Question 981117: Lily Energy Systems manufacturer's wood-burning heaters and fireplace inserts. One of its systems has
an electric blower, which is thermostatically controlled. The blower is designed to automatically turn on
when the temperature in the stove reaches 125°F and turn off at 85°F. Complaints from customers indicate
that the thermostat control is not working properly. The company feels that the thermostat is acceptable if
the variance in the cutoff temperature is less than or equal to 175. The company takes a sample of 24
thermostats and finds that the variance equals 289. The calculated chi-square test statistic and the table
value for a 0.05 significance level are
A. 37.983, 38.076.
B. 35.172, 38.99.
C. 37.983, 35.172.
D. 38.076, 38.99.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Hypothesis:
Ho: sigma^2 <= 175
Ha: sigma^2 > 175
This is a right tailed test. If the chi-square test statistic is larger than the cutoff critical value, then reject the null.
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Test Statistic:
[(n-1)*s^2]/(sigma^2)
[(24-1)*289]/(175)
37.9828571428571
37.983
The chi-square test statistic is approximately 37.983
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To find the critical value, use a table like this one. Look at the row that corresponds to df = n-1 = 24-1 = 23. Then look at the column that corresponds to the right tail having an area of 0.050. The value in this row/column intersection is 35.172
Critical value: 35.172
Because the test statistic is larger than the critical value, we reject the null. We conclude that the variance is definitely larger than 175. So the company will have to fix the issue.
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