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Alina [70]
3 years ago
13

An educational psychologist wishes to know the mean number of words a third grader can read per minute. She wants to make an est

imate at the 80% level of confidence. For a sample of 840 third graders, the mean words per minute read was 40.6. Assume a population standard deviation of 4.1. Construct the confidence interval for the mean number of words a third grader can read per minute. Round your answers to one decimal place.
Mathematics
1 answer:
nirvana33 [79]3 years ago
3 0

Answer:

The 80% confidence interval for the mean number of words a third grader can read per minute is between 40.4 wpm and 40.8 wpm.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.8}{2} = 0.1

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.1 = 0.99, so z = 1.28

Now, find M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 1.28*\frac{4.1}{\sqrt{840}} = 0.2

The lower end of the interval is the sample mean subtracted by M. So it is 40.6 - 0.2 = 40.4 words per minute.

The upper end of the interval is the sample mean added to M. So it is 40.6 + 0.2 = 40.8 words per minute.

The 80% confidence interval for the mean number of words a third grader can read per minute is between 40.4 wpm and 40.8 wpm.

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Given

1/3 = 0.333333

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a.

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Absolute Error = |Real Value - Estimated Value|

Relative Error = Absolute Error/Real Value

Assume 1/3 to be the real value and 3/4 to be the estimated value

Absolute Error = |0.333333 - 0.75|

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Absolute Error = 0.416667

Relative Error = 0.416667/0.333333

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b.

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= 0.001107

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