Competed unit 5

STAT 2000 – Unit 5 Carrie Madden Carrie Madden STAT 2000 – Unit 5 1 / 184

Carrie Madden STAT 2000 – Unit 5 2 / 184

Unit 5 – Test for Proportions and Analysis of Categorical Data and Goodness-of-fit Tests Review of Inference for a Single Population Proportion P Now suppose we are interested in the proportion ˆ p of successes: ˆ p = X n The mean and standard deviation of ˆ p are µ ˆ p = p and σ ˆ p = ttttttt p ( 1 − p ) n Carrie Madden STAT 2000 – Unit 5 4 / 184

Unit 5 – Test for Proportions and Analysis of Categorical Data and Goodness-of-fit Tests Distribution of a Sample Proportion The Central Limit Theorem says that, if a random variable X represents a sample mean, and if the sample size is high, then the sampling distribution of X is approximately normal. Specifically, Z = X − µ X σ X ∼ N ( 0 , 1 ) But we can think of ˆ p as a kind of sample mean, because we are adding up all the successes we observe and dividing by the sample size n . Carrie Madden STAT 2000 – Unit 5 5 / 184

Unit 5 – Test for Proportions and Analysis of Categorical Data and Goodness-of-fit Tests Example Example

Unit 5 – Test for Proportions and Analysis of Categorical Data and Goodness-of-fit Tests Example We calculate np = 200 ( 0 . 10 ) = 20 > 10 and n ( 1 − p ) = 200 ( 0 . 90 ) = 180 > 10 Since the population is large, we can use the normal distribution. The probability that at least 12.5% of the sampled students are left-handed is: P (ˆ p ≥ 0 . 125 ) = P 1 CCCCCCC CCCCCCC CCCCCCC CCCCCCC AAAAAA Z ≥ ˆ p − p ttttttt p ( 1 − p ) n 2 DDDDDDD DDDDDDD DDDDDDD DDDDDDD BBBBBB = P 1 CCCCCCC CCCCCCC CCCCCCC CCCCCCC AAAAAA Z ≥ 0 . 125 − 0 . 10 ttttttt 0 . 1 ( 0 . 9 ) 200 2 DDDDDDD DDDDDDD DDDDDDD DDDDDDD BBBBBB = P ( Z ≥ 1 . 18 ) = 1 − P ( Z < 1 . 18 ) = 1 − 0 . 8810 = 0 . 1190 Carrie Madden STAT 2000 – Unit 5 8 / 184

Unit 5 – Test for Proportions and Analysis of Categorical Data and Goodness-of-fit Tests Example – Airlines Example

Unit 5 – Test for Proportions and Analysis of Categorical Data and Goodness-of-fit Tests Example Solution

Unit 5 – Test for Proportions and Analysis of Categorical Data and Goodness-of-fit Tests Practice Question It is known that 20% of a certain type of lottery ticket are winners. If you buy 100 lottery tickets, what is the approximate probability that at least 25 of them are winners? A 0.1056 B 0.1539 C 0.2061 D 0.2578 E 0.3085 Carrie Madden STAT 2000 – Unit 5 13 / 184

Unit 5 – Test for Proportions and Analysis of Categorical Data and Goodness-of-fit Tests Inference for a Population Proportion We now examine inference methods for the case where the parameter of interest is some population proportion p . We estimate p by the sample proportion ˆ p , which has mean and standard deviation µ ˆ p = p and σ ˆ p = ttttttt p ( 1 − p ) n Carrie Madden STAT 2000 – Unit 5 14 / 184

Unit 5 – Test for Proportions and Analysis of Categorical Data and Goodness-of-fit Tests Confidence Interval for a Population Proportion} We take a simple random sample of n individuals and calculate the proportion ˆ p that possess some characteristic of interest. A ( 1 − α ) 100 % confidence interval for a population proportion p is given as ˆ p ± z ∗ ttttttt ˆ p ( 1 − ˆ p ) n . Ideally, we would like to use the true standard deviation of p in the formula, but we don’t know p (this is the reason for doing inference!), so we must estimate it by ˆ p , and we estimate the standard deviation by the standard error of ˆ p . Carrie Madden STAT 2000 – Unit 5 16 / 184

Unit 5 – Test for Proportions and Analysis of Categorical Data and Goodness-of-fit Tests Example – Obama Example

Unit 5 – Test for Proportions and Analysis of Categorical Data and Goodness-of-fit Tests Sample Size Determination Suppose we would like to select a sample of individuals large enough to estimate some population proportion p to within a specified margin of error m with a given level of confidence. m = z ∗ ttttttt ˆ p ( 1 − ˆ p ) n ⇒ n = z ∗ m 2 ˆ p ( 1 − ˆ p ) Carrie Madden STAT 2000 – Unit 5 19 / 184

Unit 5 – Test for Proportions and Analysis of Categorical Data and Goodness-of-fit Tests Sample Size Determination But we have a problem – we are at the stage where we have not yet selected the sample, and so we don’t know the value of ˆ p . We will estimate the value of p by some value p ∗ . We can either use an educated guess for p ∗ , or we can use a conservative estimate p ∗ = 0 . 5, which will result in a margin of error no greater than m , regardless of the sample proportion ˆ p . Carrie Madden STAT 2000 – Unit 5 20 / 184