Questions and Answers #1 Analysis

Questions and Answers Sheet 1 Analysis Question #1 We expect that students who do well on the midterm exam in a course will usually also do well on the final exam. Gary Smith of Pomona College looked at the exam scores of all 346 students who took his statistics class over a 10-year period. Assume that both the midterm and final exam were scored out of 100 points. The actual least-squares line for predicting final-exam score y from midterm-exam score x was 𝑦̂ = 46.6 + 0.41𝑥 Predict the score of a student who scored 50 on the midterm and a student who scored 100 on the midterm. Answer: Given least-squares regresion line: 𝑦̂ = 46.6 + 0.41𝑥 To predict the score of a student who scored 50 on the midterm, we replace x by 50 in least-squares regression line and evaluate: 𝑦̂ = 46.6 + 0.41𝑥 = 46.6 + 0.41(50) = 46.6 + 20.5 = 67.1 Thus the predicted final score of the student is then 67.1. To predict the score of student who scored 100 on the midterm, we replace x by 100 in the least-squares regression line and evaluate: 𝑦̂ = 46.6 + 0.41𝑥 = 46.6 + 0.41(100) = 46.6 + 41 = 87.6 Thus the predicted final score student is then 87.6. Question #2 Requires Uploaded Supporting Analysis: Find the derivative using the product rule. 𝑦 = 𝑒 𝑥 ln 𝑥 Answer: Consider the provided equation, 𝑦 = 𝑒 𝑥 𝐼𝑛 𝑥. Find the derivative of the equation. Apply the product rule of 𝑑 𝑑 [𝑦] = [𝑒 𝑥 ln 𝑥]𝑦 ′ = the derivative (𝑓 ⋅ 𝑔)′ = 𝑓 ′ ⋅ 𝑔 + 𝑓 ⋅ 𝑔′ . Here 𝑓 = 𝑒 𝑥 , 𝑔 = 𝐼𝑛(𝑥) . Therefore, 𝑑𝑥 𝑑 𝑑𝑥 (𝑒 𝑥 ) ln(𝑥) + 𝑑𝑥 𝑑 𝑑𝑥 1 (ln( x))e x = e x ln( x) + e x Hence. x Question #3 Lori Jeffrey is a successful sales representative for a major publisher of college textbooks. Historically, Lori obtains a book adoption on 36% of her sales calls. Viewing her sales calls for one month asa sample of all possible sales calls, assume that a statistical analysis of the data yields a standard error of the proportion of 0.08. (a) How large was the sample used in this analysis? That is, how many sales calls did Lori make during the month? (c) Using the sampling distribution of P, compute the probability that Lori will obtain book adoptions on 46% or more of her sales calls during a one-month period. (Round your answer to four decimal places.) Answer: Given information- Sample proportion, p=0.36 Standard error =0.08 𝑝(1−𝑝) a) We have to find the sample size (n). Now, standard error is given by- Standard error = √ √ 0.36(1−0.36) 𝑛 𝑛 0.08 = 𝑛 = 36. Hence, the sample size (n)=36. That means 36 sales calls Lori make during the month. c) Now, we have to find the probability that Lori will obtain book adoptions on 46% or more of her sales calls during a one-month period- 𝑃(𝑥 ≥ 0.46) is- 𝑃(𝑥 ≥ 0.46) = 𝑃 (𝑧 ≥ 0.46−0.36 0.08 ) 𝑃(𝑥 ≥ 0.46) = 𝑃(𝑧 ≥ 1.25), 𝑃(𝑧 ≥ 1.25) = 1 − 𝑃(𝑧 < 1.25), 𝑃(𝑧 ≥ 1.25) = 1 − 0.8944 = 0.1057. Hence, the probability that Lori will obtain book adoptions on 46% or more of her sales calls during a one-month period is 0.1057. Question #4 A small private college is interested in determining the