x
-------------------------------------------------------------
Percentiles Smallest
1% 1 1
5% 1 1
10% 1 3 Obs 11
25% 3 3 Sum of wgt. 11
50% 4 Mean 5
Largest Std. dev. 3.162278
75% 8 7
90% 9 8 Variance 10
95% 10 9 Skewness .238797
99% 10 10 Kurtosis 1.694ECN 102: Analysis of Economics Data
Midterm Exam FQ 2024 – Answer Key
This exam consists of 5 short-answer questions (with sub-parts) and 5 multiple choice questions. You will have a maximum of 50 minutes to complete this exam. No additional time may be taken without prior accommodation. Make sure to read every question thoroughly and carefully. Make sure to answer all parts of all questions completely and fully explain when asked. Show all work to receive full credit.
You may use only the calculators provided by the instructor. Graphing calculators, phone calculators, and other non-standard calculators may not be used. For questions requiring computation, it suffices to express your final answer with 2 significant figures.
There is a formula sheet and scratch paper provided at the end of this exam. You may remove these pages and discard them after the exam. If you plan to use any of these extra pages for your final answers, you must write your name and student ID at the top of each scratch page to ensure it is not lost.
This exam is worth 70 points.
Question 1: Sample Means [10 points]
Suppose we know \(Q\sim(16,16)\) and we draw 50 samples of size 121 and obtain 50 sample means.
- How would we properly denote/write the sample mean of \(Q\)? (You do not need to explain) [2 points] \(\bar{Q}\)
For the distribution of sample means of Q:
- What would you expect the population mean to be equal to? [2 points]
We know \(\mu_{\bar{Q}}=\mu_Q\), so \(\mu_{\bar{Q}}=16\).
- What would you expect the population standard deviation to be equal to? [3 points]
We know \(\sigma_{\bar{Q}}=\sigma_Q/\sqrt{n}\), so \(\sigma_{\bar{Q}}=\frac{4}{11}\).
- What range would you expect 68% of the sample means of \(Q\) to fall under? Explain. [3 points]
We expect \(\bar{Q}\sim N\) by CLT with \(n=121>30\), so by 68-95-99.7 rule for normal distributions, we expect 68% of sample means to fall between \(\pm\sigma\) of the mean: \(16\pm\frac{4}{11}\).
Question 2: Summation Notation [6 points]
Please calculate the following sums:
[3 points] \[\frac13\sum_{i=3}^5(i+2)\] \(=\frac{5+6+7}{3}=6\)
[3 points] \[\sum_{i=1}^3(2^i)\]
\(=2^1+2^2+2^3=14\)
Question 3: Summary Statistics [10 points]
Suppose you obtain the following summary for a sample of data with summarize x, detail:
- Is the median of this data higher, lower, or equal to the mean? What does this mean for our data? [2 points]
\(\bar{x}>Median;5>4\). This means we expect our data to be right-skewed.
- Interpret the skewness and kurtosis of this data. What do each mean in words? [4 points]
\(Skew>0; Kurt < 3\). Our data is indeed right-skewed and has thinner tails than a Normal distribution.
- Is this data distributed normally? Explain. [4 points]
No. As described above, the data is both skewed and has thinner tails. It cannot be normally distributed.
Question 4: Univariate Inference [11 points]
You are given the following from Stata:
invttail(64,0.05)=1.669013
invttail(64,0.10)=1.2949198
invttail(63,0.05)=1.6694022
invttail(63,0.10)=1.2951343
invttail(62,0.05)=1.6698042
invttail(62,0.10)=1.2953558Suppose you then use the summarize command to obtain the following results for the three variables below. The variables price and weight have been transformed to represent units of thousands.
(1978 automobile data)
Variable | Obs Mean Std. dev. Min Max
-------------+---------------------------------------------------------
price | 64 6.266484 3.065138 3.291 15.906
weight | 64 2.980156 .8023428 1.76 4.84
mpg | 64 21.57813 6.054789 12 41- Which of the three variables has the greatest dispersion? Which has the least? [3 points]
\(CV_{price} = \frac{3.07}{6.27}=0.49; CV_{weight}=\frac{0.80}{2.98}=0.27; CV_{mpg}=\frac{6.05}{21.58}=0.28\)
Dispersion (most-to-least): price, mpg, weight.
- Provide a 90% CI for the mean of mpg. Interpret your answer. [3 points]
\(t^*_{63,0.05}=1.67; CI_{90}= 21.58\pm1.67\left(\frac{6.05}{\sqrt{64}}\right)=(20.32,22.84)\). We are 90% sure this interval covers the true value of \(\mu_{mpg}\).
- The claim is made that the average weight of a car in 1978 was above 2.8 thousand lbs. Using your sample data, test this claim at the 5% significance level. Clearly state your null and alternate hypotheses, your test statistic, and your conclusion. [5 points]
\(H_0: \mu_{weight}\leq2.8\)
\(H_A:\mu_{weight}>2.8\)
\(t^*_{63,0.05}=1.67\); \(t=\frac{2.98-2.8}{0.80/\sqrt{64}}=1.8\)
Since \(1.8>1.67\), we reject the null and assert that the average weight of a car in 1978 was indeed above 2.8 thousand pounds.
