The heights (in inches) of 20 randomly selected adult males are listed below. Test the claim that the variance is less than 6.25. Use 伪 = 0.05. Assume the population is normally distributed.
70 72 71 70 69 73 69 68 70 71
67 71 70 74 69 68 71 71 71 72
I punched in the numbers in the TI-83 and got the mean, standard deviation. How do I test the claim?|||Hypothesis Test for population variance
If we have a sample from an underlying normal distribution and variance 蟽虏 then we can test the null hypothesis:
H0: 蟽虏 = 蟽0虏
for some fixed 蟽0虏.
If H0 is true then 围虏 = (n - 1) S虏 / 蟽0虏. Where 围虏 is the chi square with n - 1 degrees of freedom.
for the alternate hypothesis we have:
H1a: 蟽虏 %26gt; 蟽0虏
H1b: 蟽虏 %26lt; 蟽0虏
H1c: 蟽虏 鈮?蟽0虏
the test statistic is the same for all tests.
the rejection regions for the above tests are:
a) 围虏 %26gt; 围虏伪
b) 围虏 %26lt; 围虏1-伪
c) 围虏 %26lt; 围虏伪/2 or 围虏 %26gt; 围虏1-伪/2
where 围虏伪 is the value such that:
P(围虏 %26gt; 围虏伪) = 伪 where 围虏 is the chi square with n - 1 degrees of freedom.
In this question we have:
H0: 蟽虏 鈮?6.25 vs. H1: 蟽虏 %26lt; 6.25
the variance of the sample is: 2.976316
the test statistic is:
(20 - 1) * 2.976316 / 6.25 = 9.048
the p-value is: P(围虏 %26lt; 9.048) = 0.02732636
with the low p-value we reject the null and conclude the variance is less than 6.25
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