Test for Population proportions (large sample size).

General Statistics

Statistical Inference
Test for Population Proportion (large sample size)

Test for Population proportions (large sample size).

Statistics involving population proportion often have sample size that is large (), therefore the normal approximation distribution and associated statistics is used to determine a test for whether the sample proportion = population proportion.

That is, when the sample size is greater than or equal to 30 we can use the z-score statistics to compare the sample proportion against the population proportion using value of the sample standard deviation, to estimate the sample standard deviation, if it is not known.

The sample distribution of P (proportion) is approximately normal with a mean or expected value, E(P) = and standard error .

There are three questions one may ask when comparing two proportions:

Question 1:: Is ? H_a (Two-tailed test)

Question 2: : Is ? H_a (Right-tailed test)

Question 3: : Is ? H_a (Left-tailed test)

1. Know the statistics used to test for large sample size.

The test statistics is related to the standard normal distribution:

The z-score statistics for proportions
, where p = proportion estimate, q=1-p and is the population proportion.
Proportion mean is:

Standard deviation:

The confidence interval, for p at a significance level of

Decision rules:

Upper-Tailed Test ():

Accept H₀ if

Reject H₀ if

Lower-Tailed Test ():

Accept H₀ if

Reject H₀ if

Two-Tailed Test ():

Accept H₀ if

Reject H₀ if

2. Know how to use appropriate statistics to test if a sample proportion is equal to the population proportion (large sample size).

3 Types of tests in comparing sample mean to population mean:

When comparing the sample mean, to the population mean ( is known) there are 3 question to considered:

Question 1: : Is ? H_a (Two-tailed test)

Question 2: : Is ? H_a (Right-tailed test)

Question 3: : Is ? H_a (Left-tailed test)

Question 1: : Is ? H_a (Two-tailed test)

Is H_a
Is the new mean different from the population proportion?
Here we use a two-tailed test by computing the confidence interval for the test at the level of significance .
If the test statistics for z falls outside this interval we decide that the means differs, we chose the alternate hypothesis, H_a
Otherwise we have no reason to think that they differ, H₀

By Examples:

Problem 1. A sample of 1000 customers this year resulted in 791 responding favorably to a certain proposal, if the original last year's customer's favorable response to the same proposal was 90% of those responding or p = 0.90, is there a change in customer response?

This problem could be stated with either sample p exactly equal population p or sample p < population p, we will use the former.

Given , n = 1000 (large so can use normal approximation of z-score), , and

Step 1 - Hypothesis: The claim that or 0.878 = 0.90, the null hypothesis.

The alternate hypothesis is that

H₀ :

H_a or

Step 2. Select level of significance: This is given as (1 - 0.10)

So for two-tailed test:

Step 3. Test statistics and observed value.

Step 4. Determine the critical region (favors H_a)

For alpha = 0.05 at both ends of intervals: 0.05 and 0.95, z = -1.65 and z = 1.65

The critical region is and

Step 5. Make decision.

Accept the null hypothesis if or

The observed z = -11.48, and since -11.48 < -1.65 and in the critical region, we reject H₀ in favor of H_a.

So the this year's proportion of 0.791 differs from the last year's proportion of 0.90.

Question 2: : Is ? H_a (Right-tailed test)

Is H_a
Is the new mean greater than the population proportion. Here we use a right-tailed test by computing the positive z-score for the test at the level of .
And if the test statistics for z falls within the critical region (red) we decide in favor of the alternate hypothesis, H₀
Otherwise we have no reason to think that they differ, H_a

By Examples:

Problem 2. A graduate school is testing the null hypothesis that greater than ½ (p = 0.50) of all MBA's continue their formal education by taking courses within 10 years of graduation. Using a sample of 200 persons, 111 had taken course work since receiving their MBA. At the alpha = 0.05 significance level. Should the graduate school accept of reject the null hypothesis?

Given , n = 200 (large so can use normal approximation of z-score), , and

Step 1 - Hypothesis: The claim that or 0.555 = 0.50, the null hypothesis.

The alternate hypothesis is that

H₀ :

H_a or

Step 2. Select level of significance: This is given as (5%)

Step 3. Test statistics and observed value.

Step 4. Determine the critical region (favors H_a)

For alpha = 0.05 at the upper end of the acceptable region, z_0.95 = 1.65

From reference table (search for z with Pr[z=? ] = 0.95.

The critical region is

Step 5. Make decision.

Accept the null hypothesis is

The observed z = 1.56, and since 1.56 < 1.65 then it is not in the critical region, so there is no reason to reject H₀ in favor of H_a.

So the about 50% of all MBA's continue their education after graduate after graduation..

Question 3: : Is ? H_a (Left-tailed test)

Is H_a
Is the new mean less than from the population proportion. Here we use a left-tailed test by computing the negative z-score for the test at the level of .
And if the test statistics for z falls within the critical region (red) we decide in favor of the alternate hypothesis, H₀
Otherwise we have no reason to think that they differ, H_a

By Examples:

Problem 3. A professional group claims that at least 40% (p=0.40) of all engineers employed by computer companies switch jobs within three years of being hired. The alternate hypothesis is that the rate of job changing is below 40%. At a significance level of 0.01, should the claim be accepted or rejected if the sample size results show that 25 out of n = 1000 engineers changed jobs?

Given , n = 100 (large so can use normal approximation of z-score),