Calculate the confidence interval (Statistics problem)

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SUMMARY

The discussion centers on calculating the confidence interval using a TI-84 calculator for a given dataset. The user inputs a standard deviation of 8150, a mean of 26414.62, a sample size (n) of 20, and a confidence level of 0.9544. The expected confidence interval is (22769.83, 30059.41), but the user is obtaining (22772, 30058). The discrepancy is likely due to incorrect Z-value usage, as the user references a Z-value of 2 corresponding to a 95.44% confidence level.

PREREQUISITES
  • Understanding of confidence intervals and their significance in statistics.
  • Familiarity with the TI-84 calculator, specifically the STAT and TESTS functions.
  • Knowledge of standard deviation and mean calculations.
  • Basic comprehension of Z-scores and their application in statistical analysis.
NEXT STEPS
  • Review the TI-84 calculator manual for detailed instructions on using the STAT>TESTS>7>STATS function.
  • Learn how to calculate Z-scores and their impact on confidence intervals.
  • Explore the concept of sample size and its effect on confidence interval width.
  • Investigate the differences between Z-distribution and T-distribution in confidence interval calculations.
USEFUL FOR

Students studying statistics, educators teaching statistical methods, and anyone using the TI-84 calculator for statistical analysis.

RedonYellow
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I am trying to calculate the confidence interval for this problem using my T-84, and it's driving me mad because I'm so close to the correct answer. The problem is

Homework Statement



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The answer to this problem is (22769.83, 30059.41) (This is an example problem.) When I try to duplicate this in my calculator, my answer is (22772,30058).

Homework Equations


I go to STAT>TESTS>7>STATS

The Attempt at a Solution



I put the data in like this
standard deviation: 8150
mean: 26414.62
n: 20
C-Level: 0.9544

Could someone tell me what I'm doing incorrectly, please.
 
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I'm not familiar with your calculator but my guess is some problem with the Z values. If you check you will see that 95.44% corresponds to a right and left tail are each of .0228 which in turn gives a z value of 2. Why don't you just check what 2 sample standard deviations from the sample mean get you?
 

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