Statistics Probability Question. (word problem) I need help !!?

Question

A direct-mail advertiser has a continuing problem with mailing lists. Historically, about 18% of the addresses on maililing list have been bad. The advertiser has the opportunity to buy a new list, which laims to have a lower percentage of bad addresses. As a test before purchasing the new list, the advertiser takes a random sample of 200 addresses and sends catalogs by mail. After sufficient time has passes, the advertiser finds that 32 of the 200 addresses were bad.

P(32 of fewer bad addresses)= .2635

Q) Someonew commments that 32 bad addresses is certainly fewer than the expected value of 36 bad addresses, based on the historical rate of 18%, and recommends that the advertiser purchase the new list. How would you respond??

Answer 1

I would conduct the following hypothesis test and show that the there is no statistical evidence to suggest the proportion is less than 18%





Hypothesis Test for proportions:



Let X be the number of success in n independent and identically distributed Bernoulli trials, i.e., X ~ Binomial(n, p)



To test the null hypothesis of the form

H0: p = p0, or

H0: p ≥ p0, or

H0: p ≤ p0



Assuming that n*p0 > 10 and n * (1-p0) > 10 (some will say the necessary condition here is > 5, I prefer this more conservative assumption so that the approximations in the tail of the distribution are more accurate) then



find the test statistic z = (pHat - p0) / sqrt(p0 * (1-p0) / n)



where pHat = X / n



The p-value of the test is the area under the normal curve that is in agreement with the alternate hypothesis.

H1: p ≠ p0; p-value is the area in the tails greater than |z|

H1: p < p0; p-value is the area to the left of z

H1: p > p0; p-value is the area to the right of z



If the p-value is less than or equal to the significance level α, i.e., p-value ≤ α, then we reject the null hypothesis and conclude the alternate hypothesis is true. If the p-value is greater than the significance level, i.e., p-value > α, then we fail to reject the null hypothesis and conclude that the null is plausible. Note that we can conclude the alternate is true, but we cannot conclude the null is true, only that it is plausible.



The hypothesis test in this question is:



H0: p ≥ 0.18 vs. H1: p < 0.18



The test statistic is:

z = ( 0.16 - 0.18 ) / ( √ ( 0.18 * (1 - 0.18 ) / 200 )

z = -0.7362102



The p-value = P( Z < z )

= P( Z < -0.7362102 )

= 0.2308014



Since the p-value is greater than the significance level of 0.05 we fail to reject the null hypothesis and conclude p ≥ 0.18 is plausible.

Answer 2

I would respond that the data doesn't provide sufficient evidence to support that conclusion. There is a 26.35% chance that the new list was better simply to natural sampling variation. You want that percentage below 5% or at least 10%.

Answer 3

There are 40C6 ways to remove 6 marbles. There are 28C6 ways to choose 6 whites. 1 - 28C6 / 40C6 = approx .90 is prob not all 6 white