Many investors and financial analysts believe the Dow Jones Industrial Average

Question

(DJIA) gives a good barometer of the overall stock market. On January 31, 2006, 9 of the 30 stocks making up the DJIA increased in price (The Wall Street Journal, February 1, 2006). On the basis of this fact, a financial analyst claims we can assume that 30% of the stocks traded on the New York Stock Exchange (NYSE) went up the same day. A sample of 72 stocks traded on the NYSE that day showed that 25 went up. You are conducting a study to see if the proportion of stocks that went up is significantly more than 0.3. You use a significance level of a=0.02 .

What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic = What is the p-value for this sample? (Report answer accurate to four decimal places.) p-value =

how to calculate by hand ,calculator or excel?

Answer 1

My response is going to focus on how to find the test statistic and p-value by hand (I'll also include a note at the bottom of how to check these findings in a TI-83 or TI-84 graphing calculator.)

Please note: For any hypothesis test, the first things you should do are set up the hypotheses, define the variables you are working with, and then check that the conditions are met in order to actually perform the test. My answer will focus on the calculations and we will simply assume the conditions have been met, but if you would like information on any other components, please let me know!

For this problem the null and alternative hypothesis are as follows:

H₀: p = .3

H_A: p > .3

where p is the true proportion of stocks that went up

To calculate any test statistic, we use the generic formula:

(statistic - parameter)/standard error of the statistic

For this specific problem we are working with a proportion so the formula for standard error (which changes depending on if we are working with a single proportion, two proportions, means, etc.) is:

sqrt(pq/n) --> in this problem p = .3, which means q = .7 (because q = 1 - p) and n is the sample size of 72

So if we look back at the overall test statistic that we are calculating, we would be plugging in the following values: statistic = sample proportion = 25/72, parameter = proportion from the null hypothesis = .3, and standard error = sqrt((.3*.7)/72)

Ultimately, this should give you a test statistic of approximately .8744 (which is really the z-score of the sample proportion).

To find the p-value based on this statistic, we are going to use our calculator to find the probability that a z-score would be greater than the test statistic we found. *Note: The reason we are finding the probability it is greater than, rather than less than, is because we are matching the alternate hypothesis which is that p > .3.

P(z > .8744) = normalcdf(.8744, 999999, 0, 1) = .1910 = p-value

*To get to normalcdf on the TI-83 or 84, hit 2nd-->VARS-->normalcdf*

To finish the entire hypothesis test, you would compare that p-value to alpha (.02) and since that p-value is greater than alpha we would fail to reject the null and therefore, not have enough evidence to support the alternative.

Checking Your Answer in the Calculator

Using a TI-83 or 84 of any kind, hit STAT-->TESTS-->1-PropZTest--> Enter the p = .3 (from the null), x = 25, n = 72 and select the appropriate alternate hypothesis --> Hit Calculate.

This will provide you with the test statistic (z) along with the p-value (p) which should match the calculations we did above.

^That was a lot of information. If you have follow-up questions or would like more explanation on the conditions, interpretations, or conclusions please let me know!