Question

In a random sample of n = 500 families owning television sets in the city of Amman, Jordan, it is found that x = 340 subscribe to Netflix service. 1- Find a 95% confidence interval for the true proportion of families with television sets in this city that subscribe to Netflix service. 2- If ˆp is used as an estimate of p, can we be 95% confident that the error will not exceed 0.04. Explain Part (b)- The weights, in decagrams, of a sample of 10 packages of grass seed distributed by a certain company: yielded a sample mean of 46.12 and a sample standard deviation of 0.535. Find a 95% confidence interval for the true standard deviation of the weights of all such packages of grass seed distributed by this company, assuming a normal population.

Answer 1

Here Sample proportion = = x/n = 340/500= 0.68

standard error of proportion = sep = sqrt [ * (1-)]/n] = sqrt [0.68 * 0.32/500] = 0.02086

1. Here 95% confidence interval for the true proportion of families = +- Zcritical sep = 0.68 +- 1.96 * 0.02086 = (0.639, 0.721)

2. No, here error as we see is greater than 0.04 so we can't be 95% confidence that error will not exceed 0.04.

Explain Part(b)

sample size = n = 10

sample mean = = 46.12 dgm

sample standard deviation = s = 0.535 dgm

standard error =se = s/sqrt(n) = 0.535/sqrt(10) = 0.1692

95% confidence interval = +- tcritical se

dF = n -1 = 10 -1 = 9

Critical value = tcritical = TINV(0.05, 9) = 2.262

95% confidence interval = 46.12 +- 2.262 * 0.1692 = (45.74 dgm, 46.50 dgm)