Question

In: Statistics and Probability

In exercise 14.2 we saw that an SRS of 400 high school seniors gained an average...

In exercise 14.2 we saw that an SRS of 400 high school seniors gained an average of x? = 22 points in their second attempt at the SAT Mathematics exam. Assuming that the change in the score has a Normal distribution with standard deviation ? = 50, we compute a 95% confidence interval for the mean change in score µ in the population of all high school seniors.

Find a 90% confidence interval for µ based on this sample.

Expert Solution

Solution :

Given that,

= 22

= 50

n = 400

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z_/2* ( / $\sqrt$ n)

= 1.96 * (50 / $\sqrt$ 400)

= 4.9

At 95% confidence interval estimate of the population mean is,

- E < < + E

22 - 4.9 < < 22 + 4.9

17.1 < < 26.9

(17.1 , 26.9)

2)

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z_/2* ( / $\sqrt$ n)

= 1.645 * (50 / $\sqrt$ 400)

= 4.1

At 90% confidence interval estimate of the population mean is,

- E < < + E

22 - 4.1 < < 22 + 4.1

17.9 < < 26.1

(17.9 , 26.1)

orchestra answered 2 years ago

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