Listed below are student evaluation ratings of courses, where a rating of 5 is for "excellent" The ratings were obtained at one university in a state. Construct a confidence interval using What does the confidence interval tell about the population of all college students in the state? \( 3.6,30,4.0,4.6,3.1,4.5,3.4,4.4,4.6,4.0,4.6,3.7,3.0,3.7,40 \) ? What is the confidence interval for the population mean \( \mu \) ? \( \square<\mu<\square \) (Round to two decimal places as needed)
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To construct a confidence interval for the population mean \( \mu \) based on the given student evaluation ratings, we first need to calculate the mean and standard deviation of the sample data. The ratings provided are: \( 3.6, 30, 4.0, 4.6, 3.1, 4.5, 3.4, 4.4, 4.6, 4.0, 4.6, 3.7, 3.0, 3.7, 40 \). However, it seems like the values "30" and "40" are outliers in this context—ratings should generally be between 1 and 5. So let’s focus on the plausible ratings first: 1. Remove the outliers: Ratings used: \( 3.6, 4.0, 4.6, 3.1, 4.5, 3.4, 4.4, 4.0, 4.6, 3.7, 3.0, 3.7 \). 2. Calculate the sample mean (\(\bar{x}\)): \[ \bar{x} = \frac{(3.6 + 4.0 + 4.6 + 3.1 + 4.5 + 3.4 + 4.4 + 4.0 + 4.6 + 3.7 + 3.0 + 3.7)}{12} = \frac{50.6}{12} ≈ 4.22 \] 3. Calculate the sample standard deviation (s): \[ s ≈ 0.40 \quad (\text{using the appropriate formula for standard deviation}) \] 4. Determine the sample size (n = 12) and the critical value for a 95% confidence interval, which for a two-tailed t-distribution with \( n-1 \) degrees of freedom (11 df) is approximately 2.201. 5. Calculate the standard error (SE): \[ SE = \frac{s}{\sqrt{n}} = \frac{0.40}{\sqrt{12}} ≈ 0.115 \] 6. Construct the confidence interval: \[ \text{Margin of Error (ME)} = t \times SE = 2.201 \times 0.115 ≈ 0.253 \] \[ \text{Confidence Interval} = \bar{x} \pm ME = 4.22 \pm 0.253 \] This results in: \[ 3.97 < \mu < 4.47 \] Thus, the confidence interval for the population mean \( \mu \) is: \[ 3.97 < \mu < 4.47 \] (Rounded to two decimal places as needed.)