percentage of its students who live off campus and drive to class. Its claimed that less than 20% of their current students live off campus and drive to class. The college decided to take a random sample of 108 of their current students to use in the analysis. It was found that 14 of those live off campus and drive to class. Is therre enought evidence to support the claim that less than 20% of students live off campus at the 3% level of significance? Answer: The hypotheses are given below: Null hypothesis: 𝐻0 ÷ 𝑝 = 0.20 . Alternative hypothesis: 𝐻1 ÷ 𝑝 < 0.20. The sample proportion, p-cap is 0.1296 ( 14 108 ) . Asample of size 108 is considered. Here, the z test for (𝑝̂−𝑃) single proportion is used. Calculation: The test statistic is given and calculated as follows. 𝑧 = √ (0.1296−0.20) 𝑃(1−𝑃) 𝑛 = = −1.829. Computation of p-value: The p-value can be obtained by entering the formula √0.20(1−0.20) 108 “=NORM.S.DIST(-1.829,TRUE)”in EXCEL. The p-value is 0.0337. Decision rule: If p-value ≤ α, then reject the null hypothesis. Otherwise, fail to reject the null hypothesis. Conclusion: The P-value is 0.0337 and the significance level is 0.03. Here, the P-value is greater than the significance level. Hence, the null hypothesis is not rejected. Thus, there is not enough evidence to support the claim that less than 20% of students live off campus at the 3% level of significance. Question #5 Requires Uploaded Supporting Analysis Find 2 𝑑𝑦 𝑑𝑥 by implicit differentiation. 𝑦 − 2𝑥 = 4𝑦 Answer: Given: 𝑦 2 − 2𝑥 = 4𝑦 Differentiating with respect to 𝑥 2=4 𝑑𝑦 𝑑𝑦 𝑑𝑥 𝑑𝑥 (2𝑦 − 4) = 2 𝑑𝑦 𝑑𝑥 = 2 𝑑𝑦 2𝑦−4 𝑑𝑥 = 1 𝑦−2 Question #6 Requires Uploaded Supporting Analysis: Find the second derivative 𝑦 = sin 𝑡 − 𝑒 𝑡 𝑑2𝑦 𝑑𝑡 2 . 𝑑 𝑑𝑥 (𝑦 2 − 2𝑥) = 𝑑 𝑑𝑥 (4𝑦) 𝑑 𝑑𝑥 (𝑦 2 ) − 𝑑 𝑑𝑥 (2𝑥) = 4 𝑑𝑦 𝑑𝑥 2𝑦 𝑑𝑦 𝑑𝑥 − Answer: Consider the following equation: 𝑦 = [sin(𝑡) − 𝑒 𝑡 ] Differentiate y with respect to t: = 𝑑 𝑑𝑡 [𝑆𝐼𝑁(𝑇)] − = cos(𝑡) − 𝑒 𝑡 𝑑𝑦 Differentiate 𝑑2𝑦 𝑑2𝑡 = 𝑑 𝑑𝑡 𝑑𝑡 𝑑 𝑑𝑡 𝑑𝑦 𝑑𝑡 = 𝑑 𝑑𝑡 [sin(𝑡) − 𝑒 𝑡 ] (𝑒 𝑡 ) with respect to t: 𝑑 [cos(𝑡)] − 𝑑𝑡 = sin(𝑡) − 𝑒 𝑡 𝑑 𝑑𝑡 𝑑𝑦 𝑑 𝑑𝑡 𝑑𝑡 ( )− [cos(𝑡) − 𝑒 𝑡 ] (𝑒 𝑡 ) Hence,the second derivative is 𝑑2𝑦 𝑑𝑡 2 = − sin(𝑡) − 𝑒 𝑡 . Question #7 Requires Uploaded Supporting Analysis: Differentiate. 3 𝑦 = (3𝑥 2 + 5𝑥 + 1)2 Answer: Step 1 3 Let us find the differentiate of given function Given: 𝑦 = (3𝑥 2 + 5𝑥 + 1)2 Formula: 𝑑 𝑑𝑥 (𝑥 𝑛 ) = 𝑛 ⋅ 𝑥 𝑛 − 1 Step 2 3 3 Answer 𝑦1 = (3𝑥 2 + 5𝑥 + 1)2 − 1. (6𝑥 + 5) 2 3 1 𝑦 = (3𝑥 2 + 5𝑥 + 1) . (6𝑥 + 5) 2 2 Question #8 Use a Numerical approach (i,e., a table of values) to approximate lim 𝑒 2 𝑥−1 𝑥⇒0 𝑥 , Provide support for your analysis (slow the table and explain your resoning). Answer: The table for the values of 𝑒 2 𝑥−1 𝑥 x −0.5 −0.05 −0.0001 0.0001 0.05 0.5 f ( x) 1.2642 1.9033 1.9998 2.0002 2.1034 3.4366 From the table, Ax x approaches to 0 from left side right side 𝑒 2 𝑥−1 𝑥 Therefore, lim 𝑥⇒0 𝑒 2 𝑥−1 𝑥 approaches to 2. As x approaches to 0 from approaches to 2 𝑒 2 𝑥−1 𝑥 =2 Question #9 Evaluate the definite integrals.Requires a supporting analysis. 