Basics of Investing

Question: math problems?

Corporate advertising tries to enhance the image of the corporation. A study compared two ads from two sources, the Wall Street Journal and the Ntional Enquirer. Subjects were asked to pretend that their company was considering a major investment in Performax, the fictitious sportswear firm in the ads. Each subject was askedto respond to the question “how trustworthy was te source in the sportswear company ad for Performax?” on a 7-point scale. Higher values indicatedmore trustwortinss. Here is a summary of the resulets:

Ad source, n, VARx, s
Wall Street Journal, 66, 4.77, 1.50
NationalEnquirer, 61, 2.43, 1.64

(a) Compare the two sources of ads using a “t test”. Be sure t state your null and alternative hypotheses, the test statistic with degree offreedom, the P-value, and your conclusion.

(b)Give a 95% confidence interval for the difference.

(c) Write a short paragraph summarizing the results of your analyses.

Answer: MEANS TEST, INDEPENDENT SAMPLES, t DISTRIBUTION (“TWO-TAILED”) – SEVEN STEP PROCEDURE

1. Parameters of interest: “μ1″ = “Wall Street Journal trustworthy” [WSJ] population mean, “μ2″ = “National Enquirer trustworthy” [NE] population mean

2. Null Hypothesis H0: (μ1 – μ2) = 0 (no difference between means)

3. Alternative Hypothesis H1: (μ1 – μ2) <> 0 (different between means)

4. Sample Data
WSJ x1-bar = sample mean [4.77]
s1=sample standard deviation [1.50]
n1=number of samples [66]

NE x2-bar = sample mean [2.43]
s2=sample standard deviation [1.64]
n2=number of samples [61]

5. t-distribution “look-up” value df (degrees of freedom) = n1 + n2 – 2 = 66 + 61 – 2 = 125

Critical t (df = 125, α = 0.05/2 “TWO-TAILED”) = (approximately) +/-1.98

6. Computation of Test Statistic
DENOMINATOR
sigma^2 (Estimate of Difference Variance) = [(n1 - 1)*s1^2 + (n2 - 1)*s2^2] / (n1 + n2 -2) = [(65) *1.5^2 + (60)*2.43^2] / (125) = 500.544

sigma x1,x2 (Estimate of Differences of Variances) = SQRT[ sigma^2/n1 + sigma^/n2] = SQRT [ 3796.13 + 4107.28 ] = 88.9

t = (x1 – x2)/(sigma x1,x2) = (4.77 – 2.43)/(88.9) = 0.026

7. Conclusion: Computed t = 0.026 is inside the region of acceptance [Crtitical t +/-1.98] of the Null Hypothesis H0: (μ1 – μ2) = 0 (no difference between means). Therefore the Null Hypothesis must be accepted at the α = 0.05 level of signficance.

Nassim Nicholas Taleb – Air date: 12-03-01

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