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.