In: Statistics and Probability
Giving a test to a group of students, the grades and gender are
summarized below
A | B | C | Total | |
Male | 3 | 12 | 6 | 21 |
Female | 10 | 2 | 5 | 17 |
Total | 13 | 14 | 11 | 38 |
Let ππ represent the percentage of all male students who would
receive a grade of C on this test. Use a 80% confidence interval to
estimate p to three decimal places.
Enter your answer as a tri-linear inequality using decimals (not
percents).
< p <
Total number of students = 13+141+138 = 292
P[ percentage of all male students who would receive a grade of C on this test ] = 21/292
We need to construct the 80% confidence interval for the population proportion. We have been provided with the following information about the number of favorable cases:
Favorable Cases X = | 21 |
Sample Size N = | 292 |
The sample proportion is computed as follows, based on the sample size N = 292 and the number of favorable cases X = 21
The critical value for =0.2 is z_c = 1.282. The corresponding confidence interval is computed as shown below:
Therefore, based on the data provided, the 80% confidence interval for the population proportion is 0.053 < p < 0.091 , which indicates that we are 80\%80% confident that the true population proportion p is contained by the interval (0.053, 0.091)(0.053,0.091).