?1. Using the data in the file named Chapter 11 Data Set 2 (in the appendix), te

Question

?1. Using the data in the file named Chapter 11 Data Set 2 (in the appendix), test the research hypothesis at the .05 level of significance that boys raise their hand in class more often than girls. Do this practice problem by hand, using a calculator. What is your conclusion regarding the research hypothesis? Remember to first decide whether this is a one or two-tailed test.

2. Using the same data set (Chapter 11 Data Set 2), test the research hypothesis at the .01 level of significance that there is a difference between boys and girls in the number of times they raise their hand in class. Do this practice problem by hand, using a calculator. What is your conclusion regarding the research hypothesis? You used the same data for this problem as for Question 1, but you have a different hypothesis (one is directional and the other is nondirectional). How do the results differ and why?

3. Time for some tedious, by-hand practice just to see if you can get the numbers right. Using the following information, calculate the t-test statistics by hand.

X1 = 62 X2 = 60 n1 = 10 n2 = 10 s1 = 2.45 s2 = 3.16

X1 = 158 X2 = 157.4 n1 = 22 n2 = 26 s1 = 2.06 s2 = 2.59

X1 = 200 X2 = 198 n1 = 17 n2 = 17 s1 = 2.45 s2 = 2.35

4. Using the results you got from Question 3, and a level of significance of .05, what are the two-tailed critical values associated with each? Would the null hypothesis be rejected?

5. Here’s a good one to think about. A public health researcher tested the hypothesis that providing new car buyers with child safety seats will also act as an incentive for parents to take other measures to protect their children (such as driving more safely, child-proofing the home, etc.). Dr. L counted all the occurrences of safe behaviors in the cars and homes of the parents who accepted the seats versus those who did not. The findings? A significant difference at the .013 level. Another researcher did exactly the same study, and for our purposes, let’s assume that everything was the same—same type of sample, same outcome measures, same car seats, and so on. Dr. R’s results were marginally significant (remember that from Chapter 9?) at the .051 level. Whose results do you trust more and why?

6. For this question, you have to do two analyses. Using Chapter 11 Data Set 5 (in the appendix), compute the t score for the difference between two groups on the test variable. Then, use Chapter 11 Data Set 6 and do the same. Notice that there is a difference between the two t values, even though the mean scores for each group are the same. What is the source of this difference, and why—with the same sample size—did the t value differ?

Statistics for People Who (Think They) Hate Statis tics 424 Chapter 11 Data Set 1 Group: 1 treatment, 2 - no treatment Group Memory Test Group Memory Test Group Memory Test 8 12 15 10 10 8 8 4 4 Chapter 11 Data Set 2 Gender: 1 = male, 2 = female Hand Up Gender Hand Up Gender

Explanation / Answer

Q3.
a.
Given that,
mean(x)=62
standard deviation , s.d1=2.45
number(n1)=60
y(mean)=60
standard deviation, s.d2 =3.16
number(n2)=10
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.0275
from standard normal table, two tailed t /2 =2.627
since our test is two-tailed
reject Ho, if to < -2.627 OR if to > 2.627
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =62-60/sqrt((6.0025/60)+(9.9856/10))
to =1.908
| to | =1.908

b.
Given that,
mean(x)=158
standard deviation , s.d1=2.06
number(n1)=22
y(mean)=157.4
standard deviation, s.d2 =2.59
number(n2)=26
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.0275
from standard normal table, two tailed t /2 =2.369
since our test is two-tailed
reject Ho, if to < -2.369 OR if to > 2.369
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =158-157.4/sqrt((4.2436/22)+(6.7081/26))
to =0.894

c.
Given that,
mean(x)=200
standard deviation , s.d1=2.45
number(n1)=17
y(mean)=198
standard deviation, s.d2 =2.35
number(n2)=17
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.0275
from standard normal table, two tailed t /2 =2.425
since our test is two-tailed
reject Ho, if to < -2.425 OR if to > 2.425
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =200-198/sqrt((6.0025/17)+(5.5225/17))
to =2.429

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.