1- A study has indicated that the sample size necessary to estimate the average

Question

1- A study has indicated that the sample size necessary to estimate the average electricity use by residential customers of a large western utility company is 900 customers. Assuming that the margin of error associated with the estimate will be ± 20 watts and the confidence level is stated to be 95 percent, what was the value for the population standard deviation? Possible answers=

2- A regional hardware chain is interested in estimating the proportion of their customers who own their own homes. There is some evidence to suggest that the proportion might be around 0.70. Given this, what sample size is required if they wish a 96 percent confidence level with a margin of error of ± .03?

Possible answers= a. 981 b. 980 c. 452 d. 451

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

1.

As

E = Margin of error = z*sigma/sqrt(n)

Then

sigma = E*sqrt(n)/z

For 95% confidence, we know that, by table/technology, z = 1.96.

Hence,

sigma = 20*sqrt(900)/1.96 = 306.122 [ANSWER, B]
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2.

Note that      
      
n = z(alpha/2)^2 p (1 - p) / E^2      
      
where      
      
alpha/2 = (1-c)/2 = (1-0.96)/2 =   0.02  
       
      
Using a table/technology,      
      
z(alpha/2) =    2.05  
      
Also,      
      
E =    0.03  
p =    0.7  
      
Thus,      
      
n =    980.5833333  
      
Rounding up,      
      
n =    981   [ANSWER, A]

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