Question

An article published in the Washington Post claims that 45 percent of all Americans have brown eyes. A random sample of ?=76n=76 college students students found 28 who had brown eyes.
Consider testing:

?0:?=.45
??:?≠.45

a) the test statistic is z = ?

b)P-value = ?

Answer 1

Given that, x = 28. n = 76 The sample proportion is, x_28 -0.3684 n=- = - =0.300+ n 76 Null and alternative hypothesis: Ho: P=0.45 H:P+0.45 (Two tailed test) Significance level: a = 0.05 Test statistic: Zap-P P(1-P) 0.3684-0.45 10.45(1-0.45) V 76 =-1.43 P-value:

(Standard normal tables) p=2P(Z >[zel) = 2[1-P(Z 51.43)] = 2[1-0(1.43)] = 2[1 – 0.9236] = 20.0764] = 0.1527 Decision rule: Reject Ho, if psa otherwise do not reject Ho. Since the p-value is greater than the level of significance, so we fail to reject the null hypothesis at the specified significance level. Conclusion: There is sufficient evidence to rejection of the claim that population proportion is equal to 0.45.