Question

A processor of carrots cuts the green top off each carrot, washes the carrots, and inserts six to a package. Twenty packages are inserted in a box for shipment. To test the mass of the boxes, a few were checked. The mean mass was 9.2 kg, the standard deviation 0.27 kg. How many boxes must the processor sample to be 99% confident that the sample mean does not differ from the population mean by more than 0.14 kg? (Round the intermediate calculation to 2 decimal places. Round the final answer to the nearest whole number.)

Sample size      

    

Exercise 8-37 Suppose the prime minister wants an estimate of the proportion of the population who support his current policy on health care. The prime minister wants the estimate to be within 0.08 of the true proportion. Assume a 80% level of confidence. The prime ministers political advisors estimated the proportion supporting the current policy to be 0.35. (Round the intermediate calculation to 2 decimal places. Round the final answer to the nearest whole number.) a. How large of a sample is required? b. How large of a sample would be necessary if no estimate were available for the proportion that support current policy?

Answer 1

Q.1) Given that, population standard deviation $(\sigma)$ = 0.27 kg

Margin of error ( E ) = 0.14 kg

A 99% confidence level has significance level of 0.01 and critical value is,

Zo/2 = 2.575

We want to find, the sample size ( n ),

0.27 0.14 ->nN 25

Therefore, required sample size is 25

Q.2) Given that, margin of error ( E ) = 0.08

A 80% confidence level has significance level of 0.20 and critical value is,

a21.28

We want to find, sample size for following cases,

a) If estimate of proportion = p = 0.35

1.282 n = p * (1-p) * (4y2)2 = 0.35 * (1-0.35) * (しの)-58.24 0.08 > n 58

Therefore, required sample size is 58

b) If no estimate is given then we assume p = 0.5

n=ps(1-p) * (-_リー)2-0.35 * (1-0.5) * ( -)2-64 0.08 64

Therefore, required sample size is 64