12.1 Inference for A Population Proportion

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

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

 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:

 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≥10n(1-p)≥ ≥ ≥10 (or you can use # of successes and failures) 2048 ≥ ≥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  ± = ( , ) We are 95% confident that the true proportion of getting a head when flipping a coin is between 0.49 and 0.52

 Example 3: How many times would you need to toss a coin to estimate the proportion of heads within 95% confidence and a margin of error of 0.05? n≥384.16, therefore n=385

 In a test of H 0 : p = 0.3 against H a : 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)

 In a test of H 0 : p = 0.3 against H a : 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 First, your Ha is >, so we want to shade above 2.51 Try using normalcdf to get the answer Normalcdf(2.51,1000,0,1)=0.006 There’s your p-value!!!!

 In a test of H 0 : p = 0.8 against H a : 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