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Data were collected on the amount spent by 64 customers for lunch at a major Houston restaurant. These data are contained in the file named Houston. Based upon past studies the population standard deviation is known with ơ $5. Click on the datafile logo to reference the data. DATA file Round your answers to two decimal places. Use the critical value with three decimal places. a. At 99% confidence, what is the margin of error? . Develop a 99% confidence interval estimate of the mean amount spent for lunch. to s

037690 5679470 97866236 88711246 981148698239577 881130371214532 2 89300 001194768 2 1 1 2 2 2-2 3-2 2 1 3 2 3 1 3 08582167 9 212223 01234567890123456789012 234567893311965153272176758 65787130862111434883 2573196063730220517 1223212221112222311 92850577164 96381583642 04865754564 31122221122 34567890123 446 5678901234567890123 4444455555555556666

3311965153272176758 65787130862111434883 2573196063730220517 1223212221112222311 92850577164 96381583642 04865754564 31122221122 34567890123 446 5678901234567890123 4444455555555556666

55 18.38 56 26.85 57 25.10 58 27.55 9 25.87 14.37 61 15.61 62 26.46 63 24.24 64 16.66 65 20.85 60

eBook Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $20,000 and $40,000. Assume that a 95% confidence interval estimate of the population mean annual starting salary is desired a. What is the planning value for the population standard deviation? s1. b. How large a sample should be taken if the desired margin of error is $400? Round your answer to next whole number. $210? $70?How large a sample should be selected to provide a 95% confidence interval with a margin of error of 10? Assume that the population standard deviation is 20. Round your answer to next whole number.Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 85 weekly reports showed a sample mean of 17.5 customer contacts per week. The sample standard deviation was 5.8. Provide 90% and 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel 90% confidence interval, to 2 decimals: 95% confidence interval, to 2 decimals:A simple random sample of 400 individuals provides 300 Yes responses a. What is the point estimate of the proportion of the population that would provide Yes responses (to 2 decimals)? Later use p rounded to 2 decimal places. b. What is your estimate of the standard error of the proportion (to 4 decimals)? C. Compute the 95% confidence interval for the population proportion (to 4 decimals).A simple random sample of 20 items resulted in a sample mean of 15. The population standard deviation is σ = 6. Round your answers to two decimal places. a. what is the standard error of the mean, σ ? b, At 95% confidence, what is the margin of error?

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Answer #1

Given: Sample size, n =64; Std.deviation, sigma =$5

Mean amount, ar{X} =Sigma X/n =1377.28/64 =$21.52​​​​​​

At 99% confidence level, for two-tailed case, the critical value of Z is Zcrit =pm2.575

a.

Margin of error, MoE =Z. sigma/sqrt{n} = 2.575(5/V64​​​​​​) =$1.61

b.

99% confidence interval estimate =ar{X} pm MoE =21.52pm1.61 =($19.91, $23.13)

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