Q5: Bivariate Data [23 points]
- Briefly, what are the four key assumptions about our population bivariate regression model? What do they each mean? [4 points]
Linearity - our true population relationship is linear: \(\hat{y_i}=\beta_1+\beta_2x_i\) (or \(y_i=\beta_1+\beta_2x_i+u_i\))
Unbiasedness - our error term is conditional mean zero/we hit our line in expectation: \(E[u_i|x_i]=0\)
Homoskedasticity - our errors are distributed constantly across x: \(var[u_i|x_i] = \sigma^2_u\)
Independence - our errors do not influence each other: \(u_i\) is independent from \(u_j\) for \(i\neq j\)
- Assuming the four assumptions above hold and our sample is large enough, complete the statement about the distribution of \(b_2:b_2\sim\_\_(\_\_,\_\_)\). Please use Greek letters when appropriate. [3 points]
\(b_2\sim N(\beta_2,\sigma^2_{b_2});\left[\sigma^2_{b_2}=\frac{\sigma^2_u}{\sum_{i=1}^n(x_i-\bar{x})^2}\right]\)
Suppose we run the following regression and obtain these results:
Linear regression Number of obs = 64
F(1, 62) = 93.84
Prob > F = 0.0000
R-squared = 0.6350
Root MSE = 3.6874
------------------------------------------------------------------------------
| Robust
mpg | Coefficient std. err. t P>|t| [95% conf. interval]
-------------+----------------------------------------------------------------
weight | -6.013482 .6207753 -9.69 0.000 -7.254394 -4.77257
_cons | 39.49924 2.099907 18.81 0.000 35.30159 43.6969
------------------------------------------------------------------------------- Write down the estimated linear relationship between weight and mpg. [3 points]
\(\widehat{mpg}=39.50-6.01weight\)
- What does
robustmean in this regression command? Why might we have added this? [3 points]
We use robust standard errors to account for possible heteroskedasticity in our errors.
Interpret the t-statistic and p-value for the OLS slope coefficient from the regression output:
- What test is this referring to? What are its null and alternate hypotheses? [3 points]
This is our test of association between \(x\) and \(y\).
\(H_0: \beta_2=0\)
\(H_A: \beta_2\neq0\)
- What is the conclusion of this test? [2 points]
\(p\approx0.000<\alpha=0.05\Rightarrow\) Reject the null. There is a statistically significant association between mpg and weight.
- How many degrees of freedom will this test have? Why? [2 points]
We will have \(n-2\) degrees of freedom because we use 2 computed quantities, \(b_1\) and \(b_2\), to find our residual, and the standard error of our residual goes into the standard error of \(b_2\). \(df=64-2=62\).
- What would we expect the mpg to be for a car with a weight of 3.5 thousand lbs? [3 points]
\(E[mpg|weight=3.5]=39.50-6.01(3.5)=18.465\)
Multiple Choice [2 points each]
Only one answer is correct for each question. Choose the best possible answer.
MC 1
A local polling firm plans to conduct phone surveys on issues of concern for voters. They plan to call 45 new households in the county each week and ask respondents how much money they spent on groceries the previous week. This will be repeated for 6 weeks.
This data will be:
Experimental, categorical, panel
Experimental, numerical, time-series
c) Observational, numerical, repeated cross-section
Observational, categorical, cross-section
None of the above
MC 2
The OLS estimator:
minimizes the sum of horizontal deviations of actual data points from the sample regression line
minimizes the sum of horizontal deviations of actual data points from the population regression line
c) minimizes the sum of vertical deviations of actual data points from the sample regression line
- minimizes the sum of vertical deviations of actual data points from the population regression line
MC 3
For a r.v. X with population mean \(\mu\) and population variance \(\sigma^2\), the standard error is:
- An estimator for the standard deviation of \(\mu\)
b) An estimator for the standard deviation of \(\bar{X}\)
An estimator for the standard deviation of \(X\)
Always equal to the sample standard deviation of \(X\)
None of the above
MC 4
Suppose a r.v. \(Z\sim(3,25)\). We would expect the transformed r.v. \(\tilde{Z}=(Z-3)/5\) to be:
a) A z-score
Distributed identically to \(Z\)
Normally distributed as \(n\rightarrow\infty\)
All of the above
None of the above
MC 5
Suppose we regress y on x and obtain the result \(\hat{y}=0.33x+1\) with \(R^2=0.64\). If we then regress x on y, we would expect:
A slope coefficient of 0.33
A slope coefficient of \(0.33^{-1}\approx3\)
c) \(\boldsymbol{r_{yx}=0.8}\)
\(R^2\neq0.64\)
None of the above