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Unformatted text preview: d output is approximately linear, therefore we can reasonably presume that the batches of product are normally distributed. cores are approximately linear, therefore we can reasonably presume that the final exam scores of two groups are normally distributed. Chp 8 CI for one population mean u when sigma known: Margin error: Sample size for estimating n: CI for one population mean u when sigma unknown: with df=n1 Chp9: ztest for H0: u=u0 (ND or large, sigma known): ttest for H0: u=u0 (ND or large, sigma unknown): with df=n1 one sample ttest procedure when sigma known: 1) The null and alternative hypotheses are H : μ = 98.6 (mean body temperature of healthy humans equals 98.6) H a : ≠ μ 98.6 (mean body temperature of healthy humans different from 98.6) Where μ denote the mean body temperature of all healthy humans, and the hypothesis test is two tailed. 2) We are to perform the hypothesis test at the 1% significance level, or =0.01. 3) The test statistic, z= / x μ0σ n , we have μ0 =98.6, = 0.63, n=93, and x =9125.5/93= 98.12 Thus the value of the test statistic is z= / x μ0σ n =7.29 4) The test is two tailed, so the Pvalue is the probability of observing a value of z from 7.29 to 7.29 if the null hypothesis is true. Hence P=0.0000. 5) Because P<, reject H 0. The test results are (not) statistically significant at the 1% level. 6) At the 1% significance level, the data provide sufficient evidence to conclude that the mean body temperature of healthy humans differs from 98.6. The evidence against the null hypothesis is very strong. Type 2 error probabilities: bata=P(Type 2 error)= 1power one sample ttest procedure when sigma unknown: 1) The null and alternative hypotheses are H : μ = 98.6 (mean body temperature of healthy humans equals 98.6) H a : ≠ μ 98.6 (mean body temperature of healthy humans different from 98.6) Where μ denote the mean body temperature of all healthy humans, and the hypothesis test is two tailed. 2) We are to perform the hypothesis test at the 1% significance level, or =0.01. 3) The test statistic, t= / x μ0s n , we have μ0 =98.6, n=93, s=0.6468, and x =9125.5/93= 98.12 Thus the value of the test statistic is t= / x μ0s n =7.1022 4) The test is two tailed, so the Pvalue is the probability of observing a value of t from 7.1022 to 7.1022 if the null hypothesis is true. Hence P=0.0000. 5) Because P<, so reject H 0. The test results are (not) statistically significant at the 1% level. 6) At the 1% significance level, the data provide sufficient evidence to conclude that the mean body temperature of healthy humans differs from 98.6. The evidence against the null hypothesis is very strong....
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 Spring '09
 KIN
 Statistics, Null hypothesis, Statistical hypothesis testing, Statistical significance, mean body temperature

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