statistics and probability
A group of investors wants to develop a chain of fast-food restaurants. In determining potential costs for each facility, they must consider, among other expenses, the average monthly electric bill. They decide to sample some fast-food restaurants currently operating to estimate the monthly cost of electricity. They want to be 90% confident of their results and want the error of the interval estimate to be no more than $100. They estimate that such bills range from $600 to $2,500. How large a sample should they take?
The margin of error for this confidence interval for the mean is 100=Z*SD/sqrt(n).
To solve for N: sqrt N=Z*SD/1.645, now square both sides: N=[Z*SD/100]^2
The Z score for a 90% confidence interval is 1.645.
The SD standard deviation is 2500-600=1900
1900/4 N=[1.645*(1900/4)/100]^2=62
A sample of 62 is needed.
Your sample data has mean 3.1948 and standard deviation 0.0889, with a sample size of 25.
X (bar) = 3.948, s =0.0889, n = 25
Use a one-sample t-test
Conditions/assumptions for a t-test
•Random sample- our survey was a random sample of 25 stations
•Normal distribution –we are assuming gasoline prices are normally distributed
Null hypothesis: The mean regular unleaded gas prices for your region is the same as that in the study µ 0 = 3.16
Alternate hypothesis: The mean regular unleaded gas prices for your region is greater than that in the study µ A > 3.16
Find the t-test statistic: t=(x(bar) - µ 0 )/(s/sqrt(n))
t = (3.1948 – 3.16)/(0.0889/sqrt(25)) t = 1.957255343
Find the P-value:
P = 0.03102312
Write a conclusion in context. Since we are using a 1% level of significance, the -level is 0.01. Since P> 0.01, we can not reject the null hypothesis.
In context:
We do not have evidence at the 1% level (P = 0.03102312) that the mean regular unleaded gas prices for our region is greater than that in the study
The margin of error for this confidence interval for the mean is 100=Z*SD/sqrt(n).
To solve for N: sqrt N=Z*SD/1.645, now square both sides: N=[Z*SD/100]^2
The Z score for a 90% confidence interval is 1.645.
The SD standard deviation is 2500-600=1900
1900/4 N=[1.645*(1900/4)/100]^2=62
A sample of 62 is needed.
Your sample data has mean 3.1948 and standard deviation 0.0889, with a sample size of 25.
X (bar) = 3.948, s =0.0889, n = 25
Use a one-sample t-test
Conditions/assumptions for a t-test
•Random sample- our survey was a random sample of 25 stations
•Normal distribution –we are assuming gasoline prices are normally distributed
Null hypothesis: The mean regular unleaded gas prices for your region is the same as that in the study µ 0 = 3.16
Alternate hypothesis: The mean regular unleaded gas prices for your region is greater than that in the study µ A > 3.16
Find the t-test statistic: t=(x(bar) - µ 0 )/(s/sqrt(n))
t = (3.1948 – 3.16)/(0.0889/sqrt(25)) t = 1.957255343
Find the P-value:
P = 0.03102312
Write a conclusion in context. Since we are using a 1% level of significance, the -level is 0.01. Since P> 0.01, we can not reject the null hypothesis.
In context:
We do not have evidence at the 1% level (P = 0.03102312) that the mean regular unleaded gas prices for our region is greater than that in the study