COMPARING STATISTICAL TECHNIQUES 1
Chapter Nine
Problem 1) A skeptical paranormal researcher claims that the proportion of Americans that have seen a UFO is less than 1 in every one thousand. State the null hypothesis and the alternative hypothesis for a test of significance.
Solution: The null hypothesis is the proportion of Americans that have seen a UFO is 1 in every thousand. H
0
: P = 1/1000 or 0.001. The alternative hypothesis is the proportion of
Americans that have seen a UFO is more or less than 1 in every one thousand. H a
: P <>
1/1000 or 0.001
Problem 2) At one school, the average amount of time that tenth-graders spend watching television each week is 18.4 hours. The principal introduces a campaign to encourage the students to watch less television. One year later, the principal wants to perform a hypothesis test to determine whether the average amount of time spent watching television per week has decreased. Formulate the null and alternative hypotheses for the study described.
Solution: The null hypothesis is that the average amount of time that tenth-graders spend watching television each week is great than or equal to 18.4 hours. H
0
: µ ≥ 18.4. The alternative hypothesis is the average amount of time that tenth-graders spend watching television each week is less than 18.4 hours. H a
: µ < 18.4
Problem 3) A two-tailed test is conducted at the 5% significance level. What is the p-value required to reject the null hypothesis?
Solution: p-value less than 0.05, we can reject the null hypothesis, when p-value is greater than or equal to 0.05 we cannot reject the null hypothesis. p = .05/2 = .025
Problem 4) A two-tailed test is conducted at the 5% significance level. What is the right tail percentile required to reject the null hypothesis?
Solution: Since a two-tailed test requires the splitting of the significance level, the p -value on the right tail would be .05 / 2 = .025. Therefore, the right tail percentile required to reject the null hypothesis would be 1 - .025 = 97.5%.
Problem 5) What is the difference between an Type I and a Type II error? Provide an example of both.
Solution: Type I error consists in rejecting the null hypothesis when the null hypothesis is true.
Type II error is to fail to reject the null hypothesis when the null hypothesis is false.
Vitamin water: adding vitamin additive to water…..suggests that adding vitamins to water protects against infections, when in fact there is no effect. Type II null hypothesis failing to prove the existing effect adding vitamins is actually effective against fighting infections. On the other hand, if vitamin water is effecting in fighting infections, but an

experiment showed that there was no effect whatsoever, this is an example of a Type I error.
Chapter 10
Problem 1) Steven collected data from 20 college students on their emotional responses to classical music. Students listened to two 30-second segments from “The Collection from the Best of Classical Music.” After listening to a segment, the students rated it on a scale from 1 to 10, with 1 indicating that it “made them very sad” to 10 indicating that it “made them very happy.” Steve computes the total scores from each student and created a variable called “hapsad.” Steve then conducts a one-sample t-test on the data, knowing that there is an established mean for the publication of others that have taken this test of
6. The following is the scores:
5.0 5.0
10.0 3.0
13.0
7.0
5.0
14.0
8.0
10.0
3.0
4.0
13.0
5.0
15.0
18.0
12.0
7.0
15.0
3.0
A. Conduct a one-sample t-test. What is the t-test score? What is the mean? Was the test significant? If it was significant at what P-value level was it significant?
B. What is your null and alternative hypothesis? Given the results did you reject or fail to reject the null and why?
C. (Use instructions on page 437 of your textbook, under Hypothesis Tests with the t
Distribution to conduct SPSS or Excel analysis).
Solution: For null hypothesis µ = 6 and alternate hypothesis µ ≠ 6, T= (x – u) / (s / √n)
T = (8.7 – 6) / 4.7337 / √ 20

2.7/1.0585 = 2.598 = 2.6
To find P value you need three things:
T value = 2.598
Df = n -1=20 -1=19
Tail is based on alternate hypothesis p = 0.0177 which is significant at p < .05
Problem 2) Billie wishes to test the hypothesis that overweight individuals tend to eat faster than normal-weight individuals. To test this hypothesis, she has two assistants sit in a
McDonald’s restaurant and identify individuals who order the Big Mac special for lunch.
The Big Mackers as they become known are then classified by the assistants as overweight, normal weight, or neither overweight nor normal weight. The assistants identify 10 overweight and 10 normal weight Big Mackers. The assistants record the amount of time it takes them to eat the Big Mac special.

