In: Statistics and Probability
1. The results of a recent poll on the preference of shoppers
regarding two products are shown below.
Product |
Shoppers Surveyed |
Shoppers Favoring |
A |
800 |
560 |
B |
900 |
612 |
At 95% confidence, the margin of error is
a. |
.0225. |
|
b. |
.044. |
|
c. |
.025. |
|
d. |
.064. |
2. The following information was obtained from independent
random samples taken of two populations.
Assume normally distributed populations with equal
variances.
Sample 1 |
Sample 2 |
|
Sample Mean |
45 |
42 |
Sample Variance |
85 |
90 |
Sample Size |
10 |
12 |
The degrees of freedom for the t distribution are
a. |
24. |
|
b. |
20. |
|
c. |
21. |
|
d. |
22. |
3. For a sample size of 30, changing from using the standard normal distribution to using the t distribution in a hypothesis test,
a. |
will result in the area corresponding to the critical value being larger. |
|
b. |
will have no effect on the area corresponding to the critical value. |
|
c. |
will result in the area corresponding to the critical value being smaller. |
|
d. |
Not enough information is given to answer this question. |
1)
product A | product B | ||
x1 = | 560 | x2 = | 612 |
p̂1=x1/n1 = | 0.7000 | p̂2=x2/n2 = | 0.6800 |
n1 = | 800 | n2 = | 900 |
estimated difference in proportion =p̂1-p̂2 = | 0.0200 | ||
std error Se =√(p̂1*(1-p̂1)/n1+p̂2*(1-p̂2)/n2) = | 0.0225 | ||
for 95 % CI value of z= | 1.960 | ||
margin of error E=z*std error = | 0.0440 |
2) degrees of freedom for the t distribution are =n1+n2-2=10+12-2 =20
3)
will have no effect on the area corresponding to the critical value.