Testing Hypotheses about Means and Proportion

#1
Hello Everyone!

I am having trouble with one of my questions it goes:

It is claimed that fewer than 30% of Thunder Bay households have home delivery of the newspaper. To test this claim, a random sample of 390 Thunder Bay households reveals that only 95 get the paper delivered. Using an alpha value of 0.01, what conclusion can be reached about the claim? (Test Statistic z* or t*)
H0: P=0.3
H1: P not equal to 0.3

I am pretty sure I am doing t* because the sample size is greater than 30, but the formula we are to us is x-Mu/S/sqrt(n)
x=95
Mu=0.3
S=?
n=390 (I'm not sure if those are right either)
I don't know how to find the standard deviation (s) I tried using S= sqrt[(np)(1-p)] where n=390 and p=30% (0.3) but my final answer is 9.05, and when I plug it into the first formula I get a large number 206.7 and I don't think that is right.

I just don't know how to find standard deviation for this equation and what to do after I solve the equation
Thanks!
 

Karabiner

TS Contributor
#2
Your dependent variable is a categorical (binary) variable,
not an interval scaled variable. Therefore, you deal with a
proportion, not with a mean value, and t-test is inappropriate.
You need to find a test for a proportion within 1 sample.

With kind regards

K.
 
#3
Thank you, But the formula I had isn't right? My professor only gave us that equation and I am not sure how to answer the question?
 
#4
The formula needs a correction S= sqrt[p(1-p)/n]
Additionally, the thumb rule is to use z-test for sample sizes greater than 30.
Don't worry about not being able to apply t-test/z-test because of it being a categorical variable; the proportion is still the mean of a binomial distribution.
Also if the sample is greater than 10% of that of the population, you will need to use the finite population corrector.
 
#5
The formula needs a correction S= sqrt[p(1-p)/n]
Additionally, the thumb rule is to use z-test for sample sizes greater than 30.
Don't worry about not being able to apply t-test/z-test because of it being a categorical variable; the proportion is still the mean of a binomial distribution.
Also if the sample is greater than 10% of that of the population, you will need to use the finite population corrector.
Ok so I did the correction for my S and I got ~0.0232

Now when I plug that into my equation my final answer is 54,183.02 that number seems way to large, where I am I going wrong? or am I on the right track ?
 
#6
Sorry, I should have caught this earlier.
The main formula should be z=(x-p)/sqrt(S) where,
x=95/390
p=0.3
S=sqrt[p(1-p)/n]
using this, you will get z= -2.43