Answer:
0.6604
Step-by-step explanation:
Given that a market research firm knows from historical data that telephone surveys have a 36% response rate.
Sample size of random sample = 280
We know for samples randomly drawn of large size sample proportion follows a normal distribution with mean= sample proportin and std error
=
Substitute p = 0.36 and q = 1-0.36= 0.64
p follows N with mean = 0.36 and std dev =
Using normal distribution values we can find\
Answer:
The 99% confidence interval would be given (11.448;14.152).
Step-by-step explanation:
1) Important concepts and notation
A confidence interval for a mean "gives us a range of plausible values for the population mean. If a confidence interval does not include a particular value, we can say that it is not likely that the particular value is the true population mean"
represent the sample deviation
represent the sample mean
n =1003 is the sample size selected
Confidence =99% or 0.99
represent the significance level.
2) Solution to the problem
The confidence interval for the mean would be given by this formula
We can use a z quantile instead of t since the sample size is large enough.
For the 99% confidence interval the value of and
, with that value we can find the quantile required for the interval in the normal standard distribution.
And replacing into the confidence interval formula we got:
And the 99% confidence interval would be given (11.448;14.152).
We are confident that about 11 to 14 are the number of books that the people had read on the last year on average, at 1% of significance.