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

  y= -3x is the required function.

step-by-step explanation:

we have been given a table:

x             y

0             0

1             -3

3             -9

-2             6

we can see that the values of y is -3 times the value of x

hence, the function would be: y= -3x.

put the values of x will give the same y as in the table that means function is satisfying the given table.

y-y1=4(x-x1)

step-by-step explanation:

this is point-slope form since you only have 4 as the slope it goes in place of m

hope this : )