Question

Are most student government leaders extroverts? According to Myers-Briggs estimates, about 82% of college student government leaders are extroverts. Suppose that a Myers-Briggs personality preference test was given to a random sample of 72 student government leaders attending a large national leadership conference and that 57 were found to be extroverts. Does this indicate that the population proportion of extroverts among college student government leaders is different (either way) from 82%? Use a = 0.01. (a) What is the level of significance? State the null and alternate hypotheses O HG:P = 0.82; H:P = 0.82 O Ho: P = 0.82; H4: p = 0.82 O Ho: P = 0.82; H7: p > 0.82 O H:P > 0.82; H: p = 0.82 0 Hz: p = 0.82; H:p < 0.82 (b) What sampling distribution will you use? The Student's t, since np < 5 and ng < 5. The standard normal, since np > 5 and ng > 5. The standard normal, since np < 5 and ng < 5. The Student's t, since np > 5 and ng > 5. What is the value of the sample test statistic? (Round your answer to two decimal places.) (c) Find the P-value of the test statistic. (Round your answer to four decimal places.) Sketch the sampling distribution and show the area corresponding to the P-value

Answer 1

Solutim: (a) level of significance 2=0.0) The null and alternate hypothes os are Ho : P = 0. 8 2 ) H:P+0.8 ) (b) The number of trails is n = 72., and the number of sucesses is r=51 - D - 52 = 0.4919 From Ho P=0.82 and 9-1-P 1-0.7914 = 0,2083 ną= 7200.2083) + = 14.99765 np = 72 (0.7911) = 51.0024 25 The Sampling distribution will use the Standard normal, Since DP >5 and ng >5. So, we can use the normal distribution for 7 -P - 0.7914-0.82 -0.0283 -0.0283 0.82) (0.2003) 0.170 806 0.0487 ✓ 72 Do170806 12 = -0.58 .: Z=-0.58 For two tailed test, using normal distribution , we nue - 2 PL7 (-0.58)= 2(0.28096) = 0.56192 find that (C) For two tan P-value R Value 0.58 -0.58 § = 0.7917 Sample Test statistic.