chap10S4PRN

# Consider daily stock market closing prices for a

• Notes
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Consider daily stock market closing prices for a company observed in two sample periods. For Johnson & Johnson, a test of equal means in the first and second half of 1999, was developed in the lecture notes for Chapters 8.2 and 10.2. The test method assumed the same variance in the two sample periods. If this variance assumption does not describe the data then the test conclusions may be unreliable. Is the assumption of equal variance in the two sample periods a reasonable assumption ? To answer this question, test the null hypothesis of equal variances in the two samples against a two-sided alternative. Summary statistics for the closing prices in the two sample periods were reported as: January to June 124 = x n 32.54 = 2 x s July to December 128 = y n 21.45 = 2 y s Econ 325 – Chapter 10.4 12 The test statistic is the variance ratio: 1.517 21.45 32.54 = = 2 y 2 x s s Note that the larger variance is placed in the numerator to give a test statistic that exceeds one. The test statistic can be compared with an F-distribution with numerator degrees of freedom (124 – 1) = 123 and denominator degrees of freedom (128 – 1) = 127. For a two-tailed test, the p-value is found as: p-value ) F ( P ) 127 , 123 ( 1.517 2 > = The p-value calculation is illustrated in the graph below. PDF of ) 127 , 123 ( F 1.517 1 0.5 0 area = p-value / 2

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Econ 325 – Chapter 10.4 13 To find an exact p-value, with Microsoft Excel select Insert Function: FDIST(1.517, 123, 127) This gives the answer: pvalue / 2 = 0.0102 Therefore, the p-value for the two-tailed test is: 2 (0.0102) = 0.0204 At a 1% significance level the p-value exceeds α = 0.01. This gives evidence that, for the Johnson & Johnson daily closing prices, the null hypothesis of equal variance in the two sample periods is not rejected.
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