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 a 95% confidence level. what does the confidence interval tell about the population of all college students in the state?

3.63.6,

2.92.9,

3.83.8,

4.54.5,

3.23.2,

3.93.9,

3.33.3,

4.64.6,

4.14.1,

4.34.3,

4.44.4,

3.93.9,

3.23.2,

4.24.2,

3.83.8

3.573 to 4.127

Step-by-step explanation:

Given

Sample size = 15

Mean = Sum of ratings/ sample size

Mean = 57.7/15

Mean = 3.85

Degree of freedom = sample size - 1

Degree of freedom = 15 - 1 = 14

df = 14

Then we calculate the standard deviation

(x - mean)² ||

(3.6 - 3.85)² || 0.0625

(2.9 - 3.85)² || 0.9025

(3.8 - 3.85)² || 0.0025

(4.5 - 3.85)² || 0.4225

(3.2 - 3.85)² || 0.4225

( 3.9 - 3.85)² || 0.0025

( 3.3 - 3.85)² || 0.3025

( 4.6 - 3.85)² || 0.5625

(4.1 - 3.85)² || 0.0625

(4.3 - 3.85)² || 0.2025

4.4 - 3.85)² || 0.3025

( 3.9 - 3.85)² || 0.0025

(3.2 - 3.85)² || 0.4225

( 4.2 - 3.85)² || 0.1225

( 3.8 - 3.85)² || 0.0025

Total || 3.7975

Variance = 3.7975/15 = 0.253167

Standard Deviation = √0.253167 = 0.50315703314174194

Standard Deviation = 0.5 Approximated

The next step is to subtract the confidence level from 1, then divide by two.

i.e (1 - 0.95)/2 = 0.025

α = 0.025

Then we look up this answer to step in the t-distribution table.

For 14 degrees of freedom (df) and α = 0.025, my result is 2.145

The next step is to divide the sample standard deviation by the square root of the sample size.

0.5 / √15 = 0.129

Next is to multiply this result by step 2.145 (from the t table)

0.129 × 2.45 = 0.277

For the lower end of the range, subtract 0.277 from the sample mean.

3.85 – 0.277 = 3.573

Step 7: For the upper end of the range, add step 0.277 to the sample mean.

3.85 + 0.277 = 4.127

4

step-by-step explanation:

2.2°  

step-by-step explanation:

let θ = the angle of elevation.

tanθ =77/2020

tanθ = 0.038 12

      θ = tan⁻¹0.038 12

      θ = 2.2°