Normal Distribution: Hypothesis Testing

[Draggu]

Homework Statement

A cola company's long-term market share is estimated to be 45%. After an extensive advertising campaign, 68 out of 150 customers surveyed claimed to have purchased this company's cola recently. Was the advertising campaign a success? justify your assessment

Homework Equations

Let m= mean
and o=standard deviation
p=probability of success
q=probability of failure
a=alpha, significance level

The Attempt at a Solution

H0: m=0.45
H1: m<0.45
n=150
p=0.45
q=0.55

m=np
=67.5

o=sqrt npq
=6.09

x=68
a=0.01

S=o/sqrt(n)
=6.09/sqrt(150)

Z-sample:
(68-67.5)/(6.09/sqrt(150))
=P(Z<1.01)
=0.8438

I'm lost here. If this is right so far, where do I go next? I really don't even understand how I got this, basically just doing the same pattern that a note does. It's very vague though.

 

[Mark44]

I believe your alternate hypothesis should be H₁: m >= .45

This means you would be using a one-tailed test. Your value of alpha is .01, so the critical point is the z-value for which P(Z < z_c) = .99. You need to look in a table to find this number.

Your test statistic is z, which I get to be approx. 1.005. If z > z_c, you reject the null hypothesis. If z < z_c, you accept the null hypothesis. (I'm ignoring the unlikely possibility that z will be equal to z_c.)

 

[Draggu]

So my zc = 2.33
z=1.01

Zc is bigger than c so accept null hypothesis? Advertising campaign was a success?

 

[Mark44]

Yes, accept the null hypothesis. But that means the ad campaign was NOT a success, at least the way I am interpreting this problem. If the ad campaign really were a success, the market share should have been significantly larger than the known market share.

 

[Draggu]

Yes, accept the null hypothesis. But that means the ad campaign was NOT a success, at least the way I am interpreting this problem. If the ad campaign really were a success, the market share should have been significantly larger than the known market share.

So since we accept the null hypothesis (which equals .45) it wasn't really a success because they broke even, right? (68/150 = .4533~)

 

[Mark44]

As I interpret the problem, this isn't about "breaking even." The question is did the ad campaign increase the company's market share significantly, or not. The company ran the ad campaign in the hopes that their market share would be significantly larger than their long-term market share, 45%.

 

[Draggu]

Ok, well, I have a question. Why is my Zc = 2.33 and not -2.33? The main problem I am having is finding if Zs > Zc or so in hypothesis testing in other questions.

 

[Mark44]

Because your alternate hypothesis, H₁, is m > .45 versus the null hypothesis being m <= .45. The question, again, is did the ad campaign increase market share (H₁) above the long-term average of 45% or did it not (H₀). Because we're testing the hypothesis m > .45, you need to use a one-tailed test. Since alpha is .01, you are going to accept the alternate hypothesis if your test statistic falls in the region for which there is only .01 probability (the tail of the standard normal curve way off to the right), and will accept the null hypothesis (and thus reject the alternate hypothesis) if the test statistic falls in the region for which there is .99 probability (the left half of the standard normal curve plus most of the right half). You found that the z-value for which the probability (or area) to the left of it is .99 is 2.33, and that's your z_c.

Since the value of z that you found was z = 1.01, this value is less than your critical value z_c, the conclusion is to reject the alternate hypothesis, and conclude that the ad campaign did not increase market share in a significant way.

It's harder to explain this without being able to draw a graph of the standard normal (the "bell curve") showing the probability/area corresponding to the null hypothesis and to the alternate hypothesis. I can't draw one here in my response, but you certainly should do so.

 

[statdad]

You never "accept the null hypothesis" - you fail to reject it.