Assignment #3

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Take Test: Assignment #3

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QUESTION 1
What type of hypothesis test (i.e., statistical procedure) is most appropriate for this research scenario? Briefly (in a sentence or two) explain why you chose the test you did.

Independent t-test. That is because we want to compare the means of 2 unrelated groups: Happy face versus no happy face groups or conditions.

Dependent t-test. That is because we want to compare the means of 2 unrelated groups: Happy face versus no happy face groups or conditions.

One sample t-test. That is because we want to compare the mean of one group to another.

Correlation t-test. That is because we want to see if the means are related to each other.
QUESTION 2
Prior to conducting the study, what magnitude of effect size (i.e., small, medium, or large) would you argue is important in this context? In other words, how large of an effect would you have to
see before you felt that the results would have some “practical significance” to the waitresses in this population? Thoroughly explain your rationale, as this involves a value judgment on your
part.

QUESTION 3
If you were conducting the study above, would you perform a one- or a two-tailed test? Why?

You would use a regular test as your hypothesis would be directional. In other words, you would posit that the use of the happy face would be associated with different tips than no happy
face.

You would use a two-tailed test as your hypothesis would be directional. In other words, you would posit that the use of the happy face would be associated with higher tips than no happy
face.

You would use a one-tailed test as your hypothesis would be directional. In other words, you would posit that the use of the happy face would be associated with higher tips than no happy
face.

You would use a one-tailed test as your hypothesis would be directional. In other words, you would posit that the use of the happy face would be associated with different tips than no happy
face.
QUESTION 4
Write null and alternate hypotheses that correspond with your answer to question #c. If you decided to perform a one-tailed test, make sure and specify which of the two groups you predict will be
higher/lower.

Null: Drawing a happy face on customer’s check impacts the amount of tip left for the server.
Alternative: Drawing a happy face on customer’s check negatively impacts the amount of tip left for the server.

Null: Drawing a happy face on customer’s check does not impact the amount of tip left for the server.
Alternative: Drawing a happy face on customer’s check positively impacts the amount of tip left for the server.

Null: Drawing a happy face on customer’s check does not impact the amount of tip left for the server.
Alternative: Drawing a happy face on customer’s check negatively impacts the amount of tip left for the server.

Null: Drawing a happy face on customer’s check impacts the amount of tip left for the server.
Alternative: Drawing a happy face on customer’s check does not impact the amount of tip left for the server.

QUESTION 5
Staying with convention, you will perform your hypothesis test using an alpha level of .05 (5%). Suppose you had to explain this significance criterion to someone who has little experience with
statistics. Explain how alpha is used to make a decision (reject or retain) regarding the null hypothesis.

The convention in the social science of using an alpha level of .10 relates to the percent chance that study results will be described as statistically significant even though they are not.
This is also referred to as a Type I error. A lower alpha level could be selected if a lower chance of a Type I error was desired. If an alpha level of .10 is set and a p value of .15 is obtained
upon running an analysis then the null hypothesis would be rejected, meaning that the findings would be regarding as significant. A study yielding a p value of greater than .10 would result in one
“failing to reject the null hypothesis”, which is tantamount to describing the results as not statistically significant.

The convention in the social science of using an alpha level of .05 relates to the percent chance that study results will be described as statistically significant even though they are not.
This is also referred to as a Type I error. A lower alpha level could be selected if a lower chance of a Type I error was desired. If an alpha level of .05 is set and a p value of .03 is obtained
upon running an analysis then the null hypothesis would be rejected, meaning that the findings would be regarding as significant. A study yielding a p value of greater than .05 would result in one
“failing to reject the null hypothesis”, which is tantamount to describing the results as not statistically significant.

The convention in the social science of using an alpha level of .05 relates to the percent chance that study results will be described as statistically insignificant even though they are
significant. This is also referred to as a Type I error. A lower alpha level could be selected if a lower chance of a Type I error was desired. If an alpha level of .05 is set and a p value of .03
is obtained upon running an analysis then the alternative hypothesis would be rejected, meaning that the findings would be regarding as significant. A study yielding a p value of greater than .05
would result in one “failing to reject the alternative hypothesis”, which is tantamount to describing the results as not statistically significant.

