Question

Suppose that the mean PCB concentration per fish is 11.2 parts per million with a standard
deviation of 2 parts per million in a river. Assume the PCB concentration per fish have a normal distribution. A new random sample of 10 fish has been tested. The PCB concentrations are listed in the following. 11.5, 12.2, 11.4, 11.8, 10.8, 10.9, 11.9, 12.0, 12.4, 11.6 Use the 5 percent level of significance, test that the mean PCB concentration in the fish has changed at 11.2 parts per million.

Answer 1

Solution:

x	x2
11.5	132.25
12.2	148.84
11.4	129.96
11.8	139.24
10.8	116.64
10.9	118.81
11.9	141.61
12	144
12.4	153.76
11.6	134.56
∑x=116.5	∑x2=1359.67

Mean ˉx=∑xn

=11.5+12.2+11.4+11.8+10.8+10.9+11.9+12+12.4+11.6/10

=116.5/10

=11.65


Sample Standard deviation S=√∑x2-(∑x)2nn-1

=√1359.67-(116.5)210/9

=√1359.67-1357.225/9

=√2.445/9

=√0.2717

=0.5212

This is the two tailed test .

The null and alternative hypothesis is ,

H0 :    = 11.2

Ha :     11.2

Test statistic = t

= ( - ) / S / n

= (11.65- 11.2) / 0.52 / 10

= 2.736

Test statistic = t =2.736  

P-value =0.0230

= 0.05  

P-value <

0.0230 < 0.05

Reject the null hypothesis .

There is sufficient evidence to suggest that