The target activation force of the buttons on a keyless entry clicker is 1.967 newtons. Variation exists in activation force due to the nature of the manufacturing process. A sample of 9 clickers showed a mean activation force of 1.88 newtons. The standard deviation is known to be 0.145 newton. Too much force makes the keys hard to click, while too little force means the keys might be clicked accidentally. Therefore, the manufacturer's quality control engineers use a two-tailed hypothesis test for samples taken from each production batch, to detect excessive deviations in either direction. At α = .05, does the sample indicate a significant deviation from the target?
Answer to relevant QuestionsFind the t calc test statistic for each hypothesis test. a. x-bar = 14.7, μ0 = 13.0, s = 1.8, n = 12 b. x-bar = 241, μ0 = 250, s = 12, n = 8 c. x-bar = 2,102, μ0 = 2,000, s = 242, n = 17 Estimate the p-value as a range using Appendix D (not Excel): a. t = 1.457, d.f. = 14, right-tailed test b. t = 2.601, d.f. = 8, two-tailed test c. t = 21.847, d.f. = 22, left-tailed test According to J.D. Power & Associates, the mean purchase price of a smartphone device (such as an iPhone or Blackberry) in 2008 was $216. In 2009, a random sample of 20 business managers who owned a smartphone device showed a ...Calculate the test statistic and p-value for each sample. a. H0: π ≤ .60 versus H1: π > .60, α = .05, x = 56, n = 80 b. H0: π = .60 versus H1: π ≠ .60, α = .05, x = 56, n = 40 c. H0: π ≥ .60 versus H1: π < .60, ...A coin was flipped 12 times and came up heads 10 times. (a) Would we be justified in assuming that the sample proportion p is normally distributed? Explain. (b) Calculate a p-value for the observed sample outcome, using the ...
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