Question

(a). A simple random sample of 37 days was selected. For these 37 days, Parsnip was fed seeds on 22 days and Parsnip was fed pellets on the other 15 days. The goal is to calculate a 99% confidence interval for the proportion of all days in which Parsnip was fed seeds, and to do so there are two assumptions. The first is that there is a simple random sample, which is satisfied. What is the second assumption, and specific to the information in this problem, is it satisfied? Explain why or why not.

(b). If appropriate, use the data in part (a) to calculate and interpret a 99% confidence interval for the proportion of all days in his life that Parsnip has been fed seeds.

Answer 1

solution:

the given sample size = n=37

number of days on which parship was fed feed = X = 22

so proportion of sample = = 22/37 = 0.595

condition:

1: it was given that the sample is a random sample so random condition is satisfied

2: independent condition : the sample value is assumed to be independent of each other.

3: 10% conditin: the sample size should not be more than 10% of the total population so this condition also satisfied because there are many parship in the population is asssumed.

b) if our all the condition are satisfied then it is appropriate to find the confidence interval so the confidence interval are calculated as follows:

significance level = = 0-0.99 = 0.01

critical value of z =

margin of error =

confidence interval = ( - E) < p < ( + E)

upper limit = 0.595 + 0.208 = 0.803

lower limit = 0.595 - 0.208 = 0.387

interpretation:

with 99 % confident that the true proportion of parsnip was fed seed is between 0.387 and 0.803 .