The following is a random sample of the volt measurements of a solar panel taken at various times during the day:
18.9 |
17.5 |
17.2 |
18.4 |
15.7 |
18.7 |
18.1 |
18.1 |
17.9 |
18 |
18.3 |
18.1 |
16.7 |
18.8 |
17.8 |
18.4 |
18.4 |
19 |
18.3 |
Use this data to test the claim that the standard deviation of all volt measurements is less than 0.5 volts. Use a 99% confidence level. At this level, we can assume that the data are normally distributed.
What is the lower chi-square critical value for this test? Round to two decimal places.
For the record, I am posting a quiz question from an already graded assignment. Thanks
Null hypothesis
vs
Alternative hypothesis
For given data,
Sample size = n = 19,
Sample standard deviation = 0.801
Degree of freedom = n - 1 = 19 - 1 = 18
..............by using chi square critical value table or Excel command =CHIINV(1-0.01,18)
Chi square test statistic
=46.195
Chi square test statistic = 46.20 > 7.02
Therefore, we fail to reject H0 at
We do not have sufficient evidence at to say that, the standard deviation of all volt measurements is less than 0.5 volts.
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