You are teaching a class where any student with an average of 96 or more does not have to take the final and gets an A. All other students must take the final which counts for 40% of their grade. Complete parts (a) and (b) below. Click the icon to view the Excel table. (a) What IF function would you enter in cell D3 (to be filled down) for the Course Average that takes into account the two possibilities outlined above? A. = IF(B3 < 96, AVERAGE(B3,C3)*40%, C3) B. = IF(B3 < 96, B3, (B3 + C3)*40%) C. = IF(B3 >= 96, B3, C3*40%) D. = IF(B3 >= 96, B3, B3*60% + C3*40%) Excel Table A B C D E F G 1 Student Course Avg Final Exam Course Avg Course Grade Table 2 Before Final Grade 0 F 3 Jim 80 62 72.8 C 60 D 4 Bob 75 83 78.2 C 66 D+ 5 Sue 96 96 A 72 C 6 Kim 67 95 78.2 C 79 C+ 7 Jan 97 97 A 84 B 8 89 B+ 9 96 A
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If their average is less than 96, their final grade is calculated based on their course average and final exam grade. The Excel table provided has columns for Student, Course Average (Before Final), Final Exam Grade, and Course Average (After Final). The Show more…
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An investigator wishes to estimate the proportion of students at a certain university who have violated the honor code. Having obtained a random sample of $n$ students, she realizes that asking each, "Have you violated the honor code?" will probably result in some untruthful responses. Consider the following scheme, called a randomized response technique. The investigator makes up a deck of 100 cards, of which 50 are of Type I and 50 are of Type II. Type I: Have you violated the honor code (yes or no)? Type II: Is the last digit of your telephone number a $0,1,$ or 2 (yes or no)? Each student in the random sample is asked to mix the deck, draw a card, and answer the resulting question truthfully. Because of the irrelevant question on Type II cards, a yes response no longer stigmatizes the respondent, so we assume that responses are truthful. Let $p$ denote the proportion of honor-code violators (i.e., the probability of a randomly selected student being a violator), and let $\lambda=P($ yes response). Then $\lambda$ and $p$ are related by $\lambda=.5 p+(.5)(.3) .$ (a) Let $Y$ denote the number of yes responses, so $Y \sim \operatorname{Bin}(n, \lambda) .$ Thus $Y / n$ is an unbiased estimator of $\lambda .$ Derive an estimator for $p$ based on $Y .$ If $n=80$ and $y=20,$ what is your estimate? [Hint: Solve $\lambda=.5 p+.15$ for $p$ and then substitute $Y / n$ for $\lambda . ]$ (b) Use the fact that $E(Y / n)=\lambda$ to show that your estimator is unbiased for $p$ . (c) If there were 70 Type I and 30 Type II cards, what would be your estimator for $p ?$
The Basics of Statistical Inference
Point Estimation
Use the data in BENEFITS to answer this question. It is a school-level data set at the $\mathrm{K}-5$ level on average teacher salary and benefits. See Example 4.10 for background. (i) Regress lavgsal on $b s$ and report the results in the usual form. Can you reject $\mathrm{H}_{0} : \beta_{b s}=0$ against a two-sided alternative? Can you reject $\mathrm{H}_{0} : \beta_{b s}=-1$ against $\mathrm{H}_{1} : \beta_{b s}>-1 ?$ Report the $p$ -values for both tests. (ii) Define $l b s=\log (b s) .$ Find the range of values for $l b s$ and find its standard deviation. How do these compare to the range and standard deviation for $b s ?$ (iii) Regress lavgsal on lbs. Does this fit better than the regression from part (i)? (iv) Estimate the equation lavgsal $=\beta_{0}+\beta_{1} b s+\beta_{2}$ lenroll $+\beta_{3} / s t a f f+\beta_{4} / u n c h+u$ and report the results in the usual form. What happens to the coefficient on $b s ?$ Is it now statistically different from zero? (v) Interpret the coefficient on lstaff. Why do you think it is negative? (vi) Add lunch^{2} \text { to the equation from part } (iv). Is it statistically significant? Compute the turning point (minimum value) in the quadratic, and show that it is within the range of the observed data on lunch. How many values of lunch are higher than the calculated turning point? (vii) Based on the findings from part (vi), describe how teacher salaries relate to school poverty rates. In terms of teacher salary, and holding other factors fixed, is it better to teach at a school with lunch $=0$ (no poverty), lunch = 50, or lunch =100 (all kids eligible for the free lunch program)?
Exercises 71 to 74 refer to the following setting. Coaching companies claim that their courses can raise the SAT scores of high school students. Of course, students who retake the SAT without paying for coaching generally raise their scores. A random sample of students who took the SAT twice found 427 who were coached and 2733 who were uncoached.$^{44}$ Starting with their Verbal scores on the first and second tries, we have these summary statistics: Coaching and SAT scores (10.1) What proportion of students who take the SAT twice are coached? To answer this question, Jannie decides to construct a 99% confidence interval. Her work is shown below. Explain what’s wrong with Jannie’s method. $$hat{p}_{1}=\frac{427}{3160}=0.135=\underset{\text { who were coached }}{\text { proportion of students }}$$ $$\hat{p}_{2}=\frac{2733}{3160}=0.865=\underset{\text { who weren't coached }}{\text { who weren't coached }}$$ $$\mathrm{A} 99 \% \mathrm{Cl} \text { for } p_{1}-p_{2} \mathrm{is}$$ $$\begin{aligned}(0.135-0.865) \pm 2.575 \sqrt{\frac{0.135(0.865)}{3160}} &+\frac{0.865(0.135)}{2733} \\ &=-0.73 \pm 0.022=(-0.752,-0.708) \end{aligned}$$ We are 99% confident that the proportion of students taking the SAT twice who are coached is between 71 and 75 percentage points lower than students who aren’t coached.
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