Exam Two Answers
Econ 180: Quantitative Methods
Professor Robert J. Lemke
Spring 2006

  1. (10 points) Suppose X is distributed uniformly from -2 to 6.

    1. What is the probability that X takes on a negative value?
      P(X < 0) = [0 - (-2)] / [6 - (-2)] = 2 / 8 = 1/4 = 0.25.
    2. What is the probability that X is between 4 and 8?
      P(4 < X < 8) = P(4 < X < 6) = [6 - 4] / [6 - (-2)] = 2 / 8 = 1/4 = 0.25.

  2. (25 points) Suppose Z ~ N(0,1). Calculate the following probabilities.

    1. P(1.2 < Z).
      P(1.2 < Z) = 0.5 - P(0 < Z < 1.2) = 0.5 - 0.3849 = 0.1151.
    2. P(-1.55 < Z).
      P(-1.55 < Z) = 0.5 + P(-1.55 < Z < 0) = 0.5 + P(0 < Z < 1.55) = 0.5 + 0.4394 = 0.9394.
    3. P(-0.09 > Z).
      P(-0.09 > Z) = .5 - P(-0.09 < Z < 0) = .5 - P(0 < Z < 0.09) = 0.5 - 0.0359 = 0.4641.
    4. P(-0.45 < Z < 0.45).
      P(-0.45 < Z < 0.45) = 2 * P(0 < Z < 0.45) = 2 * 0.1736 = 0.3472.
    5. P(Z < 1.87).
      P(Z < 1.87) = 0.5 + P(0 < Z < 1.87) = 0.5 + 4693 = 0.9693.

  3. (15 points) Suppose X ~ N(20,100).

    1. What is the probability that a single draw on X yields a value greater than 20?
      P(X > 20) = 0.5 as 20 is the mean value of the random variable.
    2. What is the probability that a single draw on X yields a value between 16 and 22?
      P(16 < X < 22) = P([16-20]/10 < [X-20]/10 < [22-20]/10) = P(-0.4 < Z < 0.2) = 0.1554 + 0.0793 = 0.2347.
    3. What is the probability that the average value from 100 draws on X yields a value between 16 and 22?
      P(16 < X-bar < 22) = P([16-20]/[10/sqrt(100)] < [X-20]/[10/sqrt(100)] < [22-20]/[10/sqrt(100)]) = P(-4.00 < Z < 2.00) = 0.5000 + 0.4772 = 0.9772.

  4. (10 points) In a random sample of 152 union workers, 94 claim that they vote Democrat regardless of who the candidate is. Construct the 95% two-sided confidence interval for the percent of all union workers that vote Democrate regardless of who the candidate is.
    We have that n = 152 and p-bar = p = 94/152 = 0.6184. As we want the 95% two-sided confidence interval, the critical value is z0.025 = 1.96. Thus, the 95% two-sided confidence interval is p +/- z0.025 * sqrt[p * (1-p) / n] = 0.6184 +/- (1.96)*sqrt[0.6184 * 0.3816 / 152] = 0.6184 +/- (1.96)*(0.0394) = 0.6184 +/- 0.0772 = [ 0.5412 , 0.6956 ].

  5. (10 points) Twenty-four randomly chosen households in Lake Forest are asked how much money they give to charity each year. These households, on average give $12,342 to charity each year with a sample standard deviation of $6,555. Suppose charitable gifts are distributed normally. Construct the 99% two-sided confidence interval for the average amount of money given to charity annually for all households in Lake Forest.
    We have that n = 24, Xbar = 12342, and s = 6555. As we want the 99% two-sided confidence interval for a small sample, the critical value is t230.005 = 2.8188. Thus, the 99% two-sided confidence interval for the population mean of charitable giving is Xbar +/- t230.005 * s / sqrt(n) = 12342 +/- 2.8188 * 6555 / sqrt(24) = 12342 +/- 3772 = [ $8570 , $16114 ].

  6. (10 points) The Republican Party wants to know how much support there is for an anti-immigration bill that is making its way through Congress. How many people should they survey if they want to eventually construct a 97% two-sided confidence interval of the population proportion in favor of the bill to have a margin of error of at most 3.5 percentage points?
    The critical value is z0.015 = 2.17. Now, assuming the true proportion might be one-half, we need that 2.17 * 0.5 / sqrt(n <= 0.035 which requires that sqrt(n) >= 31 which requires n >= 961. Thus, they should survey 961 people.

  7. (20 points) The folks at McDonalds headquarters have dictated that its franchise managers try to serve french fries at a temperature of 115 degrees fahrenheit. To check if managers are following headquarter policy, McDonalds randomly samples the temperature of 420 different french fry orders. The average temperature of the french fries in the sample was 112.4 degrees with a standard deviation of 19.3 degrees.

    1. What are the null and alternative hypotheses that McDonalds has in mind?
      Null Hypothesis: H0: m = 115. Alternative Hypothesis: HA: m ¹ 115.
    2. Test the null hypothesis at the 5% significance level.
      The test statistic = [115 - 112.4]/[19.3/sqrt(420)] = 2.76, while the critical value is z0.025 = 1.96. Thus, we reject the null hypothesis in favor of the alternative hypothesis.
    3. Is there evidence (at the 5% level) that franchises are not following headquarter policy?
      Yes, becaue the test-statistic exceeds the critical value.
    4. What is the p-value of the test?
      As the test-statistic = 2.76, the p-value = 2* [0.5 - 0.4971] = 0.0058 = 0.58%.