The following sample of six measurements was randomly selected from a normally distributed population (unknown variance): 2 5 -2 7 2 4 a. Test the null hypothesis that the mean of the population ? ? 2 against the alternative hypothesis ? < 2. Use alpha=0.05. b. Test the null hypothesis that the mean of the population is 2 against the alternative hypothesis ??2. Use alpha=0.05. c. Construct the 95% confidence interval for mean.
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Test the null hypothesis that the mean of the population μ = 2 against the alternative hypothesis μ > 2, using alpha-0.05. We can use a one-tailed t-test for this hypothesis. The formula for the t-statistic is: t = (x̄ - μ) / (s / √n) where x̄ is the sample Show more…
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Sample size is 64, sample mean is 50. Assuming the population standard deviation is σ=16, construct 95% confidence interval for population mean. For z factor use zα/2 = 2
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We have provided a sample standard deviation and sample size. In each case, use the one-standard-deviation $x^{2}$ -test and the one-standard-deviation $x^{2}$ -interval procedure to conduct the required hypothesis test and obtain the specified confidence interval. $s=6$ and $n=26$ a. $H_{0}: \sigma=5, H_{a}: \sigma>5, \alpha=0.01$ b. $98 \%$ confidence interval
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