Question

The price to earnings ratio (P/E) is an important tool in financial work. A random sample of 14 large U.S. banks (J. P. Morgan, Bank of America, and others) gave the following P/E ratios†.24 16 22 14 12 13 17 22 15 19 23 13 11 18
The sample mean is x ≈ 17.1. Generally speaking, a low P/E ratio indicates a "value" or bargain stock.
Suppose a recent copy of a magazine indicated that the P/E ratio of a certain stock index is μ = 18.
Let x be a random variable representing the P/E ratio of all large U.S. bank stocks.
We assume that x has a normal distribution and σ = 5.1.

Do these data indicate that the P/E ratio of all U.S. bank stocks is less than 18? Use α = 0.01.(a) What is the level of significance?(b) What is the value of the sample test statistic? (Round your answer to two decimal places.)(c) Find (or estimate) the P-value. (Round your answer to four decimal places.)

Answer 1

Answer:

a) $\alpha=0.01$ is the significance level given

b) $z=\frac{17.1-18}{\frac{5.1}{\sqrt{14}}}=-0.6603$    

c) Since is a one side left tailed test the p value would be:  

$p_v =P(Z<-0.6603)=0.2545$  

Step-by-step explanation:

Data given and notation  

$\bar X=17.1$ represent the mean P/E ratio for the sample  

$\sigma=5.1$ represent the sample standard deviation for the population  

$n=14$ sample size  

$\mu_o =18$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean for the P/E ratio is less than 18, the system of hypothesis would be:  

Null hypothesis:$\mu \geq 18$  

Alternative hypothesis:$\mu < 18$  

If we analyze the size for the sample is < 30 but we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

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

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

(a) What is the level of significance?

$\alpha=0.01$ is the significance level given

(b) What is the value of the sample test statistic?

We can replace in formula (1) the info given like this:  

$z=\frac{17.1-18}{\frac{5.1}{\sqrt{14}}}=-0.6603$    

(c) Find (or estimate) the P-value. (Round your answer to four decimal places.)

Since is a one side left tailed test the p value would be:  

$p_v =P(Z<-0.6603)=0.2545$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL reject the null hypothesis, so we can conclude that the true mean for the P/E ratio is not significantly less than 18.