3- A large Midwestern university is interested in estimating the mean time that

Question

3- A large Midwestern university is interested in estimating the mean time that students spend at the student recreation center per week. A previous study indicated that the standard deviation in time is about 25 minutes per week. If the officials wish to estimate the mean time within ± 4 minutes with a 90 percent confidence, what should the sample size be? Possible answers=

a. 105.685 b. 105 c. 106. d. Cant be determined without sample mean

4- Find the critical value for a 98% interval estimate. Assume the population standard deviation is known. Possible answers= a. +/-2.05 b. +/-0.98 c. +/-2.33 d. +/-1.96

Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 25 weekly reports showed a sample mean of 19.5 customer contacts per week. The sample standard deviation was 5.2. Provide 90% confidence interval for the population mean number of weekly customer contacts for the sales personnel. Possible answers

Explanation / Answer

3.

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.05  
      
Using a table/technology,      
      
z(alpha/2) =    1.644853627  
      
Also,      
      
s = sample standard deviation =    25  
E = margin of error =    4  
      
Thus,      
      
n =    105.6852912  
      
Rounding up,      
      
n =    106   [ANSWER, C]

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4.

As c = 0.98, then

alpha/2 = (1-c)/2 = 0.01

By table/technology, then

zcrit = -2.33, 2.33 [ANSWER, C]

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