Question

Would Not Approve of Driving Drunk	Would Not Approve of Driving Drunk
n1=40	n2=25
X¯1=2.1	X¯2=8.2
s1=1.8	s2=1.9

John Worrall and colleagues (2014) found that the fear of losing the good opinion of one’s family and peers kept people from driving home drunk. Let’s say we have two independent random samples of people: those who think that their peers would disapprove of them from driving drunk, and those who think that their peers would either not care or approve of their driving drunk. We ask each person in each group to self-report the number of times that he or she has driven drunk in the past 12 months. The results are in the table above.

For the remaining questions, assume we are testing the null hypothesis that the two population means are equal against the alternative hypothesis that the group whose peers would not approve of driving drunk has a lower mean rate of driving drunk. In this hypothesis test, we are also assuming that the unknown population standard deviations are unequal (σ₁ ≠ σ₂), and we are using an alpha level of .01.

1.Which of the following is the most appropriate statement of the null hypothesis for this test?

H0: μ1 > μ2

H1: μ1 ≠ μ2

H0: μ1 = 0

H0: μ1 = μ2

H1: μ1 < μ2

2. Which of the following is the most appropriate statement of the alternative or research hypothesisfor this test? (Hint: Read the problem carefully.)

H0: μ1 > μ2

H1: μ1 ≠ μ2

H0: μ1 = 0

H0: μ1 = μ2

H1: μ1 < μ2

3. Is this a two-tailed or one-tailed test?

Two-tailed

One-tailed

4. Which of the following tests is the most appropriate for this problem?

Pooled variance t test

Separate variance t test

Dependent-samples t test

Difference between proportions z test

Chi-square test of independence

5. After stating hypotheses, determining the most appropriate test statistic and sampling distributions, and setting alpha levels, you conduct the test and calculate the following values:

Estimated df = 51,

t critical = -2.423

t obtained = -12.63

Which of the following statements is the most appropriate conclusion to draw from this test?

Since tobt < tcrit, we have proven the alternative hypothesis. Peer disapproval of drunk driving must be related to one’s own drunk driving behaviors.

Since tobt < tcrit, we fail to reject the null hypothesis. There is not enough evidence to reject the assumption that peer disapproval is unrelated to one’s own behaviors.

Since tobt > tcrit, we reject the alternative hypothesis. There is enough evidence to conclude that driving drunk leads to peer approval of driving drunk.

Since tobt < tcrit, we reject the null hypothesis. It would be highly unlikely to observe a mean difference of this magnitude if the mean rates of driving drunk were truly equal across both groups in the population.

Since tobt > tcrit, we fail to reject the alternative hypothesis. The observed difference in sample means is not highly unlikely if the null hypothesis were really true.

Answer 1

Given that,
mean(x)=2.1
standard deviation , s.d1=1.8
number(n1)=40
y(mean)=8.2
standard deviation, s.d2 =1.9
number(n2)=25
null, Ho: u1 = u2
alternate, H1: u1 < u2
level of significance, α = 0.01
from standard normal table,left tailed t α/2 =2.492
since our test is left-tailed
reject Ho, if to < -2.492
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =2.1-8.2/sqrt((3.24/40)+(3.61/25))
to =-12.849
| to | =12.849
critical value
the value of |t α| with min (n1-1, n2-1) i.e 24 d.f is 2.492
we got |to| = 12.84851 & | t α | = 2.492
make decision
hence value of | to | > | t α| and here we reject Ho
p-value:left tail - Ha : ( p < -12.8485 ) = 0
hence value of p0.01 > 0,here we reject Ho
ANSWERS
---------------
1.
null, Ho: u1 = u2
2.
alternate, H1: u1 < u2
3.
one tailed test
4.
test statistic: -12.849
critical value: -2.492
5.
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that the group whose peers would not approve of driving drunk has a lower mean rate of driving drunk.