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nawal C O z-tal &Secure; / mathel co PlayerHomework.aspx/homeworkld-486860687uquestionld-18flushed-false&cld-50478998icenterwin; yes nawaf a ghamdi 7/26/18 4:22 PM Econ 221 Summer B 2018 (46669) Save Homework: Chapter #08 Homework Assignment 70, 8 (7 complete) ? HW Score: 66.67%, 20 of 30 pts Score: 0 of 6 pts Question Help 8.2.11-T A random sample of 31 taxpayers claimed an average of $9.985 in medical expersos for the year Assume the population standard deviation for these deductions was $2.402 Construct confidence intervals to estimate the average deduction for the population wilth ho levels of signiticanoe shown below a. 2% b,5% c.10% a.The ontdene nerval with a 2%level of signfican e has a lower limitofs»and an upper init of SO Round to the nearest dollar as needed.) Enter your answer in the edit fields and then click Check Answer Clear Al POWERED BY weebIY 9953 .9955 .9956 .9957 .9959 .9960 .9961 .9962 .9963 .9964 126/2018

Explanation / Answer

Standard error = SE = sd/n^.5 = 2.402/31^.5 = 0.4314

z value for 2% level of significance is 2.33

Confidence interval-

From 9985 - 2.33(0.4314) to 9985 + 2.33(0.4314)

9983.99 to 9986.005

Lower limit is 9984 upper limit 9986

At 5% , z is 1.96

Lower limit = 9985-1.96(0.4314) = 9984

Upper limit = 9985 + 1.96(0.4314) = 9986

At 10%, z score is 1.64

Lower limit = 9985-1.64(0.4314) = 9984

Upper limit= 9985 + 1.64(0.4314) = 9986

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