In a study of children with a particular disorder, parents were asked to rate th

Question

In a study of children with a particular disorder, parents were asked to rate their child on a variety of items related to how well their child performs different tasks. One item was "Has difficulty organizing work," rated on a five-point scale of 0 to 4 with 0 corresponding to "not at all" and 4 corresponding to "very much." The mean rating for 278 boys with the disorder was reported as 2.23 with a standard deviation of 1.13. (Round your answers to four decimal places.)

Compute the 90% confidence interval.

Compute the 95% confidence interval.   

Compute the 99% confidence interval.

Explanation / Answer

Solution:

We are given n = 278, Xbar = 2.23 and SD = 1.13

The confidence interval formula is given as below:

Confidence interval = Xbar -/+ t*SD/sqrt(n)

Part 1

Compute the 90% confidence interval.

Confidence level = 90%

Degrees of freedom = n – 1 = 278 – 1 = 277

Critical t value = 1.6504

Confidence interval = 2.23 -/+ 1.6504*1.13/sqrt(278)

Confidence interval = 2.23 -/+ 0.1119

Lower limit = 2.23 – 0.1119 = 2.1181

Upper limit = 2.23 + 0.1119 = 2.3419

Confidence interval = (2.1181, 2.3419)

Part 2

Compute the 95% confidence interval.   

Degrees of freedom = n – 1 = 278 – 1 = 277

Critical t value = 1.9686

Confidence interval = 2.23 -/+ 1.9686*1.13/sqrt(278)

Confidence interval = 2.23 -/+ 0.1334

Lower limit = 2.23 – 0.1334 = 2.0966

Upper limit = 2.23 + 0.1334 = 2.3634

Confidence interval = (2.0966, 2.3634)

Part 3

Compute the 99% confidence interval.

Degrees of freedom = n – 1 = 278 – 1 = 277

Critical t value = 2.5937

Confidence interval = 2.23 -/+ 2.5937*1.13/sqrt(278)

Confidence interval = 2.23 -/+ 0.1758

Lower limit = 2.23 – 0.1758 = 2.0542

Upper limit = 2.23 + 0.1758 = 2.4058

Confidence interval = (2.0542, 2.4058)

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