In a study of children with a particular disorder, parents were asked to rate th
ID: 3217728 • Letter: I
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 283 boys with the disorder was reported as 2.35 with a standard deviation of 1.06. (Round your answers to four decimal places.) Compute the 90% confidence interval. Compute the 95% confidence interval. Compute the 99% confidence Interval. Explain the effect of the confidence level on the width of the interval. We see that the width of the interval increases with confidence level. We see that the width of the interval decreases with confidence level. We see that the width of the interval does not change with confidence level.Explanation / Answer
Solution:
We are given
Mean = 2.35
S = 1.06
Sample size = n = 283
df = n – 1 = 282
Confidence interval = Xbar -/+ t*S/sqrt(n)
First of all we have to find 90% confidence interval.
Critical t value = 1.6503
Confidence interval = 2.35 -/+ 1.6503*1.06/sqrt(283)
Confidence interval = 2.35 -/+ 0.1040
Lower limit = 2.35 – 0.1040 = 2.25
Upper limit = 2.35 + 0.1040 = 2.45
Now, we have to find 95% confidence interval.
Critical t value = 1.9684
Confidence interval = 2.35 -/+ 1.9684*1.06/sqrt(283)
Confidence interval = 2.35 -/+ 0.1240
Lower limit = 2.35 – 0.1240 = 2.23
Upper limit = 2.35 + 0.1240 = 2.47
Now, we have to find 99% confidence interval.
Critical t value = 2.5934
Confidence interval = 2.35 -/+ 2.5934*1.06/sqrt(283)
Confidence interval = 2.35 -/+ 0.1634
Lower limit = 2.35 – 0.1634 = 2.19
Upper limit = 2.35 + 0.1634 = 2.51
Explain the effect of the confidence level on the width of the interval.
We see that the width of the interval increases with confidence level.