Question

4.170 Change in Stock Prices. Standard & Poor's maintains one of the most widely followed indices of large-cap American stocks: the S&P 500. The index includes stocks of 500 companies in industries in the US economy. A random sample of 50 of these companies was selected, and the change in the price of the stock (in dollars) over the 5-day period from August 2 to 6, 2010 was recorded for each company in the sample. The data are available in StockChanges.

a. Use StatKey to calculate a 95% confidence interval for the mean change in all S&P stock prices over these dates using the bootstrap percentiles method. Include a screen shot of your Statkey output with your homework submission, and write the confidence interval below.

b. Use only the confidence interval you created (do not find a p-value) to predict the results of a hypothesis test to see if the mean change for all S&P 500 stocks over this period is different from zero.

i. Define the parameter.

ii. State the hypotheses.

iii. What significance level are you able to use based on the confidence interval?

iv. State the conclusion using nontechnical language.

c. If any error were to occur in the decision to reject or fail to reject, would it be a Type I error or a Type II error?

d. Explain what an error of this type would mean in context.

SPChange
0.29
-0.06
0.34
0.7
0.42
0.22
0.12
0.03
-0.5
0.36
0.03
0.09
-0.12
0.03
-0.47
-3.27
0.35
-0.06
0.01
0.6
0.12
4.86
-0.77
-0.03
0.39
0.1
-0.12
0.47
-0.05
0.06
-0.03
0.15
0.31
-0.15
0.32
-2.66
0.22
-0.03
0.09
0.29
0.16
0.38
0.1
0.21
0.09
0.33
0.18
1.93
0.14
0.03

Answer 1

A. Now coming to the question, let's derive the 95% CL.

Sample Mean = 0.124

Sample Std Deviation = 0.991599

Sample Size = 50

Standard Error of sample mean = Sample SD / sqrt (Sample size) = 0.991599/ sqrt(50) = 0.140233

As sample size is more than 50, we can use a std normal distribution for our analysis. (z statistics)

The critical Z stat at 95% CL from a two tailed test = 1.96

So the 95% CL around sample mean = 0.124 +/- (z*std error)

Upper limit = 0.124 + (1.96*0.140233) = 0.398857

Lower Limit = 0.124 - (1.96*0.140233) = -0.15086

B.

i. Define the parameter: Population Mean (all S&P 500 mean change) = 0

ii. State the hypotheses: Null Hypothesis: H0: Population Mean (S&P 500 mean) = 0

Alternate Hypothesis: H1: Population Mean (S&P 500 mean) <> 0

iii. What significance level are you able to use based on the confidence interval? Based on the 95% confidence level, we can use 5% significance leve (1-CL).

iv. State the conclusion using nontechnical language: As the 95% confidence level from the sample mean is between -0.15086 and 0.398857 which includes 0, we can't reject the null hypothesis that the population mean change is 0.

c. If any error were to occur in the decision to reject or fail to reject, would it be a Type I error or a Type II error?

As H0 (null hypothesis) is true, if we are rejecting the null hypothesis when it's true, it would have resulted in Type I error.

However if H0 was false, if we fail to reject the null hypothesis when it's false, it would result in Type II error.

d. Explain what an error of this type would mean in context:

Type I error: We reject the null hypothesis that the mean change in S&P 500 is 0 when actually mean change is 0.

Type II Error: Let's assume that the mean change of S7P 500 is not 0. If we fail to reject the null hypothesis that the mean change of S&P 500 is zero, then it's a Type II error.

Hope it clarifies your question.