AP Statistics12.1 Inference for A Population

Proportion

Calculate and analyze a one proportion z-test in order to generalize about an unknown population proportion.

Learning Objective:

sample proportion- the statistic that estimates the parameter

1=random sample

2=population is at least 10x sample

3=np≥10 n(1-p)≥10

Assumptions for inference about a population proportion

How do we standardize ?

The standard error of : (this is when we calculate confidence

intervals and we don’t know the true proportion so we use the sample proportion instead)

SE=

Therefore if our confidence interval has the form:

To test H₀ based on an SRS of size n from a large population with unknown proportion p of successes.

z test statistic:

Large Sample Inference For a Population Proportion

In terms of a variable z having the standard normal distribution, the p-value for a test of H₀ against

If your Ha is >, you just shade above (z-score should be positive)

If your Ha is <, you just shade below (z-score should be negative)

If your Ha is ≠, you just shade both (you need to put a 2 in front of your calculations)

Example 1: The French naturalist Count Buffon tossed a coin 4040 times. He got 2048 heads. Is there evidence that Buffon’s coin was unbalanced?

Population:p=true proportion of heads when tossing Buffon’s coin

Hypothesis: H₀: p=0.5 Ha: p≠0.5

Assumptions: -random sample -population is at least 10x sample -np≥10 n(1-p)≥10 2020 ≥10 2020≥10(or you can use # of successes and failures)2048 ≥10 1992 ≥10

Test:One proprotion z-test

Alpha level: α=0.05

Calculations:

Decision & Statement:Since p∡α, it is not statistically significant,

therefore we do not reject H₀. There is not enough evidence to say the coin is unbalanced.

Example 2: Calculate a 95% confidence interval for the proportion of heads from Buffon’s coin flip.

One proportion z interval

Assumptions: same as above

0.5069 ± 0.01545= (0.49151, 0.52235)We are 95% confident that the true

proportion of getting a head when flipping a coin is between 0.49 and 0.52

** If we are not given a value for p, use p*=0.5 as a conservative value!!***

Choosing a Sample Size

Example 3: How many times would you need to toss a weighted coin that has a 30% chance of getting a head to estimate the proportion of heads within 95% confidence and a margin of error of 0.05?

n≥322.69, thereforen=323

You want to design a study to estimate the proportion of students who like going on field trips. You will use a 90% confidence interval and want your margin of error to be within 0.06. The minimum sample required is?

Hint: we don’t know the true proportion, so we use p*=0.5

n≥187.92 therefore n=188

In a test of H0: p = 0.3 against Ha: p > 0.3, a sample of size 200 produces z = 2.51 for the value of the test statistic. Thus the P-value (or observed level of significance) of the test is approximately equal to?

***If you go to a one-prop z-test on your calculator to get the p-value, can you find it?

(try it!!)No, we don’t know what to plug in for x (the

number of successes in our sample)

Extra Example:

In a test of H0: p = 0.3 against Ha: p 0.3, a sample of size 200 produces z = 2.51 for the value of the test statistic. Thus the P-value (or observed level of significance) of the test is approximately equal to?

So what do we do? Remember if we know the z-score we can use normalcdf…. to find the area under a curve (Draw your standardized curve-it will help!!)

First, your Ha is >, so we want to shade above 2.51Try using normalcdf to get the answer

Normalcdf(2.51,1000,0,1)=0.006

There’s your p-value!!!!

In a test of H0: p = 0.8 against Ha: p ≠ 0.8, a sample of size 75 produces z = 1.15 for the value of the test statistic. Thus the P-value (or observed level of significance) of the test is approximately equal to?

Normalcdf(1.15,1000,0,1)=0.125

But be careful, your Ha is ≠, so we only found the area above. Remember we have to shade above and below when Ha is ≠.

So simply double it! 2(0.125)=0.25