Chapter 9: testing a claim Ch. 9-2 Tests About a Population Proportion

One or two sided? one! State: 𝐻 0 : 𝐻 𝑎 : 𝑝=0.8 𝑝→ true proportion of FT made by Brinkhus 𝑝 = 33 50 =0.66 𝑝<0.8 𝛼=0.01

Plan: One sample 𝑧 test for a proportion Random: “Think of these 50 shots as being an SRS” Normal: 𝑛𝑝≥10 𝑛 1−𝑝 ≥10 →50 .8 =40≥10 →50 .2 =10≥10 NOTE: we’re using 𝑝, not 𝑝 ! So the sampling distribution of 𝑝 is approximately normal. Independent: Mr. Brinkhus has shot more than 10 50 =500 free throws over the years. You can also write that the observations are already independent. The outcome of one shot does not affect the outcome of another shot.

Do: Sampling Distribution of 𝑝 𝜎 𝑝 = .8 .2 50 =.057 N 0.8, ______ .057 0.66 0.8 𝑧= 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐−𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟 𝑠𝑡. 𝑑𝑒𝑣. 𝑜𝑓 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 = 𝑝 −𝑝 𝜎 𝑝 = 0.66−0.8 0.057 =−2.47 𝑎𝑟𝑒𝑎=.007 𝑝-value normalcdf −99999, 0.66, 0.8, .057 =0.007

Conclude: Assuming 𝐻 0 is true 𝑝=.8 , there is a .007 probability of obtaining a 𝑝 value of .66 or lower purely by chance. This provides strong evidence against 𝐻 0 and is statistically significant at 𝛼=.01 level .007<.01 . Therefore, we reject 𝐻 0 and can conclude that the true proportion of free throws made by Mr. Brinkhus is less than 0.8. 1) Interpret 𝑝-value 2) evidence 3) decision with context

When a problem doesn’t specify 𝛼, use 𝛼=0.05 One or two sided? two! 𝑝→ true proportion of math teachers who are left handed State: 𝐻 0 : 𝐻 𝑎 : 𝑝=0.23 𝑝≠0.23 𝑝 = 28 100 =0.28 𝛼=0.05 Plan: One sample 𝑧 test for a proportion Random: “random sample of 100 math teachers” Normal: 𝑛𝑝≥10 𝑛 1−𝑝 ≥10 →100 .23 =23≥10 →100 .77 =77≥10 So the sampling distribution of 𝑝 is approximately normal. Independent: Sampling without replacement so check 10% condition We can assume there are more than 10 100 =1000 math teachers in the country.

Do: Sampling Distribution of 𝑝 𝜎 𝑝 = .23 .77 100 =.042 N 0.23, ______ .042 0.117 .05 .05 0.18 0.23 0.28 𝑧= 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐−𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟 𝑠𝑡. 𝑑𝑒𝑣. 𝑜𝑓 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 = 𝑝 −𝑝 𝜎 𝑝 = 0.28−0.23 0.042 =1.19 𝑎𝑟𝑒𝑎=.117 normalcdf 0.28, 99999, 0.23, .042 =0.117 𝑝-value =2 .117 =.234

Conclude: Assuming 𝐻 0 is true 𝑝=.23 , there is a .234 probability of obtaining a 𝑝 value that is .05 or more away from 𝑝 purely by chance. This provides weak evidence against 𝐻 0 and is not statistically significant at 𝛼=0.05 level (.234>.05). Therefore, we fail to reject 𝐻 0 and cannot conclude that the true proportion of math teacher who are left-handed is not 23%.

𝑝→ true proportion of current high school students who have seen the 2002 Spider-Man movie State: 𝐻 0 : 𝐻 𝑎 : 𝑝=0.3 𝑝≠0.3 𝑝 = 175 500 =0.35 𝛼=0.05 Plan: One sample 𝑧 test for a proportion Random: “SRS of 500 current high school students” Normal: 𝑛𝑝≥10 𝑛 1−𝑝 ≥10 →500 .3 =150≥10 →500 .7 =350≥10 So the sampling distribution of 𝑝 is approximately normal. Independent: Sampling without replacement so check 10% condition We can assume there are more than 10 500 =5000 current high school students.

Do: Sampling Distribution of 𝑝 𝜎 𝑝 = .3 .7 500 =.0205 N 0.3, _______ .0205 0.0073 .05 .05 0.25 0.3 0.35 𝑧= 𝑝 −𝑝 𝜎 𝑝 = 0.35−0.3 0.0205 =2.44 𝑎𝑟𝑒𝑎=.0073 normalcdf .35, 99999, 0.3, .0205 =0.0073 𝑝-value =2 .0073 =.0146

Assuming 𝐻 0 is true 𝑝=.3 , there is a .015 probability of obtaining a Conclude: Assuming 𝐻 0 is true 𝑝=.3 , there is a .015 probability of obtaining a 𝑝 value that is 0.05 or more away from 𝑝 purely by chance. This provides strong evidence against 𝐻 0 and is statistically significant at 𝛼=0.05 level (.015<.05). Therefore, we reject 𝐻 0 and can conclude that the true proportion of current high school students that have seen the 2002 Spider-Man movie is not 0.3. 1-PropZTest (STAT→TESTS→5) reject 𝐻 0 confidence interval With calculator: 𝑧=2.44 𝑝=0.015 𝑝 =0.35 𝑛=500 STAT  TESTS  1-PropZTest (5) 𝑝-value 𝑝 0 : 𝑥: n: prop: ≠ 𝑝 0 < 𝑝 0 > 𝑝 0 0.3 175 500

𝑝 = 175 500 =.35 We want to estimate the actual proportion, 𝑝, of high school students who have seen the Spider-Man movie at a 95% confidence level. One-sample 𝑧 interval for proportion Random: (same) Normal: 𝑛 𝑝 ≥10 𝑛 1− 𝑝 ≥10 →500 .35 =175≥10 →500 .65 =325≥10 So the sampling distribution of 𝑝 is approximately normal. Independent: (same)

Estimate ± Margin of Error 𝑝 1− 𝑝 𝑛 𝑝 ± 𝑧 ∗ (.35) .65 500 .35 ± 1.96 .35 ±0.042 0.308, 0.392 We are 95% confident that the interval from 0.308 to 0.392 captures the true proportion of current high school students who have seen the Spider-Man movie.

plausible strong reject 𝑝=0.3 0.308, 0.392 plausible weak fail to reject

STAT → TESTS → 1-PropZInt 𝑥=122 𝑛=500 0.206, 0.282 C-Level: 0.95 notice 𝑝=.28 is captured just barely Make a guess based off of 𝑝 in the interval. Is 𝑝-value going to be greater or less than .05? By how much? .06? 𝑝=.28 is captured by the 95% confidence interval, so it is NOT statistically significant at 𝛼=.05.

STAT → TESTS → 1-PropZTest −1.79 0.073 If 𝛼=0.05 is being used, notice 𝑧 is less than 1.96 std dev away from 𝑝, so 𝑝-value will be higher than 𝛼. However, this only gives us information about one sample. Confidence intervals give more info than significance tests. A confidence interval gives a whole range of plausible values, whereas a sig test concentrates only on the one statistic as a possibility for the population proportion. On the AP Exam, it’s acceptable to use a confidence interval rather than a sig test to address a two-sided alternative hypothesis. HOWEVER, if 𝐻 𝑎 is one-sided, you must do a sig test.