Question

Furniture Land surveyed 670 consumers and found that 472 were enthusiastic about a new home dècor it plans to show in its store in High Point.

Construct the 98% confidence interval for the population proportion. (Round the final answers to 3 decimal places.)

Confidence interval for the population proportion is  and  

2)

You need to estimate the mean number of travel days per year for outside salespeople. The mean of a small pilot study was 140 days, with a standard deviation of 21 days. If you must estimate the population mean within 4 days, how many outside salespeople should you sample? Use the 99% confidence level. (Round the intermediate calculation to 3 decimal places. Round the final answer to the nearest whole number.)                                   

Number of outside salespeople           

Answer 1

Solution :

Given that,

n = 670

x = 472

Point estimate = sample proportion = = x / n = 472/670=0.704

1 -   = 1- 0.704 =0.296

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z/2 = Z0.01 = 2.326 ( Using z table    )

Margin of error = E = Z / 2 *( (( * (1 - )) / n)

= 2.326 (((0.704*0.296) /670 )

E = 0.041

A 98% confidence interval for population proportion p is ,

- E < p < + E

0.704-0.041 < p < 0.704+ 0.041

0.663< p < 0.745

The 98% confidence interval for the population proportion p is : 0.663, 0.745

(B)

Solution :

Given that,

standard deviation = =21

Margin of error = E = 4

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = 2.576

sample size = n = [Z/2* / E] 2

n = ( 2.576*21 /4 )2

n =182.89

Sample size = n =183