Introduction to Confidence Intervals Do Now: Read the introduction below. Annotate the text as you read! Fill in any blanks and give your best answers to the questions along the way! Statistics 4/6

Agenda Do Now – 20 min Finish Notes / Practice from last class – 20 min Notes and Practice – 50 min

Finishing Up Notes from Monday 4/4

What type of distribution is this? 𝑩𝒊𝒏𝒐𝒎𝒊𝒂𝒍 𝑫𝒊𝒔𝒕𝒓𝒊𝒃𝒖𝒕𝒊𝒐𝒏 How do you know? 𝑩𝒊𝒏𝒐𝒎𝒊𝒂𝒍 𝑫𝒊𝒔𝒕𝒓𝒊𝒃𝒖𝒕𝒊𝒐𝒏 𝒈𝒊𝒗𝒆𝒏: 𝒏, 𝒑, 𝒂𝒏𝒅 𝒒 𝝈= 𝒏𝒑𝒒

𝑷( 𝒙 <𝟎.𝟕𝟓) 𝝈= 𝒏𝒑𝒒 𝒏=𝟐𝟓 𝝈=𝟏.𝟕𝟗 𝝈 𝒙 = 𝝈 𝟐𝟓 = 𝟏.𝟕𝟗 𝟓 =𝟎.𝟑𝟔 =𝒏𝒐𝒓𝒎𝒂𝒍𝒄𝒅𝒇 −𝟏𝟎𝟎𝟎, 𝟎.𝟕𝟓, 𝟎.𝟖𝟓, 𝝈 𝟐𝟓 𝑷( 𝒙 <𝟎.𝟕𝟓) =𝒏𝒐𝒓𝒎𝒂𝒍𝒄𝒅𝒇 −𝟏𝟎𝟎𝟎, 𝟎.𝟕𝟓, 𝟎.𝟖𝟓, 𝟎.𝟑𝟔 𝝈= 𝒏𝒑𝒒 =𝟎.𝟑𝟗 𝝈= (𝟐𝟓)(𝟎.𝟖𝟓)(𝟎.𝟏𝟓) 𝝈=𝟏.𝟕𝟗 𝝈 𝒙 = 𝝈 𝟐𝟓 = 𝟏.𝟕𝟗 𝟓 =𝟎.𝟑𝟔

𝒏=𝟐𝟓 𝒌= 𝟎.𝟗𝟑 𝝈=𝟏.𝟕𝟗 𝝈 𝒙 =𝟎.𝟑𝟔 𝑷( 𝒙 > ?) =𝒊𝒏𝒗𝑵𝒐𝒓𝒎 𝟎.𝟗𝟑,𝟎.𝟖𝟓, 𝟎.𝟑𝟔 𝒙 =𝟏.𝟑𝟖

𝒏=𝟐𝟓 𝒌= 𝟎.𝟔 𝝈=𝟏.𝟕𝟗 𝝈 𝒙 =𝟎.𝟑𝟔 𝑷( 𝒙 > ?) =𝒊𝒏𝒗𝑵𝒐𝒓𝒎 𝟎.𝟔,𝟎.𝟖𝟓, 𝟎.𝟑𝟔 𝒙 =𝟎.𝟗𝟒

𝑷( 𝒙 ≥𝟎.𝟔𝟎) 𝝈= 𝒏𝒑𝒒 𝒏=𝟑𝟎 𝝈=𝟐.𝟑𝟕 𝝈 𝒙 = 𝝈 𝟑𝟎 = 𝟐.𝟑𝟕 𝟓.𝟒𝟖 =𝟎.𝟒𝟑 Based on a recent census, it was found that 75% of TMA students did not travel outside of DC over spring break. You randomly select 30 students for a sample survey.   What is the probability that at least 60% of the students in your sample did not travel outside of DC over spring break? 𝒏=𝟑𝟎 =𝒏𝒐𝒓𝒎𝒂𝒍𝒄𝒅𝒇 𝟎.𝟔𝟎, 𝟏𝟎𝟎𝟎, 𝟎.𝟕𝟓, 𝝈 𝟑𝟎 𝑷( 𝒙 ≥𝟎.𝟔𝟎) =𝒏𝒐𝒓𝒎𝒂𝒍𝒄𝒅𝒇 𝟎.𝟔𝟎, 𝟏𝟎𝟎𝟎, 𝟎.𝟕𝟓, 𝟎.𝟒𝟑 𝝈= 𝒏𝒑𝒒 =𝟎.𝟔𝟒 𝝈= (𝟑𝟎)(𝟎.𝟕𝟓)(𝟎.𝟐𝟓) 𝝈=𝟐.𝟑𝟕 𝝈 𝒙 = 𝝈 𝟑𝟎 = 𝟐.𝟑𝟕 𝟓.𝟒𝟖 =𝟎.𝟒𝟑

