Question

Let x be a random variable representing dividend yield of bank stocks. We may assume that x has a normal distribution with σ = 2.1%. A random sample of 10 bank stocks gave the following yields (in percents). 5.7 4.8 6.0 4.9 4.0 3.4 6.5 7.1 5.3 6.1 The sample mean is x = 5.38%. Suppose that for the entire stock market, the mean dividend yield is μ = 4.6%. Do these data indicate that the dividend yield of all bank stocks is higher than 4.6%? Use α = 0.01

Answer 1

Answer:

$z=\frac{5.38-4.6}{\frac{2.1}{\sqrt{10}}}=1.175$  

The p value can be founded with the following probability:

$p_v =P(z>1.175)=0.120$  

Since the p value is higher than the significance level of 0.01 we have enough evidence to FAIL to reject the null hypothesis and we can't conclude that the true mean for this case is higher than 4.6%

Step-by-step explanation:

Information provided

$\bar X=5.38$ represent the sample mean

$\sigma=2.1$ represent the population standard deviation

$n=10$ sample size  

$\mu_o =4.6$ represent the value to test

$\alpha=0.01$ represent the significance level  

z would represent the statistic

$p_v$ represent the p value

Hypothesis to verify

We want to check if the true mean for the dividens is higher than 4.6%, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 4.6$  

Alternative hypothesis:$\mu > 4.6$  

The statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)  

Replacing the info given we got:

$z=\frac{5.38-4.6}{\frac{2.1}{\sqrt{10}}}=1.175$  

The p value can be founded with the following probability:

$p_v =P(z>1.175)=0.120$  

Since the p value is higher than the significance level of 0.01 we have enough evidence to FAIL to reject the null hypothesis and we can't conclude that the true mean for this case is higher than 4.6%