There are different ways you can do this kind of problem, but the formula below might be one of the easier ways:
s/[1 + (1.645/√2n)]
..to..
s/[1 - (1.645/√2n)]
...where s = standard deviation, 1.645 represents the 90% confidence interval using a z-table, and n = sample size.
You will need to calculate standard deviation from the data given, then substitute the standard deviation and sample size into the formula.
After you finish the above calculations, square the standard deviation values to find the variance values.
I hope this will help get you started.
A statistics professor is used to having a variance in his class grades of no more than 100. He feels that his current group of students is different, and so he examines a random sample of midterm grades (listed below). At 0.05 alpha level can it be conculded that the variance in the grades differs from 100? Write a 90% confidence level of the variance using the information given
92.3 89.4 76.9 65.2. 49.1. 96 7. 69.5. 72.8. 77.5 52.8. 8.85. 79.2. 72.9. 68.7. 75.8
1 answer
1 answer