A sample of 270 students who were taking online courses were asked to describe their overall impression of online learning on a scale of 1-7, with 7 representing the most favorable impression. The average score was 5.81, and the standard deviation was 0.99. Construct an 80% confidence interval for the mean score. Round the answers to two decimal places l Ask An 80% confidence interval for the mean score is [ ] <.

An 80% confidence intervals are (5.73,5.88)

Step-by-step explanation:

Given sample size (n) is = 270

The average score (μ) = 5.81

and standard deviation σ = 0.99.

80% confidence interval:-

The 80% confidence interval of the z - value is 1.28 (from z-table)

An 80% confidence interval is defined by sample mean ± 1.28 standard error

that is μ ± 1.28 σ/√n

now substitute values (5.81 ± 1.28 (0.99/√270))

(5.81 - 1.28 (0.0602),5.81 + 1.28 (0.0602)

(5.81 - 0.0770,5.81 - 0.0770)

(5.73,5.88)

Conclusion:-

An 80% confidence intervals are (5.73,5.88)

Therefore the population mean 5.81 lies between (5.73,5.88) at 80% confidence intervals