Question

Use these 30 different gas prices

2.70, 2.40, 2.58, 2.51, 2.54, 2.57, 2.50, 2.60, 2.65, 2.52, 2.30, 2.55, 2.30, 2.59, 2.54, 2.68, 2.78, 2.75, 2.71, 2.61, 2.70, 2.56, 2.58, 2.60, 2.62, 2.69, 2.55, 2.40, 2.34, 2.56

Part 1 - Estimate the mean price of regular gasoline in the Dayton area by taking a sample. What is the population you will be studying? Think carefully about how you are going to get a good sample to calculate a sample mean and collect the sample. For example, driving around your neighborhood is an example of a convenience sample and should not be done. Think of a more efficient way to collect gas prices that will also cover the whole Dayton area. Write a paragraph or two explaining how you collected your sample and why you think your sample is representative of Dayton gas prices. What is the sample mean? Construct a 95% confidence interval for the population mean price of gasoline in the Dayton area. Explain what this confidence interval tells you.

Part 2 - Compare your sample estimate to the current mean gas in the State of Ohio. Find the mean price of gasoline in the State of Ohio. using http://www.gasbuddy.com/USA. Set up a null and alternative hypothesis to see if your sample for Dayton is enough to prove that the population mean gasoline price in Dayton is different than the mean price in Ohio. Test the hypotheses. Show your work. Using a significance level of 0.05, what is your conclusion?

Part 3 - Compare your sample estimate to the current mean gas in the U.S. What is the mean price of gasoline in the U.S.? using http://www.gasbuddy.com/Charts. Set up a null and alternative hypothesis to see if your sample for Dayton is enough to prove that the population mean price in Dayton is different than the mean price in the U.S. Test the hypotheses. Show your work. Using a significance level of 0.05, what is your conclusion?

Answer 1

1.
TRADITIONAL METHOD
given that,
sample mean, x =2.566
standard deviation, s =0.123
sample size, n =30
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.123/ sqrt ( 30) )
= 0.022
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 29 d.f is 2.045
margin of error = 2.045 * 0.022
= 0.046
III.
CI = x ± margin of error
confidence interval = [ 2.566 ± 0.046 ]
= [ 2.52 , 2.612 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =2.566
standard deviation, s =0.123
sample size, n =30
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 29 d.f is 2.045
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 2.566 ± t a/2 ( 0.123/ Sqrt ( 30) ]
= [ 2.566-(2.045 * 0.022) , 2.566+(2.045 * 0.022) ]
= [ 2.52 , 2.612 ]
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interpretations:
1) we are 95% sure that the interval [ 2.52 , 2.612 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean

2.
Given that,
population mean(u)=2.566
sample mean, x =2.696
standard deviation, s =0.123
number (n)=30
null, Ho: μ=2.566
alternate, H1: μ!=2.566
level of significance, alpha = 0.05
from standard normal table, two tailed t alpha/2 =2.045
since our test is two-tailed
reject Ho, if to < -2.045 OR if to > 2.045
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =2.696-2.566/(0.123/sqrt(30))
to =5.789
| to | =5.789
critical value
the value of |t alpha| with n-1 = 29 d.f is 2.045
we got |to| =5.789 & | t alpha | =2.045
make decision
hence value of | to | > | t alpha| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 5.7889 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: μ=2.566
alternate, H1: μ!=2.566
test statistic: 5.789
critical value: -2.045 , 2.045
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that the population mean gasoline price in Dayton is different than the mean price in Ohio

3.
Given that,
population mean(u)=2.566
sample mean, x =2.598
standard deviation, s =0.123
number (n)=30
null, Ho: μ=2.566
alternate, H1: μ!=2.566
level of significance, alpha = 0.05
from standard normal table, two tailed t alpha/2 =2.045
since our test is two-tailed
reject Ho, if to < -2.045 OR if to > 2.045
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =2.598-2.566/(0.123/sqrt(30))
to =1.425
| to | =1.425
critical value
the value of |t alpha| with n-1 = 29 d.f is 2.045
we got |to| =1.425 & | t alpha | =2.045
make decision
hence value of |to | < | t alpha | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 1.425 ) = 0.1648
hence value of p0.05 < 0.1648,here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=2.566
alternate, H1: μ!=2.566
test statistic: 1.425
critical value: -2.045 , 2.045
decision: do not reject Ho
p-value: 0.1648
we do not have enough evidence to support the claim that the population mean price in Dayton is different than the mean price in the U.S