The convention in the social science of using an alpha level of .01 relates to the percent chance that study results will be described as statistically significant even though they are not.
This is also referred to as a Type II error. A higher alpha level could be selected if a lower chance of a Type II error was desired. If an alpha level of .005 is set and a p value of .15 is
obtained upon running an analysis then the null hypothesis would be rejected, meaning that the findings would be regarding as significant. A study yielding a p value of greater than ..15 would
result in one “failing to reject the null hypothesis”, which is tantamount to describing the results as not statistically significant.
QUESTION 6
Recall that one of the assumptions of the independent t-test is homogeneity of variance. If you had to explain this assumption to someone with little statistical expertise, how would you explain
it?

The t-test requires that the variances (and of course the standard deviations) of the 2 samples be similar to each other. If they are different in magnitude then an adjustment to the
calculation of the t-tests results can be conducted.

The t-test requires that the variances (and of course the standard deviations) of the 2 samples be different from each other.

The t-test requires that the variances (and of course the standard deviations) of the 2 samples be similar to each other. If they are too similar in magnitude then an adjustment to the
calculation of the t-tests results can be conducted.

The t-test requires that the variances (and of course the standard deviations) of the 2 samples be different from each other. If they are different in magnitude then the t-test is conducted
without any adjustment to the calculation.

QUESTION 7
Calculate the appropriate test statistic (i.e., z-test, t-test) using the descriptive statistics below. Note: you should check your work using the SPSS printout below. If your answer does not
match either you 1) made a mistake in your calculations, or 2) entered the data incorrectly.

t = -1.57

t = -1.99
t = -2.57

t = 1.99
QUESTION 8
Enter the data above into SPSS. You will enter in two variables for each restaurant patron: 1) which experimental group they belonged to (1 = no happy face, 2 =happy face) and 2) the tip
percentage left.
Upload a word doc of the data in SPSS.

QUESTION 9
Obtain the appropriate test statistic. From the SPSS menus choose Analyze and Compare Means, followed by the appropriate test.

t = -2.57
t = 1.57

t = -1.99

t = -1.57

QUESTION 10
Using the descriptive statistics from the SPSS printout, calculate the F-Max statistic. Also, examine the p-value (i.e., significance level) for Levene’s test. Based on the results of these two
statistics, do you feel that the homogeneity of variance assumption has been violated? Explain.

Fmax is used to determine if the means for the two groups are different from each other. To calculate it divide the larger variance by the smaller variance of the 2 groups. If the ratio is
close to 1 then you can conclude that the variances are comparable.
For our example: Fmax = (13.95 X 13.95) / (7.78 X 7.78) = 3.22
SPSS provide the Levene’s test results so the Fmax is not necessary. The Levene’s test for this example was significant and therefore equal variances could not be assumed.
Fmax is used to determine if the variances for the two groups are large in magnitude. To calculate it divide the larger variance by the larger variance of the 2 groups. If the ratio is
close to -1 then you can conclude that the variances are comparable.
SPSS provide the Levene’s test results so the Fmax is not necessary. The Levene’s test for this example was significant and therefore equal variances could be assumed.
Fmax is used to determine if the variances for the two groups are normally distributed. To calculate it divide the larger variance by the smaller variance of the 2 groups. If the ratio is
close to 1 then you can conclude that the variances are comparable.
For our example: Fmax = (13.95 X 13.95) / (7.78 X 7.78) = 3.22
SPSS provide the Levene’s test results so the Fmax is not necessary. The Levene’s test for this example was significant and therefore equal variances could be assumed.
Fmax is used to determine if the variances for the two groups are different from each other. To calculate it divide the larger variance by the smaller variance of the 2 groups. If the ratio
is close to 1 then you can conclude that the variances are comparable.
For our example: Fmax = (13.95 X 13.95) / (7.78 X 7.78) = 3.22
SPSS provide the Levene’s test results so the Fmax is not necessary. The Levene’s test for this example was not significant and therefore equal variances could be assumed (although the Levene’s
test results were very close to significant, i.e., p = .053).
QUESTION 11
What is the value of the t statistic and probability value on the SPSS printout? Your choice of statistic on the printout should reflect whether or not you felt the assumption was violated.

t = 1.57 and p = .05

t = -1.57 and p = .123

t = -1.57 and p = .20

t = -2.57 and p = .123.

QUESTION 12
Using the formula discussed in class, calculate Cohen’s d effect size measure. Provide a brief interpretation of the statistic. Note: simply saying that the effect is “small”, “medium”, or
“large” will not suffice.

