Answers S b e the mean return for a random sample of 1. Let S b e the S&P return for a randomly chosen dayand 20 days. (a) E ( S )=E (S )=:00032 p p (b) SD(S )=SD(S )= n = :00859= 20 = :00192 (c) :005 :00032 S>:005) = P Z> = P (Z>2:438) = :00074 P ( :00192 Without doing any calculations we know that it is more likely that an individual day's return will b e greater than .005 than that the mean return will b e, since the standard deviation of S is smaller than that of S . 2.(a) The graphical output for descriptive statistics gives all of the information we need: Descriptive Statistics Variable: Freshman Anderson-Darling Normality Test A-Squared: 0.794 P-Value: 0.038 Mean 82.3778 StDev 9.3983 Variance 88.3276 Skewness -3.2E-01 Kurtosis -7.8E-01 62.5 70.0 77.5 85.0 92.5 N 90 Minimum 59.0000 1st Quartile 75.0000 Median 83.0000 3rd Quartile 90.0000 95% Confidence Interval for Mu Maximum 98.0000 95% Confidence Interval for Mu 80.4093 84.3462 80 81 82 83 84 85 95% Confidence Interval for Sigma 8.1973 11.0148 95% Confidence Interval for Median 95% Confidence Interval for Median 81.2373 85.0000 So, the 95% con dence interval for the true average freshman retention rate is (80.409, 84.346). The interval requires that the variable b e reasonably normally distributed, and that the sample b e a random sample from a larger p opulation. The latter requirement is met, and the former is not to o bad. (b) The prediction interval is then r p 1 89 X t 1+ =82:378 (1:987)(9:398) 1:0111 = 82:378 18:777 = (63:601; 101:155): s :025 90 Note that the upp er limit of the interval go es past the highest p ossible value of 100, which re ects the slight left tail in the distribution. The actual value for NYU, bytheway,was 86. c 1999, Je rey S. Simono (c) Here is a graphical representation of the exp enditure variable: Descriptive Statistics Variable: Expend. Anderson-Darling Normality Test A-Squared: 6.531 P-Value: 0.000 Mean 14676.7 StDev 9801.9 Variance 96077185 Skewness 3.09663 Kurtosis 13.6810 10000 25000 40000 55000 70000 N 91 Minimum 6093.0 1st Quartile 8742.0 Median 11999.0 3rd Quartile 16923.0 95% Confidence Interval for Mu Maximum 73967.0 95% Confidence Interval for Mu 12635.4 16718.1 10000 11000 12000 13000 14000 15000 16000 17000 95% Confidence Interval for Sigma 8555.4 11476.9 95% Confidence Interval for Median 95% Confidence Interval for Median 10486.7 13416.6 The con dence interval for average (over all researchuniversities) mean (over all students within the university) exp enditure per student is thus (12635, 16718). This interval is not very useful, however, since the exp enditure variable is very long right{tailed. (d) Given the long right tail in the distribution, a prediction interval must b e constructed in the correct scale. The waytodoitistologthevariable, construct the prediction interval, and then antilog. Here is a normal plot for the logged exp enditure variable, which lo oks more Gaussian (it's still not great, b eing right{tailed, but there's no stronger transformation than the log to try): c 1999, Je rey S. Simono Normal Probability Plot for Logged e 99 Mean: 4.10586 StDev: 0.214029 95 90 80 70 60 50 40 Percent 30 20 10 5 1 3.5 3.7 3.9 4.1 4.3 4.5 4.7 4.9 Data The prediction interval in the logged scale has the form r p 1 90 =4:106 (1:987)(:214) X t 1+ 1:011 = 4:106 :4276 = (3:678; 4:534): s :025 91 3:678 4:534 Antilogging this interval gives (10 ; 10 ), or (4764; 34198). The actual value for NYU was 24135. Note that this interval is much more reasonable than the one done in the original scale, whichis 14677 (1:987)(9802)(1:0055) = 14677 19584 = (4907; 34261): (e) In the sample 30 of the 90 institutions have freshman retention rate less than 80%, so a con dence interval for the probability that a researchuniversity has retention rate less than 80% is s r 1 2 1 p(1 p) 3 3 = (1:96) = :333 :0974: p z :025 n 3 90 The continuity{corrected interval adds 1=180 = :0056 to the p ortion of the interval, giving :333 :103. 3.