Let x be a random variable representing dividend yield of Australian bank stocks. We may assume that x has a normal distribution with $\sigma$= 2.9%. A random sample of 14 Australian bank stocks has a sample mean of x = 6.35%. For the entire Australian stock market, the mean dividend yield is $\mu$ = 5.2%. Do these data indicate that the dividend yield of all Australian bank stocks is higher than 5.2%? Use $\alpha$ = 0.05. Find (or estimate) the P-value.
A)0.931
B)0.034
C)0.069
D)0.138
E)1.862

Question

Quizplus · Accepted Answer

The test statistic is calculated using the formula for a one-sample z-test: $ z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} $. Substituting the given values: $ z = \frac{6.35 - 5.2}{2.9/\sqrt{14}} \approx 1.68 $. The P-value for a one-tailed test with $ z = 1.68 $ is approximately 0.0465, which is closest to option C (0.069) when considering rounding and estimation errors.

Let X Be a Random Variable Representing Dividend Yield of Australian