9 ∫0 5√𝑥 Answer: Here f ( x) = 5 x we have to find the value of ∫0 5√𝑥 𝑑𝑥 9 9 9 ∫0 𝑓√𝑥 𝑑𝑥 = ∫0 5√𝑥 𝑑𝑥 9 = 5 ∫0 √𝑥 𝑑𝑥 9 1 = 5 ∫0 (𝑥)2 𝑑𝑥  12  ( x)  = 5 +   1 1 2  9  32  ( x)  = 5 3     2 0 9 3  10 10  3  10  32 2 =  ( x) 2  (9) − (0)   (27) 3 0 3   3 =90 9 Hence, ∫0 5√𝑥 𝑑𝑥 = 90 Question #10 Evaluate the linear equation, y=4x-7 ,at the value x=2. y=? Answer: Regression analysis: Regression analysis estimates the realationship among variables. The general form of first-order regression model is 𝑦 − 𝑐𝑎𝑝 = β0 + β1 𝑥 + ϵ, Where, the variable y is the dependent variable that is to be madelled or predicted,the variable x is the independent variable that is used to predict the dependent variable,and ϵ is the error term Given linear equation is y=4x-7 .The given value of x is 2. Thus,the value of y at given value x=2 is as follows: y=4x-7 =4×2−7 =8−7 =1 Question #11 Five samples of a ferrous-type substance were used to determine if there is a difference between a laboratory chemical analysis and an X-ray fluorescence analysis of the iron content. Each sample was split into two subsamples and the two types of analysis were applied, with the accompanying results. Assuming that the populations are normal, test at the 0.01 level of significance whether the two methods of analysis give, on the average, the same result. Lt sample 1 be the X-ay fluorescence results and let sample 2 be the laboratory chemical results. State the null and alternative hypotheses. 𝐻0 ÷ μ𝐷 = 0 𝐻1 ÷ μ𝐷 ≠ 0 Identify the critical region. Select the correct choice below and fill in the answer box(es) to complete your choice. (Round to three decimal places as needed.) A. t> B. t< C. t< or t> Answer: Step 1 In the given situation, the alternative hypothesis is not equal case. That is, the tail of the test is two tail. The sample size is n=5 . Given level of significance is 0.01. degrees of freedom= n-1=5-1=4 Step 2 From the critical value table of t distribution, the t critical values corresponding with 4 degrees of freedom and 0.01 are, -4,604 and 4.604. Thus, the correct option is, t<-4.604 or t>4.604 Question #12 In a university examination, which was indeed very tough, 50% at jeast failed in ''Statistics'', 75% at least in Topology, 82% at least in ''Functional Analysis'' and 96% at least in ''Applied Mathematics''. How many at least failed in all the four? (Ans. 3% ) Answer: Given P(failed in Statistics) =0.50 P(failed in Topology) =0.75 P(failed in Functional Analysis) =0.82 P(failed in Applied Maths) =0.96 To find: How many failed in at least all four A, B, C and D are said to be independent when : 𝑃(𝐴 ∩ 𝐵 ∩ 𝐶 ∩ 𝐷) = 𝑃(𝐵) × 𝑃(𝐶) × 𝑃(𝐷) The probability of failing in at least all four is given by: P(failed in Statistics \cap failed in Topology ∩ failed in Functional Analysis ∩ failed in Applied Maths) = 𝑃(𝑓𝑎𝑖𝑙𝑒𝑑 𝑖𝑛 𝑆𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐𝑠) × 𝑃(𝑓𝑎𝑖𝑙𝑒𝑑 𝑖𝑛 𝑇𝑜𝑝𝑜𝑙𝑜𝑔𝑦) X P(failed in Functional Analysis) × P(failed in Applied Maths) left[ ∵independent events ] = 0.50 × 0.75 × 0.82 × 0.96 =0. 