The The communications regulator of a certain country released a report refuting

Question

The The communications regulator of a certain country released a report refuting an earlier report released by the same regulator. The new report indicates that cable subscribers would save as much as 10% on their cable tevision bils. T average monthly cable prices in the country were estimated to be $31.41. Typically, such reports announce a margin of error o say $1.23 and a confidence evel o 90%. Suppose the standard deviation of the monthly cost of cable television bills was $11.00. Complete parts a through c below. a. Determine the sample size of the new study released by the regulator. The sample size is Round up to the nearest whole number as needed.) b. Calculate the sample size required to decrease the margin of error by a dollar. The sample size requiredis (Round up to the nearest whole number as needed.) c. A typical sample size used in national surveys in the country is 1,700 to 2,300. Determine a range for the margin of error corresponding to this range of sample sizes. The range for the margin oferror is from $1040 (Round up to the nearest cent. Use ascending order.)

Explanation / Answer

a) here margin of error E =1.23

std deviation =11

and for 90% CI ; critical value of z =1.6449

henc e required sample size n=(z*std deviation/E)2 =~ 217

b) for margin of error E =1

henc e required sample size n=(z*std deviation/E)2 =~ 328

c)

for margin of error =z*std error =z*std deviaiton/(n)1/2

hence for n=1700 ; margin of error E =1.6449*11/(1700)1/2 =0.44

and for n=2300; margin of error E =1.6449*11/(2300)1/2 =0.38

therefore range of the margin of error is from $ 0.38 to $ 0.44

please revert for above,

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