Question

I am looking for help with the following question:

A large distributor just received a shipment of 1000 cartons of egg. Each carton has 12 eggs and the carton is up to standard if the weight of that carton (total weight of 12 eggs) is more than 28 oz. Previous shipment shows that 96% of the egg cartons are up to standard. In this sampling problem, the setup cost to sample additional units are negligible since the weighing station is readily available whereas it does take time and resource to weigh the carton and record the data. How many cartons should one sample if one wants to estimate the proportion of up to standard egg cartons to within 1% with 99% confidence? a. If you use educated guess. b. If you use conservative method? c. Do you expect that this sample size from educated guess would guarantee the resulting confidence interval being as short as specified? Explain briefly. d. Would you choose to use the sample size from educated guess or from the conservative method? e. For this problem, can you ignore the finite correction factor?

Answer 1

Solution:-

a) We need to sample 2549 if one wants to estimate the proportion of up to standard egg cartons to within 1% with 99% confidence.

p = 0.96

$M.E = z_{\alpha /2}\times\sqrt{ \frac{p(1-p)}{n}}$

$n =\left ( \frac{z_{\alpha /2}}{M.E} \right )^{2}\times p(1-p)$

$n =\left ( \frac{2.576}{0.01} \right )^{2}\times 0.96(1-0.96)$

n = 2548.14

n = 2549

b) We need to sample 16590 if one wants to estimate the proportion of up to standard egg cartons to within 1% with 99% confidence.

p = 0.50(Conventional Method)

$M.E = z_{\alpha /2}\times\sqrt{ \frac{p(1-p)}{n}}$

$n =\left ( \frac{z_{\alpha /2}}{M.E} \right )^{2}\times p(1-p)$

$n =\left ( \frac{2.576}{0.01} \right )^{2}\times 0.50(1-0.50)$

n = 16589.44

n = 16590

c) Yes, this sample size from educated guess would guarantee the resulting confidence interval being as short as specified, because we have the exact value for the point estimate of the proportion.

d) We would choose to use the sample size from educated guess because that would decrease the costing for the sampling.