The following values are based on the random sample \$15, \$13.41, \$17, \$24, \$15, \$23, \$12, \$18, \$16, \$34, \$14, \$23, \$18, \$16, \$10 1)1)Determine Ho and Ha 2_ determine randon variable 3)determine distributon for test 4) draw the graph 5) compare the preconceived alpha with the p-value make a decision, reject or not 6) write a claer conclusion using sentence

The sample statistics give a mean of \$17.894 and a standard deviation of \$6.011. So (2) x bar=\$17.894 and s=\$6.011. Sample size n=15. (3) Normal distribution is assumed since median (\$16) is fairly close the mean (\$17.894). The outlier \$34 has skewed the distribution a little so that the mean>median.

(1) H0: µ=\$18

Ha: µ≠\$18

Significance level ɑ=0.01 (2-tail), so ɑ/2=0.005.

(4) The box plot shows that the distribution is positively skewed because of the outlier \$34. This outlier has shifted the mean to the right of the median.

Test statistic T=(17.894-18)/(6.011/√15)=-0.0683. (5) This corresponds to a probability p-value of 0.473, only a little below the mean and much greater than the significance level of 0.005.

(6) We have insufficient evidence to refute the claim that the average public offering is \$18 per unit.

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