Example 4.1: Feeling Good (cont.)

The 2004 nationwide Harris poll surveyed 1016 adults by telephone between October 14 and 17. One question that had changed appreciably from previous years was whether Americans feel good about the morals and values of Americans in general, with 55 percent saying they did.

(a) Carry out a binomial test and confidence interval to determine whether this is convincing evidence that a majority of American adults feel good about the morals and values of Americans in general. Why is it appropriate to use the binomial distribution with this study?

(b) Is the sample size large enough for the normal approximation to the binomial to be valid?

(c) If so, what results would you obtain using the normal probability model? How closely do they match the binomial results?

Analysis:

(a) Binomial analysis:

Let p = proportion of adults nationwide that say they feel good about the morals and values of Americans in general

H₀: p = .50 (no more than half of adults nationwide feel good about this issue)

H_a: p > .50 (a majority of adults nationwide feel good about this issue)

It is reasonable to use the binomial model since we have a fixed number of trials (1016), a random sample, two possible outcomes (agree or not agree), and a fixed probability of success (p) We will treat the observations as independent since the population of adult Americans is much larger than the sample size.

Minitab output:

Test and CI for One Proportion

Test of p = 0.5 vs p > 0.5

95%

Lower Exact

Sample X N Sample p Bound P-Value

1 559 1016 0.550197 0.523950 0.001

As well as determining a 95% confidence interval for p:

Exact

Sample X N Sample p 95% CI P-Value

1 559 1016 0.550197 (0.519007, 0.581095) 0.002

These results indicate that .5 is not a plausible value for p (one-sided p-value » .001) and that we are 95% confident that between 51.9% and 58.1% of adults nationwide feel good about the morals and values of Americans in 2004.

(b) To decide whether the sample size is large enough to use a normal (z) procedure, we want to check whether np> 10 and n(1-p) >10. Since we don’t know the actual value of p, when we conduct the test of significance, we use the conjectured value, .50.

1016(.5) = 508 > 10 Ö and 1016(1-.5) = 508 > 10 Ö

These conditions are easily met. In fact, we could simulate 1000 observations of a binomial random variable with parameters n=1,016 and p = .5, and we see the N(.5, .016) model fits the distribution of the sample proportions quite well:

(c) So it is valid to compute the test statistic and p-value using the normal distribution:

z = =3.20

p-value = P(Z > 3.20) = 1-P(Z<3.20) = 1-.9993 = .0007 » .001

Cumulative Distribution Function

Normal with mean = 0 and standard deviation = 1

x P( X <= x )

3.2 0.999313

This provides us with a very similar result as the exact Binomial p-value (.001) but with the added benefit of reporting a test statistic (z=3.20) as another way to measure how far (how many standard deviations) the observed sample proportion result is from the hypothesized value of the population proportion.

To decide if it is valid to calculate a confidence interval using the normal probability model as well, we would use the observed value of to check the technical conditions:

1016(.55) = 558.8 > 10 Ö and 1016(.45) = 457.2 > 10 Ö

Again, these are easily met and so the normal approximation is again valid. An approximate 95% confidence interval for p would be:

.55 + 1.96 = .55 + .03

We are (approximately) 95% confident that between 52% and 58% of adults nationwide feel good about their overall quality of life, very similar to the result from the binomial procedure.

The Minitab output corresponding to these normal procedures is below:

Test and CI for One Proportion

Test of p = 0.5 vs p > 0.5

95%

Lower

Sample X N Sample p Bound Z-Value P-Value

1 559 1016 0.550197 0.524525 3.20 0.001

Sample X N Sample p 95% CI Z-Value P-Value

1 559 1016 0.550197 (0.519607, 0.580786) 3.20 0.001