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1. Assume that a simple random sample has been selected from a normally distributed population. Find the test statistic, a range of numbers for the P-value, critical value and state the final conclusion. Claim:- The mean bottom temperature of a fish breeding pond is less than 20 degree celcius. Sample Data:- n=56, x-bar =19.3, s=1.6. The significance level is alpha = 0.05 Calculate the test statistic, p-value using t-table, critical value using t-table and conclusion. Ans. One-Sample T Test of H0: Mean N 56 Mean 19.3000 = 20 vs H1 Mean StDev 1.6000 SE Mean 0.2138 < 20 95% Upper Bound T 19.6577 -3.27 P 0.001 Test Statistic = -3.27 P value = .001 Critical value = -1.673 Conclusion :reject the null hypothesis 2. A random sample of 56 measured the accuracy of their wristwatches, with positive errors representing watches that are ahead of the correct time and negative errors representing watches that are behin the correct time. The 56 values have a mean of 94.7 sec and a standard deviation of 152.6 sec. Use a 0.01 significance level to test the claim that the population of all watches has a mean equal to 0.00 sec What is the correct conclusion? 1. There is not sufficient evidence to warrant rejection of the claim that the polpulation mean is 0.00 sec. 2. There is sufficient evidence. Ans One-Sample Z test Test of Mean = 0 vs Mean > 0 The assumed standard deviation = 152.6 99% Lower N Mean SE Mean Bound Z P 56 94.7000 20.3920 47.2610 4.64 0.000 Conclusion : Since p value is less than .01 reject the null hypothesis Thus there is sufficient evidence. 3. The birth weights are recorded for a sample of male babies born to mothers taking a special vitamin supplement. When testing the claim that the mean birth weights for all male babies of mothers given vitamins is equal to 3.39 kg, which is the mean for the population of all males. The significance level is alpha =0.05 N= 17.0000, sx = 0.6300, x bar = 3.3100 Ans One-Sample T test Test of Mean = 3.39 vs Mean < 3.39 N 17 Mean 3.31000 StDev SE Mean 0.63000 0.15280 95% Upper Bound 3.57677 T -0.52 P 0.304 Since p value is grater than 0.05 , we accept the null hypothesis 4. Find the test statistic, find the critical values of chi-square and limits containing the p-value; then determine whether there is sufficient evidence to support the given alternative hypothesis H1: 7, 0.02, n 16, s 12 Determine the test statistic, use chi-square table to determine the critical values and final conclusion. Ans (n 1) s 2 17.6326 Test Statistics = 2 Critical Value = 15.03 Since calculated value is greater than critical value reject the null hypothesis 5. Use the traditional method with alpha =0.05 to test the claim that a new method of laser cutting reduces the variation in lengths of 2x4 lumber. The standard deviation of the lengths of the sawed 2x4s is 0.33 inches. A simple random sample of 91 lasercut 2x4s, taken from a normally distributed population, was found to have a standard deviation of 0.3 inches. Calculate the test statistic and the critical value to determin the correct conclusion. Ans Test Statistics = (n 1) s 2 2 74.38 Critical Value = 113.1 d.f =90 Conclusion: Since calculated value is less than critical value accept the null hypothesis

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