Im having trouble with part 2 of this question, part 1a,1b, and 1c are all corre
ID: 3360191 • Letter: I
Question
Im having trouble with part 2 of this question, part 1a,1b, and 1c are all correct
(2 pts) A random sample 14 student cars (group 1) were found to have ages with a mean of 7.2 years and a standard deviation of 3.6 years, while second, independent random sample of 9 faculty cars (group 2) were found to have ages with a mean of 5.1 years and a standard deviation of 3.5 years. You may assume that car ages for both groups are approximately normally distributed and that the two population variances are equal. 1. Use a 0.05 significance level to test the claim that, on average, student cars are older than faculty cars. (a) The test statistic is t = 1.379 (b) The critical value is t1.721 (c) Is there sufficient evidence to support the claim that student cars are older than faculty cars? O A. Yes O B. No 2. Construct a 95% confidence interval estimate o he difference 1 2 where 1 s the mean age of student cars and h IS the mean age of ta y cars 0.01564857Explanation / Answer
given that,
mean(x)=7.2
standard deviation , s.d1=3.6
number(n1)=14
y(mean)=5.1
standard deviation, s.d2 =3.5
number(n2)=9
I.
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (13*12.96 + 8*12.25) / (23- 2 )
s^2 = 12.69
II.
standard error = sqrt(S^2(1/n1+1/n2))
=sqrt( 12.69 * (1/14+1/9) )
=1.522
III.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, = 0.05
from standard normal table, two tailed and value of |t | with (n1+n2-2) i.e 21 d.f is 2.08
margin of error = 2.08 * 1.522
= 3.166
IV.
CI = (x1-x2) ± margin of error
confidence interval = [ (7.2-5.1) ± 3.166 ]
= [-1.066 , 5.266]
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DIRECT METHOD
given that,
mean(x)=7.2
standard deviation , s.d1=3.6
sample size, n1=14
y(mean)=5.1
standard deviation, s.d2 =3.5
sample size,n2 =9
CI = x1 - x2 ± t a/2 * sqrt ( s^2 ( 1 / n1 + 1 /n2 ) )
where,
x1,x2 = mean of populations
s^2 = pooled variance
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 7.2-5.1) ± t a/2 * sqrt( 12.69 * (1/14+1/9) ]
= [ (2.1) ± 3.166 ]
= [-1.066 , 5.266]
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interpretations:
1. we are 95% sure that the interval [-1.066 , 5.266]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