𝑷( 𝒙 ≥𝟎.𝟔𝟎) 𝝈= 𝒏𝒑𝒒 𝒏=𝟑𝟎 𝝈=𝟐.𝟑𝟕 𝝈 𝒙 = 𝝈 𝟑𝟎 = 𝟐.𝟑𝟕 𝟓.𝟒𝟖 =𝟎.𝟒𝟑 Based on a recent census, it was found that 75% of TMA students did not travel outside of DC over spring break. You randomly select 30 students for a sample survey.   B) What is the cutoff for the highest 25% of sample proportions of TMA students who did not travel outside of DC over spring break? 𝒏=𝟑𝟎 =𝒏𝒐𝒓𝒎𝒂𝒍𝒄𝒅𝒇 𝟎.𝟔𝟎, 𝟏𝟎𝟎𝟎, 𝟎.𝟕𝟓, 𝝈 𝟑𝟎 𝑷( 𝒙 ≥𝟎.𝟔𝟎) =𝒏𝒐𝒓𝒎𝒂𝒍𝒄𝒅𝒇 𝟎.𝟔𝟎, 𝟏𝟎𝟎𝟎, 𝟎.𝟕𝟓, 𝟎.𝟒𝟑 𝝈= 𝒏𝒑𝒒 =𝟎.𝟔𝟒 𝝈= (𝟑𝟎)(𝟎.𝟕𝟓)(𝟎.𝟐𝟓) 𝝈=𝟐.𝟑𝟕 𝝈 𝒙 = 𝝈 𝟑𝟎 = 𝟐.𝟑𝟕 𝟓.𝟒𝟖 =𝟎.𝟒𝟑

Confidence Intervals for Proportions

sample statistic population parameter certain population parameter confidence interval

Example: We take a random sample of 30 TMA students and construct a confidence interval for the proportion of TMA students who like corn on the cob. As a result, we are 95% confident that the true proportion of TMA students who like corn on the cob is between .83 and .89 (or 83% and 89%).

Constructing a Confidence Interval Identify the population of interest and define the parameter of interest being estimated. Identify the appropriate confidence level and find the critical value. Verify any conditions (assumptions) that need to be met for that confidence interval. Calculate the confidence interval. Interpret the interval in the context of the situation.

critical value tails 𝝈 𝒏 𝒙 𝒛 ∗ standard deviation size 𝝈 𝒏 𝒙 𝒛 ∗ critical value tails standard deviation size critical value of z

Requirements for using the formula calculating the Confidence Interval for Proportions 1. The sample data is collected using a simple random sample. 2. The conditions for a binomial distribution are satisfied 3. There are at least 5 failures and 5 successes

The true proportion of adult residents of this city Example 1: Suppose a market research firm is hired to estimate the percent of adults living in a large city who have cell phones. Five hundred randomly selected adult residents in this city are surveyed to determine whether they have cell phones. Of the 500 people surveyed, 421 responded yes. Using a 95% confidence level, compute a confidence interval for the true proportion of adult residents of this city who have cell phones. What is the population parameter of interest?   Calculate the confidence interval. Interpret the confidence interval in the context of the situation. 𝒛 ∗ =1.96 The true proportion of adult residents of this city who have a cell phone 𝒏=𝟓𝟎𝟎 𝒑 ± 𝒛 ∗ 𝒑 𝒒 𝒏 𝟎.𝟖𝟒𝟐±𝟎.𝟎𝟑𝟐 𝒑 = 𝟒𝟐𝟏 𝟓𝟎𝟎 =𝟎.𝟖𝟒𝟐 𝟎.𝟖𝟒𝟐±𝟏.𝟗𝟔 (𝟎.𝟖𝟒𝟐)(𝟎.𝟏𝟓𝟖) 𝟓𝟎𝟎 𝟎.𝟖𝟏𝟎<𝒑<.𝟖𝟕𝟒 𝒒 =𝟏− 𝒑 =𝟎.𝟏𝟓𝟖 I am 95% confident that the true proportion of adult residents of this city who have a cell phone is between 𝟖𝟏% and 𝟖𝟕%.