COMPARING STATISTICAL TECHNIQUES 3
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
2.0
2.0
2.0
2.0
2.0
2.0
2.0
2.0
2.0
2.0
619.0
535.0
697.0
782.0
587.0
675.0
635.0
672.0
606.0
789.0
806.0
600.0
585.0
540.0
660.0
571.0
584.0
653.0
574.0
569.0
A. Compute an independent-samples t-test on these data. Report the t-value and the p values. Were the results significant? (Do the same thing you did for the t-test above, only this time when you go to compare means, click on independent samples t-test. When you enter group variable into grouping variable area, it will ask you to define the variables. Click define groups and place the number 1 into 1 and the number 2 into 2).
B. What is the difference between the mean of the two groups? What is the difference in the standard deviation?
C. What is the null and alternative hypothesis? Do the data results lead you to reject or fail to reject the null hypothesis? What do the results tell you?
Solution:
A. For an independent samples t-test, F = -3.275, p = .004 which is significant.
B. 684.9000 – 589.0000 = 95.9 is the difference in the means. 82.22523 – 42.61455 =
39.61 is the difference in the SD.
C. H
0
- There is no difference in the amount of time taken to eat at McDonalds by overweight individuals as opposed to normal-weight individuals.
H a
= There is a significant difference in the amount of time taken to eat at McDonalds by overweight individuals as opposed to normal-weight individuals. Since overweight individuals take on average 95.9 less seconds to eat a Big Mac special than do normal-weight individuals, this is a statistically significant difference in the amount of time take (p < .01).
Problem 3) Lilly collects data on a sample of 40 high school students to evaluate whether the proportion of female high school students who take advanced math courses in high

school varies depending upon whether they have been raised primarily by their father or by both their mother and their father. Two variables are found below in the data file: math (0 = no advanced math and 1 = some advanced math) and Parent (1= primarily father and 2 = father and mother).
Parent Math
1.0
1.0
0.0
0.0
1.0
1.0
1.0
1.0
1.0
1.0
0.0
0.0
0.0
0.0
0.0
0.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
2.0
2.0
2.0
2.0
2.0
2.0
2.0
2.0
2.0
2.0
2.0
2.0
2.0
2.0
2.0
2.0
2.0
2.0
2.0
2.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0

COMPARING STATISTICAL TECHNIQUES 5
Conduct a crosstabs analysis to examine the proportion of female high school students who take advanced math courses is different for different levels of the parent variable.
What percent female students took advanced math class
What percent of female students did not take advanced math class when females were raised by just their father?
What are the Chi-square results? What are the expected and the observed results that were found? Are they results of the Chi-Square significant? What do the results mean?
What were your null and alternative hypotheses? Did the results lead you to reject or fail to reject the null and why?
Solution:
Count
Math
Parent 1.00
Total
2.00
No
20
11
31
Yes
0
9
9
Total
20
20
40
Pearson Chi-Square
Continuity Correction b
Value
11.613
a
Likelihood Ratio
9.176
15.128
Fisher's Exact Test
Linear-by-Linear Association 11.323
N of Valid Cases 40 df
Chi-Square Tests
1
Asymp. Sig. (2sided)
.001
Exact Sig. (2sided)
1 .002
1 .000
.001
1 .001
Exact Sig. (1sided)
.001 a. 2 cells (50.0%) have expected count less than 5. The minimum expected count is 4.50. b. Computed only for a 2x2 table
B.
9 / 40 = 22.5% of female students took advanced math in this study.
C.
20 / 20 = 100.0% of female students did not take advanced math when raised by just their father.

