, which select hypnosis case:

III.C2) Draw Histogram of Success Score for Nicotine Patch Intervention Method

• Step 1:

• Step 2:

• Step 3: Click OK

• Result:

• Conclusion: Based on the above histogram, we can see the mean of the sucess score is 2.25 and the standard deviation is 1.38. And most of the data is from 1 to 3. Therefore, we have enough evidence to conclude that after the nicotine patch intervention people tend to smoke less( about 1 to 3 cigarettes)