There are many ways to measure the reading ability of children. Research designed to improve reading performance is dependent on

Question

3 years ago

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There are many ways to measure the reading ability of children. Research designed to improve reading performance is dependent on

good measures of the outcome. One frequently used test is the DRP or Degree of Reading Power. A researcher suspects that the mean score µ of all third graders in Henrico County Schools is different from the national mean, which is 32. To test her suspicion, she administers the DRP to an SRS of 44 Henrico County third-grade students and the standard deviation of scores in Henrico County Schools is known to be σ = 11 Their scores were: 40 26 39 14 42 18 25 43 46 27 19 47 19 26 35 34 15 44 40 38 31 46 52 25 35 35 33 29 34 41 49 28 42 47 35 48 22 33 41 51 27 14 54 45
a.Construct a 90% confidence interval to estimate the mean DRP score in Henrico County Schools.

b.Use the confidence interval you constructed in part (a) to comment on whether you agree with the researcher’s claim. Explain your reasoning clearly.

Mathematics

1 answer:

ELEN [110] · Answer 1 · 2022-08-26T11:26:18+03:00

Answer:

A 90% confidence interval to estimate the mean DRP score in Henrico County Schools is =32±2.7280

i.e. C.I. [29.3,34.7]

Hence the results are not significant.

Step-by-step explanation:

Given:

Total no of students =44

True mean =32

Standard deviation=11

And all 4 students score also given.

To Find:

90 % confidence interval and conclusion on it.

Solution:

Here for C.I. we required mean, standard deviation and no of students.

So mean =32,S.D.=11 and n=44

Therefore , For 90 % interval Zscore is Z=1.645

C.I.=mean±Z[Standard deviation/Sqrt(n)]

=32±1.645[11/Sqrt(44)]

=32±1.645[1.6583]

=32±2.7280

i.e. C.I. [29.3,34.7]

Now Here to commit on calculated interval we should check value of Z-score

So we need to calculate the sample mean.

Sample mean =Sum of all values /Total no of values.

=[40+ 26 +39+ 14+ 42+ 18 +25+ 43+ 46 +27 +19+ 47+ 19 +26+ 35 +34+ 15+ 44 40+ 38+ 31+ 46 +52 +25+ 35 +35+ 33 +29 +34 +41 +49+ 28+ 42+ 47 +35 +48 +22 33+ 41 +51 +27 +14+ 54 +45]/44

=34.86

Z-score=(sample mean -true mean)/[standard deviation/Sqrt(n)]

=(34.86-32)/[11/Sqrt(44)]

Z=1.72465

For ,

P(Z≥1.72465)=0.08544

For 0.05 significance level the results are not significant at p<0.01.

Researcher claimed mean is not correct upto 0.05 significant level

i.e calculated mean and Sample mean are different.

Hence the results are not significant.