In: Statistics and Probability
6. A pharmaceutical company makes a pill intended to reduce the chances of seizures
among children. The pill is supposed to contain 20.0 mg of Phenobarbital. A simple
random sample of 20 pills yielded the amounts listed here. Does this sample provide
evidence that the mean amount of Phenobarbital is different than 20.0 mg? Use a 1%
significance level.
27.5 26.0 22.9 23.4 23.0 23.9 32.6 20.9 22.9 24.3
24.8 16.1 24.3 17.3 18.9 20.7 33.0 15.6 24.3 23.3
Solution:
x | x2 |
27.5 | 756.25 |
26 | 676 |
22.9 | 524.41 |
23.4 | 547.56 |
23 | 529 |
23.9 | 571.21 |
32.6 | 1062.76 |
20.9 | 436.81 |
22.9 | 524.41 |
24.3 | 590.49 |
24.8 | 615.04 |
16.1 | 259.21 |
24.3 | 590.49 |
17.3 | 299.29 |
18.9 | 357.21 |
20.7 | 428.49 |
33 | 1089 |
15.6 | 243.36 |
24.3 | 590.49 |
23.3 | 542.89 |
∑x=465.7 | ∑x2=11234.37 |
Mean ˉx=∑xn
=27.5+26+22.9+23.4+23+23.9+32.6+20.9+22.9+24.3+24.8+16.1+24.3+17.3+18.9+20.7+33+15.6+24.3+23.3/20
=465.7/20
=23.285
Sample Standard deviation S=√∑x2-(∑x)2nn-1
=√11234.37-(465.7)220/19
=√11234.37-10843.8245/19
=√390.5455/19
=√20.555
=4.5338
This is the two tailed test .
The null and alternative hypothesis is ,
H0 : = 20
Ha : 20
Test statistic = t
= ( - ) / S / n
= (23.28-20) /4.53 / 20
= 3.238
Test statistic = t = 3.238
P-value =0.0043
= 0.01
P-value <
0.0043 < 0.01
Reject the null hypothesis .
There is sufficient evidence to suggest that