In a study of children with a particular disorder, parents were asked to rate th
ID: 3150793 • 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 282 boys with the disorder was reported as 2.04 with a standard deviation of 1.05.
(a) Do you think that these data are Normally distributed? (CHOOSE ONE:)
-The distribution cannot be Normal because the standard deviation is too large.
-The distribution is Normal because there is a mean and a standard deviation.
-The distribution is Normal because there are no outliers.
-The distribution is Normal because all values must be integers between 0 and 4.
-The distribution cannot be Normal because all values must be integers between 0 and 4.
(b) Is it appropriate to use the methods of this section to compute a 99% confidence interval? (CHOOSE ONE:)
-The sample size should make the t methods appropriate because the distribution of ratings can have no outliers.
-The sample size should make the t methods appropriate because the distribution of ratings has a large standard deviation.
-The sample size is too large to allow the use of t methods with this distribution of ratings.
-The sample size should make the t methods appropriate because the distribution of ratings has a small mean.
-The sample size is too small to allow the use of t methods with this distribution of ratings.
(c) Find the 99% margin of error and the corresponding confidence interval. (Round your answers to four decimal places.)
Explanation / Answer
(a) Do you think that these data are Normally distributed? (CHOOSE ONE:)
Normal distribution is continuous, but this is discrete. Hence,
-The distribution cannot be Normal because all values must be integers between 0 and 4. [ANSWER]
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(b) Is it appropriate to use the methods of this section to compute a 99% confidence interval? (CHOOSE ONE:)
-The sample size should make the t methods appropriate because the distribution of ratings can have no outliers. [ANSWER]
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(c) Find the 99% margin of error and the corresponding confidence interval. (Round your answers to four decimal places.)
Note that
Margin of Error E = t(alpha/2) * s / sqrt(n)
Lower Bound = X - t(alpha/2) * s / sqrt(n)
Upper Bound = X + t(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.005
X = sample mean = 2.04
t(alpha/2) = critical t for the confidence interval = 2.593438351
s = sample standard deviation = 1.05
n = sample size = 282
df = n - 1 = 281
Thus,
Margin of Error E = 0.162158857 [ANSWER, MARGIN OF ERROR]
Lower bound = 1.877841143
Upper bound = 2.202158857
Thus, the confidence interval is
( 1.877841143 , 2.202158857 ) [ANSWER, CONFIDENCE INTERVAL]