Question

An experiment was conducted to compare the alcohol content of soy sauce on two different production lines. Production was monitored eight times a day. The data are shown below. Assume both populations are normal It is suspected that production line 1 is not producing as consistently as. production line 2 in terms of alcohol content. Test the hypothesis that - against the alternative that mg. Use a Pvalue. 0.47 039 042 054 0.38 0.48 052 0.53 Production line 1 Production line 2: 0.39 037 039 041 039 0.39 041 039 Cakulate the test statistic (Type an integer or a decimal. Round to two decimal places as needed.)

Answer 1

Solution: The data given in the question statement is as follows. Computing additional columns to determine the standard deviation: Line 1 Line 2 x2)2 (x2 x1 X2. 0.0000 0.47 0.39 0.0000 0.39 0.37 0.0058 0.0005 0.42 0.39 0.0021 0.0000 0.0054 0.0003 0.54 0.41 0.38 0.39 0.0074 0.0000 0.0002 0.48 0.39 0.0000 0.52 0.41 0.0029 0.0003 0.53 0.39 0.0041 0.0000 ΣΧ13.73 x2)20.0012 (x1-x20.0280 Σ(χ Σχ3.14 = n1 = 8, n2 = 8 Σχ1 X1 3.73 0.4663 Σχ2 X = n2 3.14 0.3925 8 Σ( 1- 1)2 0.0280 = 0.0040 8 1 n1-1 Σ(x- Χ2)? 0.0012 0.0002 8 1 n21 Ho o1 Null Hypothesis: Alternative Hypothesis: Ha: 0102 = O2 Degrees of freedom for numerator df, 8 - 1 7 Degrees of freedom for numerator df2 = 8-1 7

Let us assume, level of significance a = 0.05 Test statistic 0.0040 F = = 24.337 0.0002 P-value: From F-distribution table; area under the curve to the right of (F 24.337) and for degrees of freedom (7, 7), is given by: 0.0002 The test is a two tailed test. ce, p-value 0.0002 x 2 = 0.0004 Conclusion: Since p-value <a, we reject the null hypothesis There is sufficient evidence to conclude that production 1 line is not producing as consistently as production line 2 in terms of alcohol content.