Question

The State Department of Education is developing a new test to
measure the reading

level of sixth graders. To pilot this test, they select a random
sample of 50 sixth

graders from within the state and have them take the test. The mean
score for the

students in this sample is 70.5 points. Suppose that it is already
known that the

standard deviation for all sixth graders in the state is 8.4
points.

A.Construct and interpret a 98% confidence interval to estimate
the mean score on

the new reading test for all sixth graders in the state.

B.Does this interval provide convincing evidence to support a
school district’s claim

that the mean score on this new reading test is less than 75
points?

C.Does this interval provide convincing evidence to support a
claim that the mean

score on this new reading test is more than 70 points?

Answer #1

a)

sample mean, xbar = 70.5

sample standard deviation, σ = 8.4

sample size, n = 50

Given CI level is 98%, hence α = 1 - 0.98 = 0.02

α/2 = 0.02/2 = 0.01, Zc = Z(α/2) = 2.33

ME = zc * σ/sqrt(n)

ME = 2.33 * 8.4/sqrt(50)

ME = 2.77

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))

CI = (70.5 - 2.33 * 8.4/sqrt(50) , 70.5 + 2.33 *
8.4/sqrt(50))

CI = (67.7321 , 73.2679)

2)

yes, because confidence interval does not contain 75

c)

N0, because confidence interval contain 70

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