Problem 3.docx - A producer of steel cables wants to know whether the steel cables it produces have an average breaking strength of 5000 pounds An

A producer of steel cables wants to know whether the steel cables it produces have an average breaking strength of 5000 pounds. An average breaking strength of 5000 pounds would be inadequate, and to produce steel cables with an average breaking strength of more than 5000 pounds would unnecessarily increase production costs. The producer collects a random sample of 64 steel cable pieces. The breaking strength for each is recorded. (a) Identify the null and alternative hypothesis for this situation. Solution: By testing the hypothesis average testing strength is more than 5000. The null and alternative hypothesis is given below, H 0 : µ = 5000 H a : µ ≠ 5000 (b) Using a 5% significance level, what statistical conclusion can the producer reach regarding the average breaking strength of its steel cables? Would the conclusion be any

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Unformatted text preview: different at the 1% level? Explain the answer. Solution: Mean 5158.326719 Standard Error 62.28072982 Median 5179.755 Mode #N/A Standard Deviation 498.2458386 Sample Variance 248248.9157 Kurtosis-0.287598793 Skewness 0.104559271 Range 2424.46 Minimum 3931.76 Maximum 6356.22 Sum 330132.91 Count 64 N 64 8 X 5158.326719 158.327 s 498.2458386 62.2807 t-value 2.54215 p-value 0.01349 After complete analysis we have found that p value is less than 5% it means that we will reject the null hypothesis because average breaking strength of steel is not equal to 5000. The p value is 0.013 which is greater than 0.01 so we have enough evidence to reject null hypothesis at 1%....
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