Question

please do each and every step and calculations

Coquitlam Builders is a medium size home construction company. It wants to test the hypothesis that the average size of a new home unit exceeds 2,400 square feet. A random sample of 36 newly constructed homes had a mean of 2,510 square feet. Assume that the standard deviation of the size for all newly constructed homes is 480 square feet. Toll Brothers would like to set α = 0.02. Using the confidence interval approach, what is the decision in this case?

Answer 1

a.
Given that,
population mean(u)=2400
sample mean, x =2510
standard deviation, s =480
number (n)=36
null, Ho: μ<2400
alternate, H1: μ>2400
level of significance, α = 0.02
from standard normal table,right tailed t α/2 =2.133
since our test is right-tailed
reject Ho, if to > 2.133
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =2510-2400/(480/sqrt(36))
to =1.375
| to | =1.375
critical value
the value of |t α| with n-1 = 35 d.f is 2.133
we got |to| =1.375 & | t α | =2.133
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :right tail - Ha : ( p > 1.375 ) = 0.08894
hence value of p0.02 < 0.08894,here we do not reject Ho
ANSWERS
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null, Ho: μ=2400
alternate, H1: μ>2400
test statistic: 1.375
critical value: 2.133
decision: do not reject Ho
p-value: 0.08894
we do not have enough evidence to support the claim that the average size of a new home unit exceeds 2,400 square feet.

b.
TRADITIONAL METHOD
given that,
sample mean, x =2510
standard deviation, s =480
sample size, n =36
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 480/ sqrt ( 36) )
= 80
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.02
from standard normal table, two tailed value of |t α/2| with n-1 = 35 d.f is 2.438
margin of error = 2.438 * 80
= 195.04
III.
CI = x ± margin of error
confidence interval = [ 2510 ± 195.04 ]
= [ 2314.96 , 2705.04 ]
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DIRECT METHOD
given that,
sample mean, x =2510
standard deviation, s =480
sample size, n =36
level of significance, α = 0.02
from standard normal table, two tailed value of |t α/2| with n-1 = 35 d.f is 2.438
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 = [ 2510 ± t a/2 ( 480/ Sqrt ( 36) ]
= [ 2510-(2.438 * 80) , 2510+(2.438 * 80) ]
= [ 2314.96 , 2705.04 ]
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interpretations:
1) we are 98% sure that the interval [ 2314.96 , 2705.04 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 98% of these intervals will contains the true population mean