Consumers with at least one credit card have a mean of 4.84 credit cards. You wa

Question

Consumers with at least one credit card have a mean of 4.84 credit cards. You want to test the hypothesis that the mean number of credit cards held by low-income consumers (consumers with annual incomes below $20,000) is different from the reported mean of 4.84. A random sample of 120 low-income consumers provides a sample mean number of credit cards xbar=2.95. Use significance level a=.1 for the test. Use a confidence interval estimate approach to conduct the hypothesis test. To use a confidence interval estimate approach to conduct the hypothesis test, you construct the ______________________ A. 80% B. 90% C. 10% D. 20% confidence interval estimate of the population mean. Assume that the population standard deviation is known and equal to 3.56. The confidence interval estimate is ___________ A. 3.195 B. 2.42 C. 4.31 D. 3.81 to _______________ A. 6.49 B. 3.48 C. 5.37 D. 4.87. Since the confidence interval estimate ____________________ A. contains B. does not contain ______________ A. 5.370 B. 1.645 C. 4.84 D. 2.95, the null hypothesis (that the mean number of credit cards held by low-income consumers is 4.84) is ___________________ A. rejected B. not rejected

Explanation / Answer

here confidence inerval =90% as significance level is 0.1 : option A

std error of mean =std deviation/(n)1/2 =0.325

for 90% CI, z=1.6449

hence confidence interval =mean +/- z*std error =2.42 (option B) to 5.37(option B)

does not contain (option B) 4.84 (option C)

optionA rejected

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