Question

13. (18 pts total) The U.S. Department of Agriculture (USDA) uses sample surveys to produce important economic estimates. One pilot study estimated wheat prices in July and in August using independent samples of wheat producers in the two months. Here are the summary statistics, in dollars per bushel. July: n1 = 35, x1 = $7:62, s1 = $0:28 and August: n2 = 25, x2 = $7:44, s2 = $0:35. Can we conclude that the national average price in July 1 and the national average price in August 2 are not the same? Answer this question in two ways (and use the correct df). (a) (6 pts) Calculate a 95% confidence interval for estimating 1 ? 2. (b) (6 pts) Calculate the two-sample t statistic and use Table D to compute a range of possibilities for the P-value of the test: H0 : 1 = 2 versus Ha : 1 6= 2. (c) (4 pts) What do your answers to parts (a) and (b) say about whether 1 and 2 are the same, at the = 5% level of significance? (d) (2 pts) Based on your answer to part (c), to what type of error are you now subject to (Type I or Type II)?

Answer 1

13.

a.
TRADITIONAL METHOD
given that,
mean(x)=7.62
standard deviation , s.d1=0.28
number(n1)=35
y(mean)=7.44
standard deviation, s.d2 =0.35
number(n2)=25
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
standard error = sqrt((0.078/35)+(0.123/25))
= 0.084
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.05
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 24 d.f is 2.064
margin of error = 2.064 * 0.084
= 0.174
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (7.62-7.44) ± 0.174 ]
= [0.006 , 0.354]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=7.62
standard deviation , s.d1=0.28
sample size, n1=35
y(mean)=7.44
standard deviation, s.d2 =0.35
sample size,n2 =25
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 7.62-7.44) ± t a/2 * sqrt((0.078/35)+(0.123/25)]
= [ (0.18) ± t a/2 * 0.084]
= [0.006 , 0.354]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [0.006 , 0.354] contains the true population proportion
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 proportion
b.
Given that,
mean(x)=7.62
standard deviation , s.d1=0.28
number(n1)=35
y(mean)=7.44
standard deviation, s.d2 =0.35
number(n2)=25
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.064
since our test is two-tailed
reject Ho, if to < -2.064 OR if to > 2.064
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =7.62-7.44/sqrt((0.0784/35)+(0.1225/25))
to =2.1302
| to | =2.1302
critical value
the value of |t α| with min (n1-1, n2-1) i.e 24 d.f is 2.064
we got |to| = 2.13021 & | t α | = 2.064
make decision
hence value of | to | > | t α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 2.1302 ) = 0.044
hence value of p0.05 > 0.044,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 2.1302
critical value: -2.064 , 2.064
decision: reject Ho
c.
p-value: 0.044
we have enough evidence to support the claim that difference in means between july and august.
d.
Type 1 error is possible because it reject the null hypothesis.