Answer:
To inspect a batch consisting of 20 semiconductor chips, a sample of 3 is selected. Out of these, 10 chips fail to meet customer specifications.
a) Total distinct samples possible = 20C3 = =1140
b) For exactly 2 good chips and 1 bad chip
Total samples = 10C2 * 10C1 = 45 * 10 =450
c) Combinations of 2 good 1 bad, 1 good 2 bad, and 3 bad chips
Total samples = 10C2 * 10C1 + 10C1 * 10C2 + 10C3
=
Let p denote the proportion of adults in the town who have encountered this flu strain.
According to the provided information
∵ this is a two-tailed test.
Test statistic:
, where p= denotes the population proportion
= signifies the sample proportion
n= represents the sample size
Setting n= 6 and and p=0.08
P-value for the two-tailed test:[2P(Z>|z|)
=2P(Z>|-0.415|)
=2P(Z>0.415) = 2[1-P(Z≤0.415)] [∵ P(Z>z)=1-P(Z≤z)]=2(1-0.6609) [from the z-table]
=0.6782Decision: Because the p-value(0.6782) exceeds the significance level of 0.01, we do not reject the null hypothesis.
This leads us to conclude that there is insufficient evidence to back the assertion that the percentage of all adults in this town exposed to this flu strain deviates from the national average of 8%.