Computing Confidence Intervals using the TI83 The TI83

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Computing Confidence Intervals using the TI-83 The TI-83 can compute an ENTIRE confidence interval

Computing Confidence Intervals using the TI-83 The TI-83 can compute an ENTIRE confidence interval from either summary statistics or data. These functions can be accesed by pressing STAT TEST 1

Estimating from a Large Sample If n > 30 the sample is considered to

Estimating from a Large Sample If n > 30 the sample is considered to be large, regardless of its distribution’s shape. Press STAT, choose TESTS and choose 7: ZInterval 2

Example • A study of 50 iris flowers revealed a mean petal length of

Example • A study of 50 iris flowers revealed a mean petal length of 2. 03 cm, sample standard deviation of 0. 27 cm. Compute a 90% confidence interval for the population mean petal length. • x-bar = 2. 03, s = 0. 27, n = 50 are our summary statistics. • Since n>30 okay to use Zinterval as well as s in place of • Step 1: Press STAT, selects TESTS, select Zinterval • Step 2: For method of input (Inpt: ) select STATS, since we have summary statistics. • Step 3: Enter 0. 27 for , 2. 03 for x-bar and 50 for n. • Step 4. Set C-Level( Confidence Level) to. 90 • Step 5: Select Calculate. (Double Check Entries First) 3

Screen Shots 4

Screen Shots 4

Estimating from a Small Sample If n ≤ 30 the sample is considered to

Estimating from a Small Sample If n ≤ 30 the sample is considered to be small, the population must be normal and unknown. Press STAT, choose TESTS and choose 8: TInterval 5

Example • Consumer Reports gave the following information about the life(hours) of AA batteries

Example • Consumer Reports gave the following information about the life(hours) of AA batteries in toys. 2. 3 2. 5 4. 2 6. 1 5. 7 5. 5 1. 3 1. 5 5. 3 1. 8 1. 9 5. 2 1. 8 5. 1 1. 6 5. 4 • Assume the population is normally distributed, compute a 95% confidence interval to estimate the true mean life of AA batteries in toys. 6

Example • Since n = 16 which is less than 30, is unknown and

Example • Since n = 16 which is less than 30, is unknown and the population is normal we use a TInterval. • Since we have data we must enter it into a list. • Step 1: Press STAT, choose Edit, enter the values into L 1. • Step 2: Press STAT, select TESTS, select 8: Tinterval. • Step 3: Select Data for method Input. • Step 4: Enter L 1 for List, 1 for Freq • Step 5. Set C-Level( Confidence Level) to. 95 • Step 6: Select Calculate. (Double Check Entries First) 7

Screen Shots 8

Screen Shots 8

Estimating p If the sample is a SRS, binomial and np 5 and nq

Estimating p If the sample is a SRS, binomial and np 5 and nq 5 are both satisfied. ˆ ˆ Press STAT, choose TESTS and choose A: 1 -Prop. ZInt 9

Example • Example: In a survey of 2503 men and women aged 18 to

Example • Example: In a survey of 2503 men and women aged 18 to 75 years and representative of the nation as a whole, 1927 people said the homeless are not adequately assisted by the government. Find a point estimate and a 90% confidence interval for the proportion p of adults in the general population who agree that the homeless are not adequately assisted by the government. • n = 2503, x = 1927, since we have more than 5 success and 5 failures it is okay to use a 1 -Prop. ZInt. • Step 1: Press STAT, select TESTS, select A: 1 -Prop. ZInt. • Step 2: Enter the number of success for x and sample size for n. • Step 3. Set C-Level( Confidence Level) to. 90 • Step 4: Select Calculate. (Double Check Entries First 10

Screen Shots 11

Screen Shots 11