A country clerk wishes to improve voter registration. One method under considera

Question

A country clerk wishes to improve voter registration. One method under consideration is to send reminders in the mail to all citizens in the country who are eligible to register. As part of a pilot study to determine if this method will actually improve voter registration, a random sample of 1250 potential voters was taken. Then this sample was randomly divided into two groups, in group 1, no reminders were sent to the group. The number of voters from this group who registered was 292. In group 2, reminders were sent in the mail to each member of the group, and the number who registered to vote was 350. Did the reminders improve voter registration? Suppose that a random sample of 29 college students was randomly divided into two groups. The first group consisted of 15 people who were given 1/2 liter of red wine before going to sleep. The mean and standard deviation of this group was found to be 19.65 and 1.86. The second group, consisting of 14 people, was given no alcohol before going to sleep. This group had a mean and standard deviation of 6.59 and 1.91. Everyone in both groups went to sleep at 11 pm. The average brain activity was determined for each individual in the groups. Assume the distribution of brain waves in mound-shaped and symmetric, and the variances of the two populations can be assumed to be the same. a. Find the 99% confidence interval for the difference in the true means of the groups. b. Describe what this confidence interval tells you.

Explanation / Answer

3)here p1=295/625 =0.472 ; n1=625

p2=350/625=0.56 ; n2=625

hence std error of mean =(p1(-1p1)/n1+p2(1-p2)/n2)1/2 =0.028

hence test stat z=(p1-p2)/std error =-3.125

for which one tailed p value =0.0009

as p value is significantly low; we can accept the claim that reminders improves voter registration.

4)a)

hence degree of freedom =(n1+n2-2)=27

for pooled std deviaiton =Sp=((n1-1)s12+(n2-1)s22)/(n1+n2-2))1/2 =1.8842

hence std error of difference =Sp(1/n+1/n2)1/2 =0.7002

for 27 degree of freedom and 99% CI, t=2.7707

hence confidence interval =mean difference +/- t*std error =11.12 ; 15.00

b) above confidence interval provides 99% probability to contain true differnce b/w mean of brain activity of alcohol consuming and non alcohol consuming students

S1 1.860 S2 1.910 n1 15 n2 14 X1 19.650 X2 6.590

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