In 2001 the mean household expenditure for energy was $1493. An economist wants

Question

In 2001 the mean household expenditure for energy was $1493. An economist wants to know if it has changed; to that effect he conducts a random sample of 35 households and found that in equivalent 2001 dollars, expenditure for energy is $1618 with a standard deviation of 321. Test the claim to a significance level of 0.05 The Null Hypothesis is The Alternate Hypothesis is The value of the test statistic is The Critical values is/are The P-value is The Confidence Interval is: The results of the test indicate the Null Hypothesis because (wha is/is not (specify) within (wah The final conclusion is

Explanation / Answer

PART A.
Given that,
population mean(u)=1493
sample mean, x =1618
standard deviation, s =321
number (n)=35
null, Ho: =1493
alternate, H1: !=1493
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.032
since our test is two-tailed
reject Ho, if to < -2.032 OR if to > 2.032
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =1618-1493/(321/sqrt(35))
to =2.3038
| to | =2.3038
critical value
the value of |t | with n-1 = 34 d.f is 2.032
we got |to| =2.3038 & | t | =2.032
make decision
hence value of | to | > | t | and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 2.3038 ) = 0.0275
hence value of p0.05 > 0.0275,here we reject Ho
ANSWERS
---------------
null, Ho: =1493
alternate, H1: !=1493
test statistic: 2.3038
critical value: -2.032 , 2.032
decision: reject Ho
p-value: 0.0275

PART B.
TRADITIONAL METHOD
given that,
sample mean, x =1618
standard deviation, s =321
sample size, n =35
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 321/ sqrt ( 35) )
= 54.259
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 34 d.f is 2.032
margin of error = 2.032 * 54.259
= 110.254
III.
CI = x ± margin of error
confidence interval = [ 1618 ± 110.254 ]
= [ 1507.746 , 1728.254 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =1618
standard deviation, s =321
sample size, n =35
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 34 d.f is 2.032
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 = [ 1618 ± t a/2 ( 321/ Sqrt ( 35) ]
= [ 1618-(2.032 * 54.259) , 1618+(2.032 * 54.259) ]
= [ 1507.746 , 1728.254 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 1507.746 , 1728.254 ] 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

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