Question

PLEASE ANSWER ALL QUESTIONS!

1. A manager is reviewing the number of time share tours given by employees at a sales office. She knows the standard deviation of the number of tours given is 10.5. A sample of 64 sales associates finds an average of 49.29 tours given per month. What is the standard error?

Select one:

a. 0.16

b. 1.31

c. 19.75

d. 6.16

e. 1.50

2. You are designing a study to find the average number of hotel rooms booked during bike week. If the standard deviation is estimated to be 15.8, how many samples must be taken to get an answer within 0.9 hotel rooms with 80% confidence?

Select one:

a. 503

b. 504

c. 505

d. 506

e. 507

Answer 1

Solution :

1 ) Given that,

mean = = 49.29

standard deviation = = 10.5

n = 64

= 1200

₌ / n = 10.5 64= 1.31

The standard error = 1.31

Option b ) is correct.

2 )Given that,

standard deviation = = 15.8

margin of error = E = 0.9

At 80% confidence level the z is ,

  = 1 - 80% = 1 - 0.80 = 0.20

/ 2 = 0.20 / 2 = 0.10

Z_/2 = Z_0.10 = 1.282

Sample size = n = ((Z_/2 * ) / E)²

= ((1.282* 15.8 ) / 0.9 )²

=506

Sample size = 506

Option d ) is correct.