My apologies, the figure got messed up
Paper shredders
---------------
75 115 118 287
250 300 50
Calculators
--------------
110 140 165 90 230
97 269
The figures below represent the prices (in dollars) for both paper shredders and calculators. At the 0.10 level of significance, can it be concluded that the variance in price differs between the two types of machines?
Paper shredders | Calculators
-----------------------------------
75 115 118 287 | 110 140 165 90 230
250 300 50 | 97 269
|
Thank You very much in advance for any help.
2 answers
This is a hypothesis test involving variances.
Sample size for paper shredders = 7
Variance = ? (you will need to calculate the variance from the data given)
degrees of freedom = 6 (df = n - 1)
Sample size for calculators = 7
Variance = ? (you will need to calculate the variance from the data given)
degrees of freedom = 6
Determine the critical or cutoff value to reject the null using an F-distribution table at 0.10 level of significance using the above information.
Test statistic = variance of paper shredders divided by variance of calculators
Once you have the test statistic, compare to the critical value you found from the table. If the test statistic exceeds the critical value, reject the null (difference). If the test statistic does not exceed the critical value, do not reject the null (no difference).
I hope this will help.
Sample size for paper shredders = 7
Variance = ? (you will need to calculate the variance from the data given)
degrees of freedom = 6 (df = n - 1)
Sample size for calculators = 7
Variance = ? (you will need to calculate the variance from the data given)
degrees of freedom = 6
Determine the critical or cutoff value to reject the null using an F-distribution table at 0.10 level of significance using the above information.
Test statistic = variance of paper shredders divided by variance of calculators
Once you have the test statistic, compare to the critical value you found from the table. If the test statistic exceeds the critical value, reject the null (difference). If the test statistic does not exceed the critical value, do not reject the null (no difference).
I hope this will help.