Question

The following frequency distribution shows the monthly stock returns (in percent) for Home Depot for the years 2003 through 2007. (You may find it useful to reference the appropriate table: chi-square table or F table)

Monthly Returns	Observed Frequency
Less than ?5	13
?5 up to 0	16
0 up to 5	20
5 or more	11
	n = 60

SOURCE: www.yahoo.finance.com.

Over this time period, the following summary statistics are provided:

Mean	Median	Standard Deviation	Skewness	Kurtosis
0.31%	0.43%	6.49%	?0.15	?0.38

a-1. Choose the null and alternate hypotheses.

H₀: The monthly stock returns follow a normal distribution with a mean of 0.31% and a standard deviation of 6.49%.; H_A: The monthly stock returns do not follow a normal distribution with a mean of 0.31% and a standard deviation of 6.49%.

H₀: The monthly stock returns do not follow a normal distribution with a mean of 0.31% and a standard deviation of 6.49%.; H_A: The monthly stock returns follow a normal distribution with a mean of 0.31% and a standard deviation of 6.49%.

a-2. Calculate the value of the test statistic. (Round intermediate calculations to at least 4 decimal places and final answer to 3 decimal places.)

Test statistic _____________?

a-3. Find the p-value.

p-value < 0.01

0.01 ? p-value < 0.025

0.025 ? p-value < 0.05

0.05 ? p-value < 0.10

p-value ? 0.10

a-4. Can you conclude that monthly returns do not follow the normal distribution at the 5% significance level?

Yes since the p-value is more than ?.

Yes since the p-value is less than ?.

No since the p-value is less than ?.

No since the p-value is more than ?.

b-1. Using the Jarque-Bera test, calculate the value of the test statistic. (Round intermediate calculations to at least 4 decimal places and final answer to 3 decimal places.)

Test Statistic = 0.586

b-2. Find the p-value.

p-value < 0.01

0.01 ? p-value < 0.025

0.025 ? p-value < 0.05

0.05 ? p-value < 0.10

p-value ? 0.10

b-3. Can you conclude that monthly returns do not follow the normal distribution at the 5% significance level?

No since we do not reject H₀.

Yes since we reject H₀.

Yes since we do not reject H₀.

No since we reject H₀.

I actually only need help with the test statistic below a-2.   Already figured out the rest of the question but wanted to post the whole question just in case. As for the a-2 test statistic I keep on getting 27.047, but showing up incorrect.  

Answer 1

Helo for test statistics below a-2

Here the test is chi-square goodness of fit since we have to check the given data follows a normal distribution with given parameters or not.

So the formula to find the Chi-square test statistics is:

Where O - Observed frequencies which are given in table

E - Expected frequencies have to find using the given information.

To find Expected values first have to find the probability for the given monthly returns.

The mean and standard deviation are:

Let us find the probabilities for all groups of monthly return.

Less than -5 that first has to find P(X < -5)

The z score formula is,

That is P(X < -5) becomes P(Z < -0.82)

by using the z table the probability is, 0.2061

-5 up to 0 means P(-5 < X < 0)

First, find the z scores for -5 and 0

The z score for -5 is -0.82 and the z score for 0 is

That is P(-5 < X < 0) becomes P(-0.82 < z < -0.05)

By using z table find the probability for both z scores and then subtract small from large to get between

The probability for z = -0.82 is 0.2061 and for z = -0.05 is 0.4801

The subtraction of bot is 0.274

0 up to 5 means P(0 < X < 5)

The z score for 0 is -0.05

and z score for 5 is,

That is P(0 < X < 5) becomes P(-0.05 < Z < 0.72)

The probability for z = -0.05 is 0.4801 and the probability for 0.72 is 0.7642

So the subtraction of small from large is, 0.2841

5 or more means P(X > 5)

The z score for x = 5 is 0.72

The less than probability for 0.72 is 0.7642, so to get more than just subtract this less from 1

P(Z > 0.72) = 1 - 0.7642 = 0.2358

Now multiply all these probabilities by n that is 60 will get the expected values.

Less than -5 0.2061 * 40 = 12.366

-5 up to 0 0.274 * 60 = 16.44

0 up to 5 0.2841 * 60 = 17.046

5 or more 0.2358 * 60 = 14.148

And the observed frequencies are given as 13, 16, 20 and 11