This Question: 1 pt 8 of 15 (0 complete) This Quiz: 15 pts possible Question Hel

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This Question: 1 pt 8 of 15 (0 complete) This Quiz: 15 pts possible Question Help Arandom sample of students at a college reported what they believed to be their heights in inches. Then the students measured each others' heights in centimeters without shoes. The data provided are for the men, with their believed heights converted from inches to centimeters. Assume that conditions for t-tests hold. Complete parts a and b below EEB Click the icon to view the data. a. Find a 95% confidence interval for the mean difference as measured in centimeters. Does it capture 0? What does that show? The 95% confidence interval is (Round to three decimal places as needed.) The interval | include 0, so a hypothesis that the means are equal | be rejected b. Perform a t-test to test the hypothesis that the means are not the same. Use a significance level of 0.05 Determine the hypotheses for this test. Let difference be the population mean difference between measured and believed height in centimeters. Choose the correct answer beloW A. Ho : difference #0 a: difference-o C. Ho: difference:0 0 DE, Ho-difference:0 sn Click to select your answer(s)

Explanation / Answer

a.
TRADITIONAL METHOD
given that,
mean(x)=177.7333
standard deviation , s.d1=7.6669
number(n1)=15
y(mean)=175.9373
standard deviation, s.d2 =8.5041
number(n2)=15
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((58.781/15)+(72.32/15))
= 2.956
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, =
from standard normal table, two tailedand
value of |t | with min (n1-1, n2-1) i.e 14 d.f is 2.145
margin of error = 2.145 * 2.956
= 6.341
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (177.7333-175.9373) ± 6.341 ]
= [-4.545 , 8.137]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=177.7333
standard deviation , s.d1=7.6669
sample size, n1=15
y(mean)=175.9373
standard deviation, s.d2 =8.5041
sample size,n2 =15
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 = [( 177.7333-175.9373) ± t a/2 * sqrt((58.781/15)+(72.32/15)]
= [ (1.796) ± t a/2 * 2.956]
= [-4.545 , 8.137]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-4.545 , 8.137] 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)=177.7333
standard deviation , s.d1=7.6669
number(n1)=15
y(mean)=175.9373
standard deviation, s.d2 =8.5041
number(n2)=15
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.14
since our test is two-tailed
reject Ho, if to < -2.14 OR if to > 2.14
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =177.7333-175.9373/sqrt((58.78136/15)+(72.31972/15))
to =0.61
| to | =0.61
critical value
the value of |t | with min (n1-1, n2-1) i.e 14 d.f is 2.14
we got |to| = 0.6075 & | t | = 2.14
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 0.6075 ) = 0.553
hence value of p0.05 < 0.553,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 0.61
critical value: -2.14 , 2.14
decision: do not reject Ho
p-value: 0.553
we do not have enough evidence to support the claim that difference between means of measured and believed heights

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