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Question
Two independent samples of sizes 20 and 25 are randomly selected from two normal populations with equal variances. In order to test the difference between the population means, the test statistic is:a. a standard normal random variable
b. approximately standard normal random variable
c. t-distributed with 45 degrees of freedom
d. t-distributed with 43 degrees of freedom
Answer
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Related questions
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NARRBEGIN: SA_108_112The following values have been calculated using the TDIST and TINV functions in Excel. These values come from a t- distribution with 15 degrees of freedom.These values represent the probability to the right of the given positive values.Valuet -probability1.000.16661.200.12441.400.0909These values represent the positive t- value for a given probability in both tails (sum of both tails).Probabilityt -value0.201.34060.101.75310.052.1315NARRENDWhat is the probability of a t-value between "1.40 and +1.40?
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NARRBEGIN: SA_101_107The personnel department of a large corporation wants to estimate the family dental expenses of its employees to determine the feasibility of providing a dental insurance plan. A random sample of 12 employees in 2004 reveals the following family dental expenses (in dollars): 115, 370, 250, 93, 540, 225, 177, 425, 318, 182, 275, and 228. Use StatTools for your calculations.NARREND(A) Construct a 90% confidence interval estimate of the mean family dental expenses for all employees of this corporation.(B) What assumption about the population distribution must be made to answer (A)?(C) Interpret the 90% confidence interval constructed in (A).(D) Suppose you used a 95% confidence interval in (A). What would be your answer?(E) Suppose the fourth value were 593 instead of 93. What would be your answer to (A)? What effect does this change have on the confidence interval?(F) Construct a 90% confidence interval estimate for the standard deviation of family dental expenses for all employees of this corporation.(G) Interpret the 90% confidence interval constructed in (E).
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NARRBEGIN: SA_95_96An automobile dealer wants to estimate the proportion of customers who still own the cars they purchased six years ago. A random sample of 200 customers selected from the automobile dealer's records indicates that 88 still own cars that were purchased six years earlier.NARREND(A) Construct a 95% confidence interval estimate of the population proportion of all customers who still own the cars they purchased six years ago(B) How can the result in (A) be used by the automobile dealer to study satisfaction with cars purchased at the dealership?
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NARRBEGIN: SA_79_80
Auditors of Independent Bank are interested in comparing the reported value of all 1775 customer savings account balances with their own findings regarding the actual value of such assets. Rather than reviewing the records of each savings account at the bank, the auditors randomly selected a sample of 100 savings account balances from the frame. The sample mean and sample standard deviations were $505.75 and 360.95, respectively.
NARREND
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NARRBEGIN: SA_75_78The percent defective for parts produced by a manufacturing process is targeted at 4%. The process is monitored daily by taking samples of sizes n = 160 units. Suppose that today's sample contains 14 defectives.NARREND(A) Determine a 95% confidence interval for the proportion defective for the process today.(B) Based on your answer to (A), is it still reasonable to think the overall proportion defective produced by today's process is actually the targeted 4%? Explain your reasoning.(C) The confidence interval in (A) is based on the assumption of a large sample size. Is this sample size sufficiently large in this example? Explain how you arrived at your answer.(D) How many units would have to be sampled to be 95% confident that you can estimate the fraction of defective parts within 2% (using the information from today's sample)?
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