D.
chi-square = 11.6, degrees of freedom = 1, p = 0.001. The results of the Chi-Square signify that the frequency of female students who take advanced math are significant
(p ≤ .001) demonstrating that females raised in two-parent households are more likely to take advanced math, and that students raised by their fathers alone are unlikely to take advanced math.
E.
H
0
– There are no significant differences for female students taking advanced math between students raised in two-parent households, and students raised by their father alone.
H a
– There is a significant difference for female students taking advanced math between students raised in two-parent households, and students raised by their father along.
The results support rejecting the null hypotheses in that there appears to be a significant difference between the two groups at a p ≤ .001 level.
Problem 4) This problem will introduce the learner into a technique called Analysis of Variance.
For this course we will only conduct a simple One-Way ANOVA and touch briefly on the important elements of this technique. The One-Way ANOVA is an extension of the independent –t test that can only look at two independent sample means. We can use the
One-Way ANOVA to look at three or more independent sample means. Use the following data to conduct a One-Way ANOVA:
4
5
6
2
3
4
Scores
1
2
3
3
3
3
2
2
2
Group
1
1
1
Notice the group (grouping) variable, which is the independent variable or factor is made up of three different groups. The scores are the dependent variable.
Use the instructions for conduction an ANOVA on page 438 of the text for SPSS or Excel.
What is the F-score; Are the results significant, and if so, at what level (P-value)?
If the results are significant to the following: Click analyze, then click Compare Means, and then select one-way ANOVA like you did previously. Now click Post Hoc. In this area check
Tukey. If there is a significant result, we really do not know where it is. Is it between group 1 and 2, 1 and 3, or 2 and 3? Post hoc tests let us isolate where the level of significance was. So if the results come back significant, conduct the post hoc test as I mentioned above and explain where the results were significant.
What do the results obtained from the test mean?

COMPARING STATISTICAL TECHNIQUES 7
Solution:
A.
F = 8.225, p = .015. The results are significant between the groups.
B.
According to the Tukey HSD there is statistically significant differences only between groups one and three.
C.
There are no significant differences between group one and group two, or between group two and group three. The only significant difference is between the values of group one and group three (p = .013).
ANOVA
Scores
Sum of Squares df Mean Square F Sig.
Between Groups 14.100 2 7.050 8.225 .015
Within Groups 6.000 7
Total 20.100 9
.857
Multiple Comparisons
Scores
Tukey HSD
(I) Group (J) Group
1.00
2.00
3.00
2.00
3.00
1.00
3.00
1.00
Mean
Difference (I-J) Std. Error Sig.
.70711 .385 -1.00000
-3.00000
*
1.00000
-2.00000
3.00000
*
.75593
.70711
.70711
.75593
.013
.385
.059
.013
2.00 2.00000 .70711 .059
*. The mean difference is significant at the 0.05 level.
Scores
Tukey HSDa,b
Group
1.00
N
3
Subset for alpha = 0.05
1
2.0000
2
95% Confidence Interval
Lower Bound Upper Bound
-3.0825
-5.2263
-1.0825
-4.0825
.7737
-.0825
1.0825
-.7737
3.0825
.0825
5.2263
4.0825
2.00
3.00
Sig.
4
3
3.0000
.400
3.0000
5.0000
.064

Means for groups in homogeneous subsets are displayed. a. Uses Harmonic Mean Sample Size = 3.273. b. The group sizes are unequal. The harmonic mean of the group sizes is used. Type I error levels are not guaranteed.

COMPARING STATISTICAL TECHNIQUES
Reference
Bennett, J. O., Briggs, W. L., & Triola, M. F. (2009). Statistical reasoning (3rd ed.). Boston,
MA: Pearson Addison Wesley.
9