Unformatted text preview: h numerator degrees of freedom (n x − 1) = 3 and denominator degrees of freedom (n y − 1) = 6 the F‐distribution critical value is: Fc = 4.76 The calculated test statistic of 7.095 exceeds the critical value and therefore, there is evidence to reject the null hypothesis of equal variance in the two sample periods. The data suggests that the variance of sales is significantly higher in the period of active price competition. 18 Chapter 11 The graph below illustrates that the calculated test statistic of 7.095 is in the rejection region for a 5% level one‐tailed test. PDF of F( 3 , 6 ) upper tail area = 0.05 0 Fc=4.76 7.095 d o n o t r e j e c t ← → Reject H0 19 Chapter 11 Now approach a decision rule by calculating and interpreting a p‐value for the test. The p‐value of the test is the smallest significance level at which a null hypothesis can be rejected. In the above exercise, the null hypothesis was rejected at a 5% level and so the p‐value must be smaller than 5%. An e...
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- Spring '10
- Null hypothesis, FC, S2 Yachts