there was a people had already helped me with this but only one answer was still

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there was a people had already helped me with this but only one answer was still incorecct

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This part of your analysis looks at condos that do not have a river view and have PMI. Find the appropriate variable in the data set. To check if the sampling distribution is normal, which of the following would you check? Pick the BEST choice. That the histogram is bell-shaped and symmetrical That np 10 and n(1 p) 10 That np 10 and n(1 p 100 That n 30 That p 30 None of the above Compute the 89% confidence interval for the proportion of those condos without a river view that have PMI. Ifyou don't have enough information to compute the confidence interval, enter "DNE" in both boxes below. (0.4546 0.4954 (Round your answers to 4 decimal places) Compute the 88% confidence interval for the proportion of those condos without a river view that have PMI. Ifyou don't have enough information to compute the confidence interval, enter "DNE" in both boxes below. (0.4552 0*, 0.4948 as) (Round your answers to 4 decimal places) For the 82% confidence interval, compute the margin of error. If you don't have enough information to compute the margin of error, enter "DNE" in the box below. 0.01248

Explanation / Answer

here total number without a river view n=40 and out of which x=19 have PMI

hence estimated proportion p=x/n=19/40=0.475

for std error =(p(1-p)/n)1/2 =0.079

and for 89% CI, z=1.5982

hence 89% confidence interval =p -/+ z*std error =0.3488 ; 0.6012

for 88% CI, z=1.5548

hence 88% confidence interval =p -/+ z*std error =0.3522 ; 0.5978

for 82% CI, z=1.3408

hence margin of error =z*std error =0.1059

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