Question

Recall that your economic consultancy firm has been hired by a law firm that is representing the owners of a restaurant. The restaurant burnt down because of a neighbouring business’ negligence. The law firm’s client is suing the neighbour’s insurance company for lost sales. Suppose the insurance company and the law firm have agreed that any payment should be based on the average sales among restaurants operating in similar cities. If S is defined as the random variable representing sales at a restaurant in similar cities then the law and insurance firms are interested in knowing E [S]. Both firms have agreed to use a data set, where the sampling was random, and where n = 1024. In this data set, the insurance company has estimated a sample mean of 1.28 (in millions of dollars) and a sample variance of 0.64. You calculated a 90% confidence interval for the mean of X that was: 1.24 < E [S] < 3.2 Now, you use the same information to test the null hypothesis that E [S] = 3 at the .1 significance level. Show your work. Use three non-zero decimal places in your calculations. Make sure you: • Write down the null and alternative hypotheses. • Write down the formula and calculate the test-statistic. • Write down the decision rule • Write down the appropriate critical value(s) • Reach a conclusion about your test.

Answer 1

a.
TRADITIONAL METHOD
given that,
standard deviation, σ =0.8
sample mean, x =1.28
population size (n)=1024
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 0.8/ sqrt ( 1024) )
= 0.03
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.645
since our test is two-tailed
value of z table is 1.645
margin of error = 1.645 * 0.03
= 0.04
III.
CI = x ± margin of error
confidence interval = [ 1.28 ± 0.04 ]
= [ 1.24,1.32 ]
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DIRECT METHOD
given that,
standard deviation, σ =0.8
sample mean, x =1.28
population size (n)=1024
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.645
since our test is two-tailed
value of z table is 1.645
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 1.28 ± Z a/2 ( 0.8/ Sqrt ( 1024) ) ]
= [ 1.28 - 1.645 * (0.03) , 1.28 + 1.645 * (0.03) ]
= [ 1.24,1.32 ]
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interpretations:
1. we are 90% sure that the interval [1.24 , 1.32 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean

b.
Given that,
population mean(u)=3
standard deviation, sigma =0.8
sample mean, x =1.28
number (n)=1024
null, Ho: μ=3
alternate, H1: μ!=3
level of significance, alpha = 0.1
from standard normal table, two tailed z alpha/2 =1.645
since our test is two-tailed
reject Ho, if zo < -1.645 OR if zo > 1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 1.28-3/(0.8/sqrt(1024)
zo = -68.8
| zo | = 68.8
critical value
the value of |z alpha| at los 10% is 1.645
we got |zo| =68.8 & | z alpha | = 1.645
make decision
hence value of | zo | > | z alpha| and here we reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -68.8 ) = 0
hence value of p0.1 > 0, here we reject Ho
ANSWERS
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null, Ho: μ=3
alternate, H1: μ!=3
test statistic: -68.8
critical value: -1.645 , 1.645
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that mean is 3