A random sample of students at a college reported what they

A random 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. Click the icon to view the data. 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 The interval include 0: so a hypothesis that the means are equal be rejected. 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 mu difference be the population mean difference between measured and believed height Find the test statistic for this test.

Solution

Consider the table:

A)

Getting the mean and standard deviation of the third column, (difference column),

X = -0.794666667
s = 2.979544228

Note that              
              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    -0.794666667          
t(alpha/2) = critical t for the confidence interval =    2.144786688          
s = sample standard deviation =    2.979544228          
n = sample size =    15          
df = n - 1 =    14          
Thus,              
              
Lower bound =    -2.444683263          
Upper bound =    0.855349929          
              
Thus, the confidence interval is              
              
(   -2.444683263   ,   0.855349929   ) [ANSWER, 95% CONFIDENCE]

Yes, this captures 0.

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[If you want to write this the other way round,

(-0.855349929,   2.444683263)]

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b)

Formulating the null and alternative hypotheses,              
              
Ho:   ud   =   0  
Ha:   ud   =/=   0   [OPTION B]

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At level of significance =    0.05          
As we can see, this is a    two   tailed test.      
              
Calculating the standard deviation of the differences (third column):              
              
s =    2.075062792          
              
Thus, the standard error of the difference is sD = s/sqrt(n):              
              
sD =    0.535778909          
              
Calculating the mean of the differences (third column):              
              
XD =    0.794666667          
              
As t = [XD - uD]/sD, where uD = the hypothesized difference =    0   , then      
              
t =    -1.483198859 [ANSWER, TEST STATISTIC]          
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Additional info:

As df = n - 1 =    14          
              
Then the critical value of t is              
              
tcrit =    +/-   2.144786688      
              
Thus, comparing t and tcrit, we   WE FAIL TO REJECT THE NULL HYPOTHESIS.          

165	165.1	-0.1
174	175.26	-1.26
184	185.42	-1.42
165	167.64	-2.64
189	195.58	-6.58
171	175.26	-4.26
172	172.72	-0.72
192	195.58	-3.58
177	175.26	1.74
176	175.26	0.74
183	177.8	5.2
175	177.8	-2.8
173	172.72	0.28
168	167.64	0.36
186	182.88	3.12