Question

1. 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 274 boys with the disorder was reported as 2.34 with a standard deviation of 1.07. (Round your answers to four decimal places.)

Compute the 90%, 95%, and 99% confidence interval

Answer 1

Here n = 274

90% confidence interval for population mean

confidence level = c =0.90

alpha = 1 - c = 1 - 0.90 = 0.1

where tc is t critical value for degrees of freedom = n-1 = 274 - 1 = 273 and alpha = 0.1 can be calculated from excel using command:

=TINV(0.1,273)

= 1.65                         (Round to 2 decimal)

                 (Round to 4 decimal)

90% confidence interval is (2.2333, 2.4467)

95% confidence interval for population mean

confidence level = c =0.95

alpha = 1 -c = 1 -0.95 = 0.05

where tc is t critical value for degrees of freedom = n-1 = 274 - 1 = 273 and alpha = 0.05 can be calculated from excel using command:

=TINV(0.05,273)

= 1.97                      (Round to 2 decimal)

            (Round to 4 decimal)

95% confidence interval is (2.2127, 2.4673)

99% confidence interval for population mean

confidence level = c =0.99

alpha = 1 -c = 1 -0.95 = 0.01

where tc is t critical value for degrees of freedom = n-1 = 274 - 1 = 273 and alpha = 0.01 can be calculated from excel using command:

=TINV(0.01,273)

= 2.59                     (Round to 2 decimal)

               (Round to 4 decimal)

99% confidence interval is (2.1726, 2.5074)