Question

QUESTIONS: ( 1 )(a)A government agency receives many consumer complaints that the boxes of detergents sold by a company contain less than the 20 oz. of detergents advertised. To check these consumers complaints, the agency purchases 49 boxes of the detergent and finds that the mean is 18 oz. with a standard deviation of 3 oz Conduct an appropriate statistical test of significance at the 5% level to determine if the firm could be held culpable of false advertisement. (b)Would the results in part (a) be different if the government agency purchased 36 boxes and found the mean to be 19 oz with a standard deviation of 2 oz.? (2)A lightbulb producing firm wants to determine if it can claim that the lightbulbs it produces last 1000 burning hours. Officials of the Quality Control Department of the firm take a sample of 100 lightbulbs from the firm's production line and find that the sample mean and standard deviation are 980 hours and 80 hours respectively. Show how the officials should go about the process of conducting this test at the 5% level of significance. (3)A purchaser of electronic components wants to test the hypothesis that they last less than 100 hours. To do this she takes a random sample of 36 such components and finds that, on average, they last 96 hours with a standard deviation of 8 hours. If the purchaser knows that the life time of the components is normally distributed, should she accept the hypothesis that they last less than 100 hours at the 1% level of significance? (4)A producer of steel cables wants to test if the steel cables it produces have a breaking strength of 5000 lb. A breaking strength of less than 5000 lb. would not be adequate and cables with strength of more than 5000 lb. would unnecessarily increase production costs. The producer takes a sample of 64 cables and finds the average breaking strength is 5100 lb. with a standard deviation of 480 lb. Should the producer accept the hypothesis that its steel cable has a breaking strength of 5000 lb. at the 1% level of significance? NOTE: We did # (4) in class but we did the test at the 5% level of significance! So try it at the 1% level

Answer 1

a) b) 20 x bar18 x bar S. n 49 n. 36 0.05 0.05 H0 : μ220 Ha:μ< 20 Ha : μ < 20 0.05 0.05 This is one-tail (left) test This is one-tail (left) test ( we used s as 49) is unknown) (we used s as ơ is unknown) sx = 0.4286 ← ( 3 sx =0.3333 ←(21136) a0.05 ZC1.645 0.05 ZC1.645 (x bar-u) / s: ( 18-20 ) / 0.4286 =-4.67 (x bar-u) / sx = ( 18-20 ) / 0.4286 - 3.00 Probability for z-score: 0.0000 (Using normal table) Probability for z-score: 0,0013 (Using normal table) p-value 0.0000 < 0.05 p-value 0.0013 0.05 Ho can be rejected Ho can be rejected There is sufficient evidence to conclude that the first could be held culpable of false advertisement. There is sufficient evidence to conclude that the first could be held culpable of false advertisement We got the same result in part (a) also