Probability 1. According to Investment Digest (”Diversification and the Risk/Reward Relationship”, Winter 1994, 1-3), the mean of the annual return for common stocks from 1926 to 1992 was 15.4%, and the standard deviation of the annual return was 24.5%. During the same 67-year time span, the mean of the annual return for long-term government bonds was 5.5%, and the standard deviation was 6.0%. The article claims that the distributions of annual returns for both common stocks and long-term government bonds are bell-shaped and approximately symmetric. Assume that these distributions are distributed as normal random variables with the means and standard deviations given previously. Find the probability that the return for common stocks will be greater than 0%. Find the probability that the return for common stocks will be less than 20%.
Let X denote the return for common stocks. Here it is assumed that X follows a Normal distribution with mean 15.4% and standard deviation 24.5%.
Thus, Z=(X-15.4)/24.5 follows a Standard Normal distribution.
1) Now, the probability that the return for common stocks will be greater than 0% is given by,
P[X > 0] = P[(X-15.4)/24.5 >(0-15.4)/24.5]
= P[Z > -0.62857]
= 1 - P[Z < -0.62857]
= 1 - 0.2648
= 0.7352
2) The probability that the return for common stocks will be less than 20% is given by,
P[X < 20] = P[(X-15.4)/24.5 < (20-15.4)/24.5]
= P[Z < 0.18776]
= 0.5745
Note the probability P[Z < a] can be calculated using the Excel formula =NORMSDIST(a)
Confidence Interval Estimation 2. Compute a 95% confidence interval for the population mean, based on the sample 10, 12, 13, 14, 15, 16, and 49. Change the number from 49 to 16 and recalculate the confidence interval. Using the results, describe the effect of an outlier or extreme value on the confidence interval.
A 95 % confidence interval for the population mean is
(xbar – t*s/√n, xbar + t*s/√n),
where xbar is the sample mean , s is the sample standard deviation and t* is the tabled value from Student’s t distribution with (n-1) = 6 degrees of freedom.
i) Here, xbar = 18.4286, s= 13.6242, t* = 2.447
Thus a 95 % confidence interval for the population mean is
(18.4286 – 2.447*13.6242/√7, 18.4286 + 2.447*13.6242/√7)
= (5.8283, 31.0289)
ii) Here, xbar = 13.7143, s= 2.2147, t* = 2.447
Thus a 95 % confidence interval for the population mean is
(13.7143 – 2.447*2.2147/√7, 13.7143 + 2.447*2.2147/√7)
= (11.6661, 15.7625)
Thus the presence of an outlier or an extreme value, increases the width of the confidence interval and give meaningless confidence limits.
Hypothesis Testing 3. The director of admissions at the University of Maryland, University College is concerned about the high cost of textbooks for the students each semester. A sample of 25 students enrolled in the university indicates that X (bar) = $315.4 and s = $43.20. a. a. Using the 0.10 level of significance, is there evidence that the population mean is above $300? b. b. What is your answer in (a) if s = $75 and the 0.05 level of significance is used? c. c. What is your answer in (a) if X (bar) = $305.11 and s = $43.20? d. d. Based on the information in part (a), what decision should the director make about the books used for the courses if the goal is to keep the cost below $300?
a) Here the null hypothesis is Ho: μ = 300 and the alternative hypothesis is Ha: μ >300.
The test Statistic, t = (Xbar- 300)/(s/√n) = (315.4 -300)/(43.2/√25) = 1.7824
Degrees of freedom = n-1 = 25 -1 = 24
Level of significance = 0.10
The critical value = 1.318
Critical region is t > 1.318.
Here, t = 1.7824 > 1.318. So we reject the null hypothesis Ho.
Thus we can conclude that the population mean is above $300.
b) Here the null hypothesis is Ho: μ = 300 and the alternative hypothesis is Ha: μ >300.
The test Statistic, t = (Xbar- 300)/(s/√n) = (315.4 -300)/(75/√25) = 1.0267
Degrees of freedom = n-1 = 25 -1 = 24
Level of significance = 0.05
The critical value = 1.711
Critical region is t > 1.711.
Here, t = 1.0267 < 1.711. So we fail to reject the null hypothesis Ho.
Thus we can conclude that the population mean is not above $300.
c) Here the null hypothesis is Ho: μ = 300 and the alternative hypothesis is Ha: μ >300.
The test Statistic, t = (Xbar- 300)/(s/√n) = (305.11 -300)/(43.2/√25) = 0.5914
Degrees of freedom = n-1 = 25 -1 = 24
Level of significance = 0.10
The critical value = 1.318
Critical region is t > 1.318.
Here, t = 0.5914 < 1.318. So we fail to reject the null hypothesis Ho.
Thus we can conclude that the population mean is not above $300.
d) Here the null hypothesis is Ho: μ = 300 and the alternative hypothesis is Ha: μ <300.
The test Statistic, t = (Xbar- 300)/(s/√n) = (315.4 -300)/(43.2/√25) = 1.7824
Degrees of freedom = n-1 = 25 -1 = 24
Level of significance = 0.10
The critical value = -1.318
Critical region is t < -1.318.
Here, t = 1.7824 > -1.318. So we fail to reject the null hypothesis Ho.
Thus we can conclude that the population mean is below $300.
4. A large candy manufacturer is concerned that the mean weight of their bag of Gooey Sour Worms is not greater than 7.3 ounces. It can be assumed that the population standard deviation is .5 ounces based on past experience. A sample of 169 gummy worms is selected and the sample mean is 7.35 ounces. Using a level of significance of .10, is there evidence that the population mean weight of the candy bars is greater than 7.3? Fully explain your answer.
Let μ denote the mean weight of the candy bars.
Here the null hypothesis is Ho: μ ≤ 7.3 and the alternative hypothesis is Ha: μ >7.3.
The test Statistic, z = (Xbar- 7.3)/(σ/√n) = (7.35 -7.3)/(0.5/√169) = 1.3
Level of significance = 0.10
The critical value = 1.282
Critical region is z > 1.282.
Here, t = 1.3 > 1.282. So we reject the null hypothesis Ho.
Thus we can conclude that the population mean weight of the candy bars is greater than 7.3.
