Question

5. I took a random sample of people from the 2000 U.S. Census and recorded the incomes. The sample size was n = 30 (with replacement). The average of the sample was $23606 and the standard deviation of the sample was $24757. Answer these questions using the statistics that I found from my sample. a) Find a 95% confidence interval for the mean income of the population. b) What assumptions must you make? c) Your confidence interval is of the form (LB, UB) where LB and UB are both numbers. Suppose I'm going to take a new sample of size 30. Can I conclude that the probability the average of my sample is between LB and UB is 95%? (where LB and UB are the values you found in your interval.) Why or why not? d)A politician claims that the mean income of US residents (in 2000) was $25,000. Test this claim with an appropriate hypothesis test using a significance level of 5%. e) For part (d), what's the smallest significance level you could have used so that the result would be that you would reject the null hypothesis?

Answer 1

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ton! Given mean: 23606 standard deviabion 24757 n- 30 a) 95% confidene interval will be d: 0.05 df n-1= 30 1-29 takdf = to 05,292.045 >) 23606± 2.045% The 95% CI is between 14361-58 and 32850-42 b) The one-ample t-test has feur main asiumptions. The dependent vauiakle must be continuous Gntenallatio). The obserabiong me indeperdent of cne anothen . The dependent varade shoad be appoximately reamally dtabales! The dependert variabe sdould not contain any cultiers. c) we ae 95% confdent tat a 95% confdence interual coill contaon the average of tte sample ehen the new sample is taken from the game popelablon. d) Ho:M 26000 V/s Haill725000 Test statistic8 (b) = S/5R 23606 -25000 24757/030 - -0.308407822 t -0.308 The prvalue i6 0:7600 Since the p-value (0.76) is goreater tan the signfcaneo bevel 16 005, we camot Jeject the null hypotheis Ho. These, we can condude ttat the mean inconme is $25000 c) The significace level duld be 077

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