Answer:
The 95% confidence interval for the true proportion of Gastown residents living below the poverty line is (0.133, 0.2766).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of , and a confidence interval , we have the following confidence interval of proportions.
In which
Z is the zscore that has a pvalue of .
For this problem, we have that:
He takes a simple random sample of 122 people living in Gastown and finds that 25 have an annual income that is below the poverty line. This means that and
We have , z is the value of Z that has a pvalue of = 0.975, so .
The lower limit of this interval is:
The upper limit of this interval is:
The 95% confidence interval for the true proportion of Gastown residents living below the poverty line is (0.133, 0.2766).
I don’t really know but here’s an example of another one maybe it’ll help
C because I had this yesterday
Answer:
Step-by-step explanation:
given that we are interested in finding out the proportion of adults in the United State who cannot cover a $400 unexpected expense without borrowing money or going into debt.
Sample size = 765
Favour = 322
a) The population is the adults in the United State who cannot cover a $400 unexpected expense without borrowing money or going into debt
b) The parameter being estimated is the population proportion P of adults in the United State who cannot cover a $400 unexpected expense without borrowing money or going into debt.
c) point estimate for proportion = sample proporiton =
d) We can use test statistic here as for proportions we have population std deviation known.
d) Std error = 0.01785(
Test statistic Z = p difference / std error
f) when estimated p is 0.50 we get Z = -4.43
g) Is true population value was 40% then
Z = 1.17 (because proportion difference changes here)