To calculate the effect size, Cohen’s d, you take the difference of the 2 means and subtract by the standard deviation. In this case the standard deviations of the 2 groups are not
equivalent. You can calculate the pooled standard deviation or simply take the average of the 2 standard deviations (the number of participants in the 2 groups is almost the same).
s = (s1 + s2) / 2 = 10.86
d = x1 – x2 / s = 0.49
A d of 0.49 is considered to be a medium effect size.

To calculate the effect size, Cohen’s d, you take the difference of the 2 means and subtract by the standard deviation. In this case the standard deviations of the 2 groups are not
equivalent. You can calculate the pooled standard deviation or simply take the average of the 2 standard deviations (the number of participants in the 2 groups is almost the same).
s = (s1 + s2) / 2 = 10.86
d = x1 – x2 / s = 0.49
A d of 0.49 is considered to be a very small effect size.

To calculate the effect size, Cohen’s d, you take the difference of the 2 standard deviations and subtract by the overall standard deviation. In this case the standard deviations of the 2
groups are not equivalent. You can calculate the pooled standard deviation or simply take the average of the 2 standard deviations (the number of participants in the 2 groups is almost the same).
s = (s1 + s2) / 2 = 10.86
d = x1 – x2 / s = 0.49
A d of 0.49 is considered to be a null effect size.

To calculate the effect size, Cohen’s d, you take the difference of the 2 means and subtract by the standard deviation. In this case the standard deviations of the 2 groups are not
equivalent. You can calculate the pooled standard deviation or simply take the average of the 2 standard deviations (the number of participants in the 2 groups is almost the same).
s = (s1 + s2) / 2 = 10.86
d = x1 – x2 / s = 0.97
A d of 0.49 is considered to be a very large effect size.

QUESTION 13
Suppose you had to explain the probability value (i.e., significance level) on the printout to someone with limited statistical experience. Provide a brief interpretation of the p-value.

The p value is the value that you compare with the beta level that you set a priori for a study. The beta level is the Type II error rate that you are willing to tolerate. In the social
sciences this is usually 0.05, which corresponds to a 5 percent chance that the results will indicate that there is a statistically significant difference between the groups when there is in fact
no significant difference between the groups. In our study we obtained a one-tailed p value of 0.01. Since this was not less than 0.001 we would fail to reject the null hypothesis, which means
that we would not be able to regard the tip percent means of the 2 groups as different from each other.

The p value is the value that you compare with the alpha level that you set afterwardifor a study. The alpha level is the Type I error rate that you are willing to tolerate. In the social
sciences this is usually 0.10, which corresponds to a 10 percent chance that the results will indicate that there is a statistically significant difference between the groups when there is in fact
a significant difference between the groups. In our study we obtained a one-tailed p value of 0.62. Since this was not less than 0.05 we would reject the null hypothesis, which means that we would
not be able to regard the tip percent means of the 2 groups as similar.

The p value is the value that you compare with the alpha level that you set a priori for a study. The alpha level is the Type II error rate that you are willing to tolerate. In the social
sciences this is usually 0.05, which corresponds to a 5 percent chance that the results will indicate that there is a statistically significant difference between the groups when there is in fact
no significant difference between the groups. In our study we obtained a one-tailed p value of 0.49. Since this was less than 0.05 we would fail to reject the null hypothesis, which means that we
would not be able to regard the tip percent means of the 2 groups as different from each other.

The p value is the value that you compare with the alpha level that you set a priori for a study. The alpha level is the Type I error rate that you are willing to tolerate. In the social
sciences this is usually 0.05, which corresponds to a 5 percent chance that the results will indicate that there is a statistically significant difference between the groups when there is in fact
no significant difference between the groups. In our study we obtained a one-tailed p value of 0.62. Since this was not less than 0.05 we would fail to reject the null hypothesis, which means that
we would not be able to regard the tip percent means of the 2 groups as different from each other.
QUESTION 14
What is your decision concerning the null hypothesis? Did you reject or retain?

Fail to not reject the null hypothesis

Reject the null hypothesis

Fail to reject the null hypothesis

Fail to accept the null hypothesis
QUESTION 15
Considering both the probability value and effect size measure, what interpretations would you make about the findings? That is, what are your conclusions about the effects of leaving happy faces
on checks?

The p value that is calculated for the t-test is dependent on several factors including sample size and effect size. It is very likely that, given the medium effect size obtained, if the
sample size was larger the study would have yielded statistically significant results. Therefore, it appears that the low statistical power (i.e., high Type II error) of the study resulting from
the small sample size was mainly responsible for the lack of significant findings.