2952 ≅ 0.3 Therefore, 3% of them at least failed in all the four. Question #13 Quick decision analysis Please pick the best outcome after calculation, a) Council would accept (𝑃2 = 0.6) and the mayor would not veto (𝑃3 = 0.8), The probability of the compound event is b) Council would accept (𝑃2 = 0.6) and the mayor would veto (1 − 𝑃3 = 0.2) . The probability of the compound event is c) Council would reject (1 − 𝑃2 = 0.4) and the council would return the same budget (𝑃4 = 0.6) . The probability of this compound event is Answer: For two independent events A and B, 𝑃(𝐴 and 𝐵) = 𝑃(𝐴) ⋅ 𝑃(𝐵) Also note that Mayor's decision for veto or not veto, is independent of the council's decision once it is accepted by the council. a) The probability of the compound event is calculated as, 𝑃(compound event) = 𝑃2 × 𝑃3 = 0.6 × 0.8 =0.48 =48% b) The probability of the given compound event is calculated as, 𝑃(compound event) = 𝑃2 × (1 − 𝑃3 ) = 0.6 × 0.2 =0.12 =12% c) The probability of the given compound event is calculated as, 𝑃(compound event) = (1 − 𝑃2 ) × 𝑃4 = 0.4 × 0.6 =0.24 =24% Thus, the most probable event is that the council accepts and mayor would not veto. Question #14 Graph the sets of points whose polar coordinates satisfy the given polar equation. The given polar equation is written as follows: π θ = ,𝑟 ≤ 0 2 Answer: Description: Consider the polar coordinates as P(r,6) . Here, r is the directed distance from origin O to point P and \theta is the directed angle from initial ray to OP. π From the given equation, the range of r is from −∝ to 0 and 0s fixed at . Therefore, the polar point traces a straight line at an angle of π 2 2 from the initial ray (x-axis) and the line also satisfies the range of r. From the analysis, draw the sets of points whose polar coordinates satisfy the given polar equation as shown in Figure 1. π θ= 2 𝑟≤0 Question #15 Sketch the line for a given polar equation and find the corresponding Cartesian equation. The given polar equation is 𝑟 = 2 sec θ. Answer: Formula used: Write the expression for relation between Cartesian and polar equations as follows: x = r cos  y = r sin  y tan  = x Calculation: Rewrite the given polar equation as follows: 𝑟 = 2 𝑠𝑒𝑐 θ 2 𝑟= cos θ 𝑟 cos θ = 2 From relation between Cartesian and polar equations, substitute x for rcos 0 to find the Cartesian equation for the given polar equation. x=2 From the obtained Cartesian equation, it is clear that, the graph is a vertical line which passes through (2,0). Question #16 Graph the sets of points whose polar coordinates satisfy the given polar equation. The given polar equation is written as follows: θ= 2π 3 , 𝑟 ≤ −2 Answer: Description: Consider the polar coordinates as 𝑃(𝑟. θ). Here, r is the directed distance from origin Oto point Pand θ is the directed angle from initial ray to OP. 2π From the given equation, the range of r is from -2 to −∝ and 0s fixed at . Therefore, the polar point traces a straight line at an angle of 2π 3 3 from the initial ray (x-axis) and the line also satisfies the range of r. From the analysis, draw the sets of points whose polar coordinates satisfy the given polar equation as shown in Figure 1. 