A novel you are reading had more than 500 pages. A random sample of 25 pages of

Question

A novel you are reading had more than 500 pages. A random sample of 25 pages of the novel was taken, and the number of words on each page counted. The mean number of words for the sample was 323 words. and the standard deviation was 38.4 words.A) (1 point) Calculate the 95% confidence limits for the mean number of words for the pages of the entire novel. B) (1 point) Interpret your confidence interval. What does this mean? C) (1 point) Is your confi- dence interval exact? Explain your reasoning. D) (2 points) Calculate the 99% confidence interval. Explain the tradeoff between a lower and higher confidence interval. E) (1 point) What can you do to make a more precise estimate of the range of values for the population mean number of words? F) (1 point) If you wanted a margin of error of 2, what would be the sample size required to achieve this?

Explanation / Answer

n= 25 , mean= 323 words and sample standard deviation= 38.4 words

A) 95% confidence interval

Mean = 323
t critical = 2.06
sM = ((38.4)^2/25) = 7.68

= M ± t(sM)
= 323 ± 2.06*7.68
= 323 ± 15.85

M = 323, 95% CI [307.15, 338.85].

You can be 95% confident that the population mean () falls between 307.15 and 338.85.

A 95% confidence interval is a range of values that you can be 95% certain contains the true mean of the population.

My confidence interval is in decimal not in exact values.

d) 99% confidence interval

Mean = 323
t critical = 2.8
sM = ((38.4)^2/25) = 7.68

= M ± t(sM)
= 323 ± 2.8*7.68
= 323 ± 21.48

M = 323, 99% CI [301.52, 344.48].

You can be 99% confident that the population mean () falls between 301.52 and 344.48.

95% confidence interval width was 338.85-307.15= 31.7

99% confidence interval width was 344.48-301.52= 42.96

As confidence level increased the width of confidence interval also increased.

E) We can make confidence interval more precise by increasing sample size and reducing variability.

f)Step1: Find t a/2 by dividing the confidence interval by two, and looking that area up in the t-table:

.99/2 = 0.495. The closest z-score for 0.495 with 24 d.f is 2.8

Step 2: Multiply step 1 by the standard deviation.
38.4 * 2.8 = 107.52

Step 3: Divide Step 2 by the margin of error. Our margin of error (from the question), is 2
107.52/2 = 53.76

Step 4: Square Step 3.
53.76*53.76= 2890.1376

Sample size is 2890 required to achieve this.

PLEASE NOTE: I ASSMED HERE STANDARD DEVIATION AS SAMPLE STANDARD DEVIATION.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.