A company is deciding whether to renew its ad buy with a local TV station. It wi

Question

A company is deciding whether to renew its ad buy with a local TV station. It will renew the ad if a 2-sided hypothesis test at the 10% level concludes that at least 17.25% of the local residents remember the ad. They decide to test this by contacting 500 randomly selected local residents. Ninety-nine (99) of the 500 remember the ad.

a. Conduct the requested hypothesis test based on this sample using both the exact method, then the normal approximation. (Use Minitab) Report your conclusions, stated in terms of the sampling scenario.

b. The station advocated that the decision be made based on a 1-sided hypothesis test at the 10% level using this sample. Why? (Hint: if you conduct the appropriated 1-sided test based on this sample using the normal approximation to the binomial, what will the p-value be? Confirm your answer with Minitab by conducting this 1-sided test.)

c. If you conduct the 1-sided test based on this sample using the exact method, will the p-value be exactly one-half of the 2-sided exact method p-value? Why or why not?

d. To check your answer from part c, conduct the 1-sided test based on this sample using the exact method. Is the p-value exactly one-half of the 2-sided exact method p-value?

e. Compute the power of the 2-sided hypothesis test at the 10% level. If the a level is reduced to 5%, will the power increase or decrease?

f. Compute the power of the 1-sided hypothesis test at the 10% level. Comparing this to the power from part f, which test appears to be more useful?

Explanation / Answer

I don't have minitab. I am helping you to solve in R. It's equivalent.

Exact Test :

If you write the following in r :

binom.test(99,500,0.1725, alternative = c("two.sided", "less", "greater"),
conf.level = 0.9)

then you will get the result as :

Exact binomial test

data: 99 and 500
number of successes = 99, number of trials = 500, p-value = 0.1387
alternative hypothesis: true probability of success is not equal to 0.1725
90 percent confidence interval:
(0.1690708 0.2296150)
sample estimates:
probability of success
0.198

As you can see that 0.1725 lies within the confidence interval and and p-value is greater than 0.1 , So we can conclude that the atleast 17.25% remember the ad.

Normal Test :

If you write the following in R
phat = 99/500 # sample proportion
p = 0.1725 # hypothesized value
n = 500 # sample size
sd = sqrt(p*(1-p)/n) # population standard deviation
z = (phatp)/sd
z # test statistic
pval = 2*(1-pnorm(z))
pval

Then you will get the output as

> phat = 99/500 # sample proportion
> p = 0.1725 # hypothesized value
> n = 500 # sample size
> sd = sqrt(p*(1-p)/n) # population standard deviation
> z = (phatp)/sd
> z # test statistic
[1] 1.509199
> pval = 2 * (1-pnorm(z))
> pval
[1] 0.1312478

So the p-value is greater than 0.1 we can conclude that actual percentage is 17.25%, after accepting our null.

(b) For (b) alternative is p < 17.25 , So it will be a left tailed test. So for the normal approximation code will be change as :

phat = 99/500 # sample proportion
p = 0.1725 # hypothesized value
n = 500 # sample size
sd = sqrt(p*(1-p)/n) # population standard deviation
z = (phatp)/sd
z # test statistic
pval =pnorm(z)
pval

the output is

> phat = 99/500 # sample proportion
> p = 0.1725 # hypothesized value
> n = 500 # sample size
> sd = sqrt(p*(1-p)/n) # population standard deviation
> z = (phatp)/sd
> z # test statistic
[1] 1.509199
> pval =pnorm(z)
> pval
[1] 0.9343761

P-value= 0.9343761 . Now the p-value is greater than 0.1 so we will again accept null to claim p=17.25%

(c) For exact binomial left tail test, the r-code will be :

binom.test(99,500,0.1725, alternative ="less",
conf.level = 0.9)

The output is :

Exact binomial test

data: 99 and 500
number of successes = 99, number of trials = 500, p-value = 0.9396
alternative hypothesis: true probability of success is less than 0.1725
95 percent confidence interval:
0.000000 0.229615
sample estimates:
probability of success
0.198

Note that 0.1725 is well within confidence interval and p-value = 0.9396 > 0.1, So we will accept null to claim p=17.25%. Note that p-value is not excatly half of the two -sided test because under null the parametr of the binomial distribution is 0.1725 not 0.5, so under null the distribution of the test statistics is not symmetric.

calculate the remanining of your own. I think you got the whole idea to do it

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.