**Problem 27: Confidence Interval Calculation for Mobile Device Repair Costs**In a random sample of six mobile devices, the mean repair cost was $65.00 and the standard deviation was $14.00. Assume the population is normally distributed, and use a t-distribution to find the margin of error and construct a 90% confidence interval for the population mean. Interpret the results.**Solution Steps:**1. **Identify the given information:** - Sample size (n): 6 - Sample mean ($\bar{x}$): $65.00 - Sample standard deviation (s): $14.00 - Confidence level: 90%2. **Determine the appropriate t-score:** - Degrees of freedom (df) = n - 1 = 6 - 1 = 5 - For a 90% confidence level and 5 degrees of freedom, use a t-table or calculator to find the critical t-value (t*).3. **Calculate the margin of error (E):** \[ E = t^* \times \left(\frac{s}{\sqrt{n}}\right) \]4. **Construct the confidence interval:** \[ \text{Lower bound} = \bar{x} - E \] \[ \text{Upper bound} = \bar{x} + E \]5. **Interpret the results:** - The interval will provide the range in which we are 90% confident the true mean repair cost for the population of mobile devices falls.**Detailed Steps:**1. Given: - $\bar{x} = 65$ - s = 14 - n = 6 - Confidence level = 90%2. Degrees of freedom (df) = 6 - 1 = 5 Using a t-table or calculator for a 90% confidence level with df = 5, the critical t-value (t*) is approximately 2.015.3. Margin of error (E): \[ E = 2.015 \times \left(\frac{14}{\sqrt{6}}\right) \] \[ E = 2.015 \times \left(\frac{14}{2.4495}\right) \] \[ E = 2.

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Answered: 27. In a random sample of six mobile…

MATLAB: An Introduction with Applications

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27. In a random sample of six mobile devices, the mean repair cost was $65.00 and the standard dev $14.00. Assume the population is normally distributed and use a t-distribution to find the margin o construct a 90% confidence interval for the population mean. Interpret the results.