This article shows you how to use the TI-83 to find confidence intervals for a population proportion, p. You may also want to check out these other articles:
Using TI-83 to Find Confidence Interval: Overview
Confidence intervals give you a range of values which is likely to include an unknown parameter for the population. A simple example: let’s say a poll shows that that 75 out of 100 people drink alcohol. The proportion is 75/100. Although that’s probably close to what you would find in the entire population, you might find it’s actually somewhere between 70/100 and 80/100.
Using TI-83 to Find Confidence Interval: Steps
Watch the video or read the steps below:
Sample problem: A recent poll shows that 879 of 1412 Americans have had at least one caffeinated beverage in the last week. Construct a 90% confidence interval for p, the true population proportion.
- “x” is the number of successes and must be a whole number. Successes in this question is how many Americans have had at least one caffeinated beverage (879). If you are given p̂ instead (the sample proportion), multiply p̂ by n to get x (because x=n* p̂).
- “n” is the number of trials.
Step 1: Press STAT.
Step 2: Right arrow over to “TESTS.”
Step 3: Arrow down to “A:1–PropZInt…” and then press ENTER.
Step 4: Enter your x-value: 879.
Step 5: Arrow down and then enter your n value: 1412.
Step 6: Arrow down to “C-Level” and enter .90. This is your confidence level and must be entered as a decimal.
Step 7: Arrow down to calculate and press ENTER. The calculator will return the range (.6013, .64374)
That means the 98 percent CI for the population proportion is between 0.6013 and .64374.
Tip: Instead of arrowing down to select A:1–PropZInt…, press Alpha and MATH instead.
That’s using TI-83 to find confidence interval in easy steps!
Lost your guidebook? You can download one from the TI website here.
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