10.59
A sample random sample of 300 items is selected from a large shipment,
and testing reveals that 4% of the sampled items are defective. The
supplier claims that no more than 2% of the items in the shipment are
defective. Carry out an appropriate hypothesis test and comment on the
credibility of the supplier’s claim. (Hypothesis Testing)

Hi, I need clear step by step work answers. However, not long
explanations or professional answers. Thank you very much.

Hi wta2k!
Once again, it's me answering your questions. I've already worked on
your questions #185196 and #185198, so please refer to those first,
since I will draw on some concepts I outlined in those answers.

As in the previous exercises, what you want to find here is the
p-value of the test; in this case, what is the probability of
observing 12 defective items (that's 4% of 300) or more if the claim
that no more than 2% of the items is defective is true. Thus, first we
define the null and alternative hypothesis:

Call 'p' the proportion of defective items
Null hypothesis: p<=0.02
Alternative hypothesis: p>0.02

Second, note that the number of defective items follows a binomial
distribution. Check the following page, which describes what a
binomial experiment looks like to see why is it so.

http://stat.tamu.edu/stat30x/notes/node66.html

Now that you know the distribution, it's easy to calculate the
probability you're looking for. Define X to be "number of defective
items observed". You're looking for

Prob( X >= 12 )

given p=0.02 and n=300 (n is number of observations). Now, it's
difficult to calculate this probability using the binomial
distribution. Fortunately, for a large n, we can approximate the
binomial by using a normal distribution, with mean equal to n*p and
standard deviation (SD) equal to sqrt( n*p*(1-p) )(recall that sqrt
means square root). Thus, we can say that X has approximately a normal
distribution with
Mean = 300*0.02 = 6
SD = sqrt(300*0.02*0.98) = 2.42

And now, we calculate the probability just like in the other
questions. We have to substract the mean and divide by the SD to have
a *standard* normal distribution in order to use the table.

Prob( X >= 12 )
=Prob( X-6 >= 12-6 )
=Prob( (X-6)/2.42 >= (12-6)/2.42 )
=Prob( (X-6)/2.42 >= (12-6)/2.42 )
=Prob( z >= 2.47 )

(z is (X-6)/2.42, i.e. a random variable with standard normal
distribution)
In order to find that probability, again, we use the table. You may
use the link I provided in question #185198. The value of this
probability is 0.006, and is also called the p-value of this test.

Based on the p-value, we could conclude that the supplier's claim is
not credible. If the true proportion of defective items was 2%, we
would observe 12 defective items in 300 with a probability of only
0.006. Thus we reject the null hypothesis. Even if we set the level of
significance as low as 0.01 (rejecting only p-values that are smaller
than 0.01), we still reject the null hypothesis.

I hope this explanation was clear to you. As in the previous
questions, please don't hesitate to request further clarification
before rating my answer.


Best wishes,
elmarto

Thank you for your quick help!!