The p value that is calculated for the t-test is dependent on one factor: sample size. It is very likely that, given the large effect size obtained, if the sample size was larger the study
would have yielded statistically significant results. Therefore, it appears that the strong statistical power of the study resulting from the small sample size was mainly responsible for the lack
of significant findings.

The p value that is calculated for the t-test is dependent on several factors including sample size and effect size. It is very likely that, given the small effect size obtained, if the
sample size was smaller the study would have yielded statistically significant results. Therefore, it appears that the low statistical power (i.e., high Type II error) of the study resulting from
the large sample size was mainly responsible for the lack of significant findings.

The p value that is calculated for the t-test is dependent on several factors including sample size and effect size. It is very likely that, given the medium effect size obtained, if the
sample size was smaller the study would have yielded statistically significant results. Therefore, it appears that the low statistical power (i.e., high Type II error) of the study resulting from
the large sample size was mainly responsible for the lack of significant findings.
QUESTION 16
If you were conducting the study above, would you perform a one- or a two-tailed test? Why?

One-tailed. Because the hypothesis would be directional.

Two-tailed. Because the hypothesis would be non-directional.
QUESTION 17
Write null and alternate hypotheses that correspond with your answer to question #a. If you decided to perform a one-tailed test, make sure and specify which of the two groups you predict will be
higher/lower.

Null Hypothesis:
Overweight individuals complete a meal quicker compared with normal weight individuals.
Alternative Hypothesis:
Overweight individuals have no difference compared with normal weight individuals on the amount of time they take to complete a meal.
Null Hypothesis:
Overweight individuals have no difference compared with normal weight individuals on the amount of time they take to complete a meal.
Alternative Hypothesis:
Overweight individuals complete a meal quicker compared with normal weight individuals.

Null Hypothesis:
Overweight individuals have no difference compared with normal weight individuals on the amount of time they take to complete a meal.
Alternative Hypothesis:
Overweight individuals complete a meal no differently in terms of amount of time compared with normal weight individuals.

QUESTION 18
Compute an independent-samples t test on these data. Report the t values and the p values assuming equal populations variances and not equal population variances.

t(38) = -3.88, p = .01 when assuming equal variances and t(30.83) = -5.50, p = .01 when not assuming equal variances.

t(38) = -3.98, p < .01 when not assuming equal variances and t(30.83) = -5.40, p < .01 when assuming equal variances.

t(38) = 3.98, p < .001 when assuming equal variances and t(30.83) = -3.40, p < .001 when not assuming equal variances.

t(38) = -3.98, p < .01 when assuming equal variances and t(30.83) = -5.40, p < .01 when not assuming equal variances.
QUESTION 19
On the output, identify the following:
a. Mean eating time for overweight individuals
b. Standard deviation for normal weight individuals
c. Results for the test evaluating homogeneity of variances

a) 81.95 seconds
b) 591 seconds
c) F(1,38) = 2.74, p = .11
a) 590 seconds
b) 82.05 seconds
c) F(1,38) = 2.74, p = .11
a) 589 seconds
b) 82.95 seconds
c) F(1,38) = 2.74, p = .11
a) 593 seconds
b) 84.95 seconds
c) F(1,38) = 2.74, p = .119

QUESTION 20
Compute an effect size that describes the magnitude of the relationship between weight and the speed of eating Big Mac meals.

QUESTION 21
Write a brief results section based on your analyses.

QUESTION 22
1. Conduct a one-way ANOVA to investigate the relationship between hair color and social extroversion. Be sure to conduct appropriate post-hoc tests. On the output identify the following:
a. F ratio for the group effect
b. Sums of squares for the hair color effect
c. Mean for redheads
d. p value for the hair color effect
a) F(2, 15) = 3.31
b) SS Hair Color = 24.01
c) Mean for redheads = 2.03
d) p = .04
a) F(2, 15) = 3.51
b) SS Hair Color = 24.11
c) Mean for redheads = 2.33
d) p = .06
a) F(2, 15) = 2.51
b) SS Hair Color = 23.11
c) Mean for redheads = 1.33
d) p = .12
a) F(2, 15) = 3.50

b) SS Hair Color = 14.11
c) Mean for redheads = 3.23
d) p = .07
QUESTION 23
What is the effect size for the relationship between hair color and extroversion?

Eta squared = -.30

Eta squared = 1.31

Eta squared = .32

Eta squared = .29
QUESTION 24
Create a boxplot to display the differences among the distributions for the three hair color groups.
Upload a word doc graphic file of the boxplot.


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