The true proportion of burritos that are made within 2 minutes Practice 1: The manager of the Chipotle in Eastern Market wishes to know if burritos orders are being completed within 2 minutes of the customer’s initial order. A random sample of 75 burrito orders found that 60 were completed within 2 minutes of the customer’s order. Find the 95% confidence interval for the true proportion of burrito orders that are completed within 2 minutes. What is the population parameter of interest?   B) Calculate the confidence interval. C) Interpret the confidence interval in the context of the situation. 𝒛 ∗ =1.96 The true proportion of burritos that are made within 2 minutes of a customer’s order at the Chipotle in Eastern Market. 𝒏=𝟕𝟓 𝒑 ± 𝒛 ∗ 𝒑 𝒒 𝒏 𝟎.𝟖±𝟎.𝟎𝟗𝟏 𝒑 = 𝟔𝟎 𝟕𝟓 =𝟎.𝟖 𝟎.𝟖±𝟏.𝟗𝟔 (𝟎.𝟖)(𝟎.𝟐) 𝟕𝟓 𝟎.𝟕𝟎𝟗<𝒑<.𝟖𝟗𝟏 𝒒 =𝟏− 𝒑 =𝟎.2 I am 95% confident that the true proportion of burritos that are made within 2 minutes of a customer’s order at the Chipotle in Eastern Market is between 𝟕𝟏% and 𝟖𝟗%.

finding the critical values that separate the middle 90% of the area of the standard normal distribution 90% 5% 5% 𝒛 ∗ −𝒛 ∗ 𝟎 𝒛 ∗ =𝒊𝒏𝒗𝑵𝒐𝒓𝒎 𝟎.𝟗𝟓,𝟎,𝟏 = 𝟏.𝟔𝟓 −𝒛 ∗ =𝒊𝒏𝒗𝑵𝒐𝒓𝒎 𝟎.𝟎𝟓,𝟎,𝟏 = −𝟏.𝟔𝟓

finding the critical values that separate the middle 95% of the area of the standard normal distribution 95% 2.5% 2.5% 𝟎 𝒛 ∗ −𝒛 ∗ 𝒛 ∗ =𝒊𝒏𝒗𝑵𝒐𝒓𝒎 𝟎.𝟗𝟕𝟓,𝟎,𝟏 = 𝟏.𝟗𝟔 −𝒛 ∗ =𝒊𝒏𝒗𝑵𝒐𝒓𝒎 𝟎.𝟎𝟐𝟓,𝟎,𝟏 = −𝟏.𝟗𝟔

finding the critical values that separate the middle 99% of the area of the standard normal distribution 99% 0.5% 0.5% 𝟎 𝒛 ∗ −𝒛 ∗ 𝒛 ∗ =𝒊𝒏𝒗𝑵𝒐𝒓𝒎 𝟎.𝟗𝟗𝟓,𝟎,𝟏 = 𝟐.𝟓𝟖 −𝒛 ∗ =𝒊𝒏𝒗𝑵𝒐𝒓𝒎 𝟎.𝟎𝟎𝟓,𝟎,𝟏 = −𝟐.𝟓𝟖

The true proportion of Americans who had a cold on November 29th 2012. Confident level: 90% Critical Value: 𝐳 ∗ =𝟏.𝟔𝟓

We can assume that the sample data was collected using a simple random sample. The conditions for a binomial distribution are satisfied (2 possible outcomes for each trial independent trial; set number of trials) With a sample size of1000, there were at least 5 failures and 5 successes

𝐳 ∗ =𝟏.𝟔𝟓 𝒏=𝟏𝟎𝟎𝟎 𝒑 =𝟎.𝟎𝟕𝟏 𝟎.𝟖𝟏𝟎<𝒑<.𝟖𝟕𝟒 𝒒 =𝟏− 𝒑 =𝟎.𝟗𝟐𝟗 𝒑 ± 𝒛 ∗ 𝒑 𝒒 𝒏 𝐳 ∗ =𝟏.𝟔𝟓 𝟎.𝟖𝟒𝟐±𝟎.𝟎𝟑𝟐 𝒏=𝟏𝟎𝟎𝟎 𝒑 =𝟎.𝟎𝟕𝟏 𝟎.𝟖𝟏𝟎<𝒑<.𝟖𝟕𝟒 𝟎.𝟎𝟕𝟏±𝟏.𝟔𝟓 (𝟎.𝟎𝟕𝟏)(𝟎.𝟗𝟐𝟗) 𝟏𝟎𝟎𝟎 𝒒 =𝟏− 𝒑 =𝟎.𝟗𝟐𝟗 I am 95% confident that the true proportion of adult residents of this city who have a cell phone is between 𝟖𝟏% and 𝟖𝟕%.

What is the population parameter of interest? Identify the appropriate confidence level and find the corresponding critical value. Verify any conditions that need to be met for the confidence interval.

D. Calculate the confidence interval.   E. Interpret the confidence interval in the context of the situation.