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Statistics 12 Online
OpenStudy (anonymous):

all steps for hypothesis testing and calculations. A new herbal supplement claims to improve physical fitness. A sample of n=16 average score for the sample is M=39, the distribution of test scores is normal with mu = 35 and a standard deviation of =12 use a one-tailed test with an alpha level of =.05

OpenStudy (anonymous):

1. Establish your null and alternative hypotheses. The null hypothesis takes on the given average in this case, such that \(\mu_0\) (the mean) is \(35\). The alternative hypothesis claims that the calculated mean, \(M=39\) in this case, is greater than the mean under the null hypothesis. 2. Determine the rejection region. We're given a significance level of \(\alpha=0.05\) and a sample size of \(n=16\). The latter tells us that we have \(15\) degrees of freedom. The critical value \(t_{1-\alpha,n-1}\) is thus \(t_{0.95,15}\approx1.753\) (value taken from this table: ` http://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf `). This means if our computed \(t\) statistic satisfies \(t>1.753\), we reject the null hypothesis. 3. Compute the test statistic. \[t=\frac{M-\mu_0}{\sigma/\sqrt n}=\cdots\] 4. State the conclusion. Is \(t>1.753\) ?

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