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Learn how to compute confidence intervals on the TI-83 calculator for different sample sizes and scenarios. Includes step-by-step instructions and examples for large and small samples.
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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 STATTEST
If n > 30 the sample is considered to be large, regardless of its distribution’s shape. Press STAT, choose TESTSand choose 7:ZInterval Estimating from a Large Sample
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)
If n ≤ 30 the sample is considered to be small, the population must be normal and unknown. Press STAT, choose TESTSand choose 8:TInterval Estimating from a Small Sample
Example • Consumer Reports gave the following information about the life(hours) of AA batteries in toys. • Assume the population is normally distributed, compute a 95% confidence interval to estimate the true mean life of AA batteries in toys.
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 L1. • Step 2: Press STAT, select TESTS, select 8:Tinterval. • Step 3: Select Data for method Input. • Step 4: Enter L1 for List, 1 for Freq • Step 5.Set C-Level( Confidence Level) to .95 • Step 6: Select Calculate. (Double Check Entries First)
If the sample is a SRS, binomial and np 5 and nq 5 are both satisfied. Press STAT, choose TESTSand choose A:1-PropZInt Estimating p ˆ ˆ
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-PropZInt. • Step 1: Press STAT, select TESTS, select A:1-PropZInt. • 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