Question

1. A personnel officer found that the mean score of the company's overseas branches on a stress inventory was 44.6 with a standard deviation of 10 based on a random sample of 400 observations taken from the population. There is a 0.99 probability that the sample mean will provide a margin of error of (using a t-value)

Select one:

a. None of the other answers are correct

b. .98 or less

c. 4 or more

d. 13.16 or less

e. 1.456 or less

2. A random sample of a company’s monthly operating expenses for a sample of 36 months produced a sample mean of $5,474 with a standard deviation of $764. The 95% confidence interval for the average operating expenses using the t-table is

Select one:

a. $5,213.99 to $5,734.01

b. None of the other answers are correct

c. $5,346.67 to $5,601.33

d. $5,215.50 to $5,732.

e. $5,264.54 to $5,683.46

3. The t critical value with a 99.5% confidence with 24 degrees of freedom is

Select one:

a. 2.576

b. 2.797

c. 2.064

d. 2.069

e. None of the other answers are correct

4. Read the t statistic from the table of t distributions and find the correct answer. In a one-tailed test (lower tail), a sample size of 27 at a 0.05 level of significance; t =

Select one:

a. -1.703

b. 1.703

c. -1.706

d. 1.706

e. None of the other answers are correct

Answer 1

1)
margin of error = z * sigma/sqrt(n)

for 0.99 confidence ,

t = 2.588   {df = n-1 = 399} =T.INV.2T(0.01,399) in Excel

= 2.588* 10/sqrt(400)
= 1.294

A) None of the other answers are correct   is correct answer

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