The heights of UNC sophomores are approximately normally distributed. The height
ID: 3155167 • Letter: T
Question
The heights of UNC sophomores are approximately normally distributed. The heights in inches of 8 randomly selected sophomores are shown below. Use these heights to find a 95% confidence interval for the average height p of UNC sophomores. Give the endpoints of your interval to one decimal place. 72, 69, 70, 68, 70, 67, 75, 72 Use these heights to find a 95% confidence interval for the average height p of UNC sophomores. Give the endpoints of your interval to one decimal place. 95% confidence interval: Based on your answer to part a), conduct a level alpha =0.05 test of the hypotheses H_0: mu = 68; Ha: mu 68. No additional work should be required! Reject the null hypothesis at alpha =0.05. Fail to reject the null hypothesis at a=0.05. Using the 8 data points above, conduct a test of following hypotheses at level of significance alpha = 0.05: H0: mu = 67; Ha: mu > 67 Find the sample mean x, the sample standard deviation s, the appropriate t-statistic and p -value. Use t distribution calculator mentioned in class and give all answers to four decimal places.Explanation / Answer
a)
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 = 70.375
t(alpha/2) = critical t for the confidence interval = 2.364624252
s = sample standard deviation = 2.55999442
n = sample size = 8
df = n - 1 = 7
Thus,
Lower bound = 68.23479111
Upper bound = 72.51520889
Thus, the confidence interval is
( 68.23479111 , 72.51520889 ) [ANSWER]
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c)
Formulating the null and alternative hypotheses,
Ho: u <= 67
Ha: u > 67
As we can see, this is a right tailed test.
df = n - 1 = 7
Getting the test statistic, as
X = sample mean = 70.375 [ANSWER, SAMPLE MEAN]
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uo = hypothesized mean = 67
n = sample size = 8
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s = standard deviation = 2.55999442 [ANSWER, SAMPLE STANDARD DEVIATION]
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Thus, t = (X - uo) * sqrt(n) / s = 3.728891544 [ANSWER, t]
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Also, the p value is
p = 0.003684726 [ANSWER, P VALUE]
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