Question

: Chapter 21: Confidence Intervals Example of a Confidence Interval for a population proportion, p: The public television station BPBS wants to find the percent of its viewing population that gives donations to the station. . 300 randomly selected viewers were surveyed, and it was found that 123 viewers made contributions to the station. . Find a 95% confidence interval for the probability that a viewer of BPBS selected at random contributes to the station To find a confidence interval for a proportion, press the STAT key and select TESTS. Use option A:1-PropZint under TESTS. Notice that the normal distribution will be used . . PTeti J .35434,.46566) Calculate 2-SamPZInt nt . The letter x is used to count the number of successes. Enter 123 for x and 300 for n. e Use 0.95 for the C-level . Highlight Calculate and press ENTER Your Turn: 2. Many types of error will cause a computer program to terminate or give incorrect results. One type of error is punctuation. For instance, if a comma is inserted in the wrong place, the program might not run. A study of programs written by students in an introductory programming course showed that 75 out of 300 errors selected at random were punctuation errors. (a) Find a 99% confidence interval for the proportion of errors made by beginning programming students that are punctuation errors. Interpret the results in a clear sentence. (b) Find a 95% confidence interval. Is this interval longer or shorter? (c) Find the margin of error using SD for each confidence interval. (Hint: you do have enough information from the calculator output! When using 1/sqrt(n) for the MOE we will call this the Quick Method) (d) Compare the margin of error from part (b) with the quick method we learned early in the semester. Is it the same Why or why not? 2I Page

Answer 1

Answer

(A) Using TI-84, press STAT, then TESTS, then 1-PropZTest

enter the data

x: 75

n: 300

c = 0.99

Click on CALCULATE

we get

(0.1856, 0.3144)

We are 99% confident that the true proportion of erros made by beginning programming students that are punctuation erros is between 0.1856 and 0.3144

(B) Using TI-84, press STAT, then TESTS, then 1-PropZTest

enter the data

x: 75

n: 300

c = 0.95

Click on CALCULATE

we get

(0.201,0.299)

this interval is shorted as compared to the 99% confidence interval

(C) z score for 99% confidence interval is 2.58, sample size is n = 300 and p = 75/300 = 0.25

Margin of error for 99% =

$z*\sqrt{((1-\hat{p})*\hat{p})/n} \\ = 2.58*\sqrt{((1-0.25)*0.25)/300} \\ = 0.0645$

z score for 95% confidence interval is 1.96, sample size is n = 300 and p = 75/300 = 0.25

Margin of error for 99% =

$z*\sqrt{((1-\hat{p})*\hat{p})/n} \\ = 1.96 * \sqrt{((1-0.25)*0.25)/300} \\ = 0 .049$

(D) Margin of error from part (B) = (Upper limit- lower limit)/2 = (0.299-0.201)/2 = 0.049

Yes, both the margin of errors are same. This is because both the methods providing us exactly same value using same input data