Each child in a sample of 65 low-income children was administered a language and

Question

Each child in a sample of 65 low-income children was administered a language and communication exam. The sentence complexity scores had a mean of 7.627.62 and a standard deviation of 8.94. Complete parts a through d

a. From the sample, estimate the true mean sentence complexity score of all low-income children

b. Form a 90% confidence interval for the estimate, part a.

c.Give a practical interpretation of the interval, part b.

A. We are 90% confident that the mean sentence complexity score of all low-income children is between the endpoints of the confidence interval.

B. We are 90% confident that the mean sentence complexity score of all low-income children is outside the confidence interval.

C.We are 10% confident that the mean sentence complexity score of all low-income children is outside the confidence interval.

D. We are 10% confident that the mean sentence complexity score of all low-income children is between the endpoints of the confidence interval.

E. An interpretation cannot be determined.

d- Suppose the true mean sentence complexity score of middle-income children is known to be 15.55. Is there evidence that the true mean for low-income children differs from 15.55?

( ) Yes

( ) No

Explanation / Answer

a)

True mean sentence complexity score of all low-income children = 7.627

b)

CI for = 90%

n = 65

mean = 7.627

z-value of 90% CI = 1.6449

std. dev. = 8.94

SE = std.dev./sqrt(n) = 1.10887

ME = z*SE = 1.82393

Lower Limit = Mean - ME = 5.80307

Upper Limit = Mean + ME = 9.45093

90% CI (5.8031 , 9.4509 )

c)

A. We are 90% confident that the mean sentence complexity score of all low-income children is between the endpoints of the confidence interval.

d)

Yes, because it does not lie in the confidence interval

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