Question

11. A pharmaceutical company makes pills that should contain 20.0 mg of the drug Phenobarbital. A random sample of 25 pills yielded a mean of 20.5 mg and standard deviation of 1.5 mg. At significance level001 test the claim that the pills are acceptable (H 20.0 mg). 603

Answer 1

Q.11) Given that, sample size ( n ) = 20

sample mean $(\bar x)$ = 20.5 mg

sample standard deviation ( s ) = 1.5 mg

significance level $(\alpha)$ = 0.03

The null and the alternative hypotheses are,

$H_0: \mu = 20.0$

$H_a: \mu \neq 20.0$

Test statistic is,

$t = \frac {\bar x - \mu}{\frac { s }{\sqrt { n }}} = \frac { 20.5 - 20}{\frac { 1.5}{\sqrt { 20}}}= 1.491$

Degrees of freedom = n - 1 = 20 - 1= 19

p-value = 0.1524

Here, p-value = 0.1524 > $\alpha = 0.03$

So, we fail to reject the null hypothesis ( H₀).