(a) From the marginal distribution determined earlier, the con dence interval has the form s 203 203 1 203 362 362 (1:645) ; 362 362 c 1999, Je rey S. Simono or :561 :043 = (:518;:604). The continuity{corrected interval is :561 (:043 + :001) = (:517;:605). (b) Using the joint distribution of employment opp ortunities and salary size, and counting the num- b er of resp ondents who rated salary as more imp ortant than employment opp ortunities gives the con dence interval s 58 58 1 58 362 362 (1:645) ; 362 362 or :16 :032 = (:128;:192). The continuity{corrected interval is :16 (:032 + :001) = (:127;:193). 4.(a) Here is MINITAB output: Confidence Intervals Variable N Mean StDev SE Mean 95.0 % CI Weekly p 88 0.880 8.186 0.873 ( -0.855, 2.614) The con dence interval for true average price change includes zero, so the observed value is not signi cantly di erent from zero. The price change variable is reasonably Gaussian here. p (b) A prediction interval for the price change of a randomly chosen sto ckis:88 (1:988)(8:186) 1:011 = :88 16:363. The wide interval re ects the large variability of sto ck prices. (c) Here are values separated bymarket: Descriptive Statistics Variable Market N Mean Median Tr Mean StDev SE Mean Weekly p OTC 51 0.64 0.00 0.09 9.43 1.32 NYSE 24 -0.03 -0.85 -0.58 5.96 1.22 ASE 13 3.49 1.30 2.79 6.18 1.71 Variable Market Min Max Q1 Q3 Weekly p OTC -19.40 34.20 -5.00 4.40 NYSE -7.60 19.70 -4.55 1.38 ASE -4.10 18.80 -1.05 6.30 The con dence intervals for eachmarketareasfollows: p 51 = :64 2:65 OTC ::64 (2:009)(9:43)= p 24 = :03 2:52 NYSE : :03 (2:069)(5:96)= p ASE :3:49 (2:179)(6:18)= 13 = 3:49 3:73 All three intervals contain zero, so the average returns in eachwere not signi cantly di erentfrom zero. Note that the ASE interval was close, but didn't quite make it b ecause of the small sample size. A sample of size 15 with the same mean and standard deviation would havegiven an interval that did not contain zero. The distributions of price change within each group are reasonably Gaussian here. (d) Here are prediction intervals: OTC ::64 (2:009)(9:43)(1:01) = :64 19:134 NYSE : :03 (2:069)(5:96)(1:021) = :03 12:59 ASE :3:49 (2:179)(6:18)(1:038) = 3:49 13:978 Note that the intervals for NYSE and ASE are narrower than that for all of the sto cks together, which re ects the gains in treating the markets separately (despite the reduction in e ective sample size). c 1999, Je rey S. Simono 5.(a) The mean of X is just the p opulation mean , or .00039. p p (b) The standard deviation (or standard error) of X is = n,or:00721= 30 = :00132. (c) :01 :00039 P (X>:01) = P Z> :00132 = P (Z>7:28) 0 It is more likely that the sample mean return will b e less than .01 than an individual return, since the mean is much more tightly centered around the p opulation mean. In fact, the probabilityof individual return b eing less than .01 is ab out .91. 6. Before we do anything, we should recognize that we're assuming that our data are a random sample from a roughly Gaussian distribution. Normal plots for return and yield show that the while the yield variable is reasonably Gaussian (with one p ossibly unusual lowvalue), returns are not, b eing bimo dal. In fact, as the structure of the data in the le suggests, there are twowell{de ned subgroups here: no load funds, and load funds. Normal Probability Plot for Yield ML Estimates Mean: 4.75179 99 StDev: 0.501942 95 90 80 70 60 50 40 30 Percent 20 10 5 1 3 4 5 6 Data Normal Probability Plot for Return ML Estimates Mean: 4.22628 99 StDev: 2.22312 95 90 80 70 60 50 40 30 Percent 20 10 5 1 0 5 10 Data Stem-and-leaf of Return N = 78 Leaf Unit = 0.10 c 1999, Je rey S. Simono 1 -1 7 2 -0 4 3 0 5 7 1 5789 27 2 33333445556777888889 (20) 3 00112223344444467889 31 4 445689 25 5 36667 20 6 023456 14 7 023444489 5 8 11278 p (n1) (n1) X t s= n. Since n =78, t =1:99. (a) Wewant a con dence interval, which has the form =2 =2 Here is the answer from Minitab: Variable N Mean StDev SE Mean 95.0 % CI Return 78 4.226 2.238 0.253 ( 3.722, 4.731) The nonnormality of returns here is less worrying, since the sample size is somewhat large, so this interval mightbeokay.