In a random sample of six mobile devices, the mean repair cost was $60.00 and the standard deviation was $13.50. Assume the population is normally distributed and use a t-distribution to find the margin of error and construct a 90% confidence interval for the population mean. Interpret the results. The 90% confidence interval for the population mean p is ( . ). (Round to two decimal places as needed.)
In a random sample of six mobile devices, the mean repair cost was $60.00 and the standard deviation was $13.50. Assume the population is normally distributed and use a t-distribution to find the margin of error and construct a 90% confidence interval for the population mean. Interpret the results. The 90% confidence interval for the population mean p is ( . ). (Round to two decimal places as needed.)
MATLAB: An Introduction with Applications
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ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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Problem 1P
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Also need 90% interval

Transcribed Image Text:**Transcription with Explanation**
In a random sample of six mobile devices, the mean repair cost was $60.00 and the standard deviation was $13.50. Assume the population is normally distributed and use a t-distribution to find the margin of error and construct a 90% confidence interval for the population mean. Interpret the results.
**The 90% confidence interval for the population mean μ is (__, __).**
(Round to two decimal places as needed.)
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**Explanation (for Educational Context):**
This text is part of a statistical analysis exercise designed to teach students how to calculate a confidence interval using a t-distribution. The exercise provides a scenario involving mobile device repair costs and asks students to apply appropriate statistical methods to estimate the population mean.
- **Mean Repair Cost**: $60.00
- **Standard Deviation**: $13.50
- **Sample Size**: 6
- **Confidence Level**: 90%
Students are expected to use these values to compute the confidence interval, which involves determining the margin of error using the t-distribution (likely with a t-value lookup for the given confidence level and degrees of freedom). The results will provide a range in which the true mean repair cost for the population is likely to fall with 90% confidence. The final step is to interpret this interval in the context of the data provided.
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