/4 points A web based software company is interested in estimating the proportio

Question

/4 points A web based software company is interested in estimating the proportion of i sample of 200 of for the proportion of all individuals that use Firefox. My Notes ndividuals who use the Firefox browser. In a individuals, 31 users stated that they used Firefox. Using this data, construct a 99% confidence interval a) what is the lower limit on the 99% confidence interval? Give your answer to three decimal places. I b) what is the upper limit on the 99% confidence interval? Give your answer to three decimal places. c) Google states that the proportion of all individuals that use Firefox is 0.2. Based on the interval above, does the Google claim seem reasonable? No because o.2 is not inside the interval. Yes because 0.2 is not inside the interval. No because 0.2 is inside the interval. Yes because 0.2 is inside the interval. d) If you hadn't already taken the sample listed above, what sample size would be required so that the width of the 99% confidence interval would be at most o.02 units wide? Be as conservative as possible with your answer!

Explanation / Answer

Using MINITAB

The command for one sample proportion z test in minitab is

Stat>>>Basic statistics>>>1-proportion...

Then click on summarized data

number of events = x = 31

Number of trials = n = 200

Click on "Perform hypothesis test

then click on option

Level of confidence in percentage = c = ( 1- lpha)*100 = = (1 -0.01)*100 = 99.0

so put "Confidence level " = 99.0

Alternative = Not equal

then click on "Use test and interval based on normal approximation"

Then click on OK and again click on OK

So we get the following output

MTB > POne 200 31;
SUBC>   Confidence 99;
SUBC>   UseZ.

Test and CI for One Proportion

Sample   X    N     Sample p         99% CI
1     31 200 0.155000 (0.089083, 0.220917)

Using the normal approximation.

a]

Lower limit of 99% CI is 0.089

b]

Upper limit of 99% CI is 0.221

c]

here we test population proportion is 0.2

MTB > POne 200 31;
SUBC>   Test 0.2;
SUBC>   Confidence 99;
SUBC>   UseZ.

Test and CI for One Proportion

Test of p = 0.2 vs p not = 0.2

Sample   X    N     Sample p         99% CI          Z-Value     P-Value
1      31 200   0.155000   (0.089083, 0.220917)    -1.59       0.112

Using the normal approximation.

Here P-value is greater than alpha value so we do not reject null hypothesis.

Conclusion:

Yes, the Google claim seem reasonable because 0.2 is inside the interval.

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