# A random sample of five resulted in the following values: 18, 15, 12, 19 and 21.

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#### A random sample of five resulted in the following values: 18, 15, 12, 19 and 21. Using the 0.01 significance level, can we conclude the population mean $\mu$ is less than 20? State the null and alternative hypothesis. Compute the value of the test statistic. What is your decision regarding the null hypotheses. Estimate the p value.

asked Apr 1, 2013

## 1 Answer

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##### The hypothesis testing problem is $H0: \mu \geq 20$ against $H1:\mu < 20$. From the five data values $\overline{X} = 17$, $s=3.5355$. The test statistics is $t =\dfrac{\overline{X}-\mu_0}{s/\sqrt{n}} =\dfrac{17-20}{3.5355/\sqrt{5}}=-1.8974$. The critical value of $t$ for 4 degrees of freedom and at 0.01 level of significance is $t_{4, 0.01} = -3.7469$. The p-value is $P(t<-1.8974) = 0.0653$. As the p-value 0.0653 is not less than 0.05, we do not reject the null hypothesis. Thus the population mean is greater than or equal to 20. That is $\mu \geq 20$.
answered Apr 2, 2013 by (19,690 points)