In a study of children with a particular disorder, parents were asked to rate th
ID: 3149936 • 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 2.26 with a standard deviation of 1.04.
(a) Do you think that these data are Normally distributed? Explain why or why not. The distribution is Normal because there are no outliers. 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 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? Explain why or why not. The sample size is too small to allow the use of t methods with this distribution of ratings. 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 can have no outliers. The sample size should make the t methods appropriate because the distribution of ratings has a small mean. The sample size should make the t methods appropriate because the distribution of ratings has a large standard deviation.
(c) Find the 99% margin of error and the corresponding confidence interval. (Round your answers to four decimal places.) ( , )
(d) The boys in this study were all evaluated at a particular clinic at a university. To what extent do you think the results could be generalized to boys with this disorder in other locations?
Explanation / Answer
A)
The distribution cannot be Normal because all values must be integers between 0 and 4. [ANSWER]
[Note: Normal distribution is continuous, not discrete.]
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b)
The sample size should make the t methods appropriate because the distribution of ratings can have no outliers. [ANSWER]
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c)
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 = 2.26
t(alpha/2) = critical t for the confidence interval = 2.593438351
s = sample standard deviation = 1.04
n = sample size = 282
df = n - 1 = 281
Thus,
Margin of Error E = 0.160614487
Lower bound = 2.099385513
Upper bound = 2.420614487
Thus, the confidence interval is
( 2.099385513 , 2.420614487 ) [ANSWER]
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d)
We can only generalize it to the other locations to a limited extent [or not at all] because it is not a random sample from different locations, rather, just from 1 location.