Homework: Independent t-tests

1. For each of the following, please use the t-table in the back of your book to provide the appropriate critical t-value. (2.5 points)

a. df = 7, α = .01, one tailed2.9980

b. df = 14, α = .02, two-tailed2.6245

c. df = 12, α = .10, two-tailed1.7823

d. df = 12, α = .05, one-tailed1.7823

e. df = 26, α = .05, two-tailed2.0555

2. For each of the following decision criteria, make a decision to either reject H0 or fail to reject H0. Star the ones that indicate statistical significance. (3 points)

a. α = .05, p = .25fail to reject, p is higher than a

b. α = .01, p = .25fail to reject, p is higher than a

c. α = .01, p = .05fail to reject, p is higher than a

d. tcrit = 6.97, tobs = 8.45***reject, obs is higher than crit

e. tcrit = 2.90, tobs = 1.78fail to reject, obs is lower than crit

f. tcrit = -3.25, tobs = -2.75***reject, obs is higher than crit

3. Write out an example of a research question that could be tested using an independent-samples t-test (feel free to be creative). Make sure to demonstrate how each of the various pieces of the research question meets the assumptions for conducting an independent-samples t-test. (5 points)If I wanted to know whether golden retriever owners or Labrador retriever owners spent more on average for veterinary services over the course of the dogs’ lives, I could do an independent samples t-test. Instead of asking every single dog owner, I could take a sample (say 250 of each of the aforementioned groups) to test this. The assumption of independence in maintained, because the cost of each group is not related to the answers of another. If the owner of a golden spends $3000 on vet costs, that does not mean that it will cause the owner of a lab to spend $4000. This scenario has qualitative categories as independent variables, of lab owners and golden retriever owners, and the dependent variables are quantitative, which would be the cost each spends in the span of their dogs’ lives with an assumed normal distribution. The null hypothesis would be that there is no difference in the spending of these groups ( ), while the alternative hypothesis would be that there is a difference in them ( ).

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4. 20 students are given a statistics quiz on white paper and 15 students are given the same statistics quiz on blue paper. Researchers want to see if the color of paper affects students’ performance. The mean score for the white-paper group is 25 and the variance is 18, and the mean score for the blue-paper group is 21 with a variance of 16. Does color of paper impact performance? Set up and execute your hypothesis test, as well as calculate your answer and draw your conclusions. Please use the steps we’ve been learning about, and make sure to

show all of your work. (Hint: Although you can solve for ∑ ( x− x̄ )2 algebraically, it might

be quite challenging. To help you out, the values are: ∑ ( x− x̄ )2 for the white paper group =

342; ∑ ( x− x̄ )2 for the blue paper group = 224). (10 points)

The null hypothesis is that the color of paper has no effect (), while the alternative hypothesis is that the color of paper does have an effect (

). I will use an alpha (a) of 0.05.

(25-21)

( 342+22420+15−2) (1/25+1/21)

(4)(566/33)(0.09) = 4 /√1.54 = 4/1.24= 3.23t=3.23degrees of freedom is n1+n2-2= 20+15-2=33I will use a two tailed test because the direction of the scores, based on the color of the paper, was not specified so it is non-directionalWith a DF of 33 for a two tailed test to a significance level of 0.05, we get a critical value of 2.0345. Since the critical value of 2.0345 is less than the t statistic of 3.23, so we reject the null hypothesis that the color of paper has no effect on test scores. This gives us a 5% margin of error for rejecting the null hypothesis when it could be true. From this, we can conclude with 95% confidence that the color of paper has an effect on the test scores that the students got, although we don’t exactly know in which direction.

5. Calculate and interpret an effect size for the problem in question 4. (3 points)S1= 342/(20-1)= 342/19= 18S2= 224/(15-1)= 224/14 = 16S pooled= √ (n 1−1 ) s12+( n2−1 ) s22/n1+n2-2(20-1)(18)+(15-1)(16)/ (20+15-2)(19)(18)+(14)(16)/ 33(342)+ (224)/33=√17.15=4.14

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d= |25−21|/4.14

= |4|

4.14d= 0.97This is a large effect size, which means there is almost a whole standard deviation in between the means. However, it also means that it is pretty plain to the naked eye, without the use of a study, that paper color does affect test scores.

6. A researcher identified a new drug that may suggest a more effective treatment for a genetic disorder that appears at birth and causes mental retardation. 12 infants were randomly assigned to the drug condition (but 5 dropped out, leaving a total of 7 infants), and 12 infants were randomly assigned to the control condition. After 6 months, all the infants were given a cognitive abilities test (where higher scores indicate better performance). Does the new drug improve cognitive abilities? Drug group: mean = 58.40, standard deviation = 5.60; Control group: mean = 52.60, standard deviation =5.20. Set up and execute your hypothesis test, as well as calculate your answer and draw your conclusions. Please use the steps we’ve been learning about, and make sure to show all of your work. (Hint: Although you can solve for

∑ ( x− x̄ )2 algebraically, it might be quite challenging. To help you out, the values are:

∑ ( x− x̄ )2 for the drug group = 188.16; ∑ ( x− x̄ )2

for the control group = 297.44). (10 points)The drug group values are: mean = 58.4, SD= 5.6 and will be labeled u1The control group: mean = 52.6, SD= 5.2 and will be labeled u2We will use an alpha (a) of 0.01 as the significance, because this is about infant health which requires a higher degree of confidence than something not as important. We will use a one tailed test since we are testing whether the drug is more effective than the control group, which implies a positive direction and therefore is a directional hypothesis. The null

hypothesis is that there is no difference ( ) and the alternative hypothesis is that the drug is more effective so better performance scores (H0: u1>u2)

T= (58.40-52.60)/ √188.16− 297.447+12−2

( 17+ 1

12)

(188.16-297.44)= -109.28 / (17) = -6.43(1/7 + 1/12)= (.14+ .08) = .22

58.40-52.60= 5.805.80/ √ (−6.43 ) .22¿¿5.80/ √−1.425.80/1.19t= 4.87Degrees of freedom is n1+n2-1 = (7+12-2)= 17So the critical value for a one tailed test with a DF of 17 and an alpha of 0.01 is 2.5669. Since the critical value of 2.5669 is lower than the t statistic of 4.87, we reject the null hypothesis and say that there is a difference between the drug group and the control. From this, we can

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state with 99% certainty that the drug improves test scores when compared to those not taking the drug.

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