Question

A school principal is interested in assessing the performance of her students on the Totally Oppressive Standardized test (TOST). She selects a simple random sample of 16 of her students and finds the following set of test scores:

6	5	6	12	5	10	11	13
12	10	9	20	23	20	28	18

Assume that the sample is drawn from a population with a standard deviation σ = 7.20.

a. (1 point) What is the mean test score for the sample of students?

13

b. (4 points) Calculate and write a sentence to interpret the 95% confidence interval for the mean test score.

c. (3 points) Calculate and write a sentence to interpret the 99% confidence interval for the mean test score.

13 - 2.58 * (1.8) 13 + 2.58 * (1.8)= 8.3,17.6????

d. (4 points) Explain why the margin of error for the 95% confidence interval in question (b) is smaller than the margin of error for the 99% confidence interval in question (c).

Answer 1

the necessary calculation table:-

test score(xi)
6
5
6
12
5
10
11
13
12
10
9
20
23
20
28
18
sum=208

sample size (n)=16

population sd () =7.20

a).the mean test score for the sample of students be:-

b). z critical value for 95% confidence level, both tailed test be:-

the 95% confidence interval for the mean test score be:-

interpretation:-

we are 95% confident that the true mean test score will lie within 9.472 and 16.528

c).z critical value for 95% confidence level, both tailed test be:-

the 99% confidence interval for the mean test score be:-

interpretation:-

we are 99% confident that the true mean test score will lie within 8.363 and 17.637

d).the margin of error for the 95% confidence interval in question (b) is smaller than the margin of error for the 99% confidence interval in question because,

we know that as confidence level increase in order to be more confident our confidence interval becomes wider...so, the margin of error increases with increase in confidence level..an increase in margin of errpr cause the confidence interval to be more wide in the higher confidence level.

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