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 282 boys with the disorder was reported as 2.26 with a standard deviation of 1.05.

(a) Do you think that these data are Normally distributed? Explain why or why not.

The distribution is Normal because all values must be integers between 0 and 4.

The distribution is Normal because there are no outliers.    

The distribution cannot be Normal because all values must be integers between 0 and 4.

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.

(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 should make the t methods appropriate because the distribution of ratings has a large standard deviation.

The sample size should make the t methods appropriate because the distribution of ratings can have no outliers.   

The sample size is too large to allow the use of t methods with this distribution of ratings.

The sample size is too small 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.

(c) Find the 99% margin of error and the corresponding confidence interval. (Round your answers to four decimal places.)

study to rate thei tems related to how well their child performs different tasks. One corresponding to "not and 4 correspon g to "very muc The meat rating fo 262 boys with the disorder w reported a 2.26 with a ndard viation of (a) Do you think that these data are Normally distributed? Explain why or why not. O The distribution is Normal because all values must be integers between 0 and 4. The distribution is Normal harause there ara nn outliers. The distribution cannot be Normal because all values must be integers between and 1. O The distribution cannot be Normal because the standard deviation is too large. The distributi standard (b) Ts it appropriate to use the methods of this section to compute a 99% confidence interval? Fxplain why or why not. The sample size should make the t methods appropriate because the distribution of ratings has a large standard deviation. The sample size should make the r methods appropriate because the distribution of ratings can have no outiers. The sample size is large uo allow the use uf methods with Uhis distribution of ratings. The sample size is too small to allow the use of t methods with this distribution of ratings The sample size should make the r methods appropriate because the distribution of ratings has a small mean. (c Find tha ogse margin of error and the corresponding confidence interval. (Round your answers to four decimal places.) diffi ty organizing

Explanation / Answer

a.
The distribution is Normal because there are no outliers.
b.
The sample size should make the t methods appropriate because the distribution of ratings has a small mean.
c.
Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=2.26
Standard deviation( sd )=1.05
Sample Size(n)=282
Confidence Interval = [ 2.26 ± t a/2 ( 1.05/ Sqrt ( 282) ) ]
= [ 2.26 - 2.593 * (0.063) , 2.26 + 2.593 * (0.063) ]
= [ 2.098,2.422 ]
d.
Margin of Error = t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
Mean(x)=2.26
Standard deviation( sd )=1.05
Sample Size(n)=282
Margin of Error = t a/2 * 1.05/ Sqrt ( 282)
= 2.593 * (0.063)
= 0.162

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