Answer to Question #145207 in Statistics and Probability for SANGARI A/P KANNADASAN

Question

The score of driving test has a normal distribution with mean 70 if given the standard deviation of sample is eight. A driving school's instructor claimed that if the candidate learned more than three hours per week, the mean score would be different than 70. A driving test was given to a random sample of 50 candidates with the mean score was 78.

a) state the null and alternative hypothesis
b) identify the type I error and type II error that correspond to the hypothesis above
c) test the claim at 5% level of significance

Answer 1

$a)\ H_0: a=a_0=70, H_1: a\neq a_0=70\\ \overline{x}=78\\ N=50\\ s=8\\ c)\ \alpha=0.05\\ \text{We will use the following random variable:}\\ T=\frac{(\overline{X}-a_0)\sqrt{n}}{S}\\ T\text{ has t-distribution with } k=n-1\text{ degrees of freedom}.\\ \text{Observed value:}\\ t_{obs}=\frac{(78-70)\sqrt{50}}{8}\approx 7.07\\ \text{Critical value:}\\ t_{cr}=t_{cr}(\alpha;k)\approx 2.0096\\ \text{Critical region: }(-\infty,-2.0096)\cup (2.0096,\infty)\\ t_{obs}\text{ falls into the critical region. So we reject } H_0.\\ b)\text{ Type I Error}=\alpha=0.05.\\ \text{We will find Type II Error }\beta.\\ \frac{(\overline{X}-70)\sqrt{50}}{8}> -2.0096\\ \overline{X}> 67.73\\ \frac{(\overline{X}-70)\sqrt{50}}{8}< 2.0096\\ \overline{X}< 72.27\\ \beta=P\{67.73<\overline{X}< 72.27|a=78\}=\\ =P\{\frac{(67.73-78)\sqrt{50}}{8}<T< \frac{(72.27-78)\sqrt{50}}{8}\}=\\ =P\{-9.08<T< -5.06\}\approx 0\text{ (we use t-distribution table here)}.$