2π θ= , 3 𝑟 ≤ −2 Question #17 At HD Sport and Fitness gym, analysis shows that, as the demand of the gym, the number of members is 86 when annual membership fee is $25 per member and the number of members is 84 when annual membership fee is $33 per member. As the operator of the gym, you found out that number of members (q) and membership fee (p) have a linear relationship. A) At what membership price, p, is the revenue maximized? p= Round your answer to 2 decimal places. B) What is the maximum annual revenue? R= . Round your answer to 2 decimal places. Answer: We find for (a) 𝑅(𝑝) = −5 2 −5 2 ⋅ −5 2 (86)2 + 240𝑝 + 240𝑝 ⋅ 7396 + 240 We find for (b) 2 𝑅 = (𝑞) = ⋅ 𝑞 2 + 96𝑝 2 5 = ⋅ (84)2 + 96𝑞 2 5 5 ⋅ 7056 + 96𝑞 Question #18 Prove the following limit. lim 5𝑥 − 4 = 6 𝑥⇒2 SOLUTION 1. Preliminary analysis of the problem (guessing a value for δ). Let ϵ be a given positive number. We want to find a number δ such that if 0 < |𝑥 − 2| < δ 𝑡ℎ𝑒𝑛 |(5𝑥 − 4) − 6| < ϵ. But |(5𝑥 − 4) − 6| = |5𝑥 − 10| = 5 | | − . Therefore, we want δ such that if 0 < |𝑥 − 2| < δ 𝑡ℎ𝑒𝑛 5|| < ϵ ϵ that is, if 0 < |𝑥 − 2| < δ 𝑡ℎ𝑒𝑛 5|| < 5 ϵ This suggests that we should choose δ = . 5 ϵ 2. Proof (showing that δ works). Given ϵ > 0 , choose δ = . If 0 < || < δ , then 5 |(5𝑥 − 4) − 6| = | | = 5|| < 5δ = 5(||) =ϵ Thus if 𝑂 < |𝑥 − 2| < 6 then |(5𝑥 − 4) − 6| < ϵ. Therefore, by the definition of a limit lim 5𝑥 − 4 = 6 𝑥⇒2 Answer: We have to prove that lim 5𝑥 − 4 = 6 𝑥⇒2 Let \epsilon be a given positive number, we want to find a number δ such that If 0 < |𝑥— 2| < δ 𝑡ℎ𝑒𝑛 |(5𝑥 − 4) − 6| < Now solve |(5x-4)-6| |(5x-4)-6|=|5x- 4-6| |5x -10| =5|x-2| ⇒ |(5𝑥 − 4) − 6| = 5|𝑥 − 2| Therefore, we want to δ such that if 0 < |𝑥 − 2| < δ 𝑡ℎ𝑒𝑛 5|𝑥 − 2| < ϵ if 0 < |𝑥 − 2| < δ 𝑡ℎ𝑒𝑛 |𝑥 − 2| < Hence, we choose δ = ϵ 5 5 Step 2 Since, given \epsilon>0 and we choose \delta=\frac{\epsilon}{5} if 0<|x-2|<\delta then |(5x-4)-6|=|5x- 4-6| |5x —10| =5|x—2| <5\delta (Since, 0<|x-2|<\delta) Substitute \delta=\frac{\epsilon}{5} 5\delta=5\cdot \frac{\epsilon}{5} Thus, if 0<|x-2|<\delta then (5x-4)-6<\epsilon Therefore, by definition of limit \lim_{x \Rightarrow 2}(5x-4)=6 Question #19 Why is it very easy to conclude that \int_{1}^{\propto}x^{2}dx diverges without making any integration calculations? Answer: Step 1 This question is taken from the calculus and sub-topic is improper integral in which we have to give the reason that given integral is diverges in nature without making any integration calculation. We can write the integral below consider u=\int_{1}^{\propto}x^{2}dx now to solve we move to next step-2 Step 2 the given integral is improper integral in which we directly say about the integral is converge or diverges by simply checking the limit of the integral function at the upper limit of the integral so first we write it as. f(x)=x^{2} u=\int_{1}^{\propto}f(x)dx now if the limit of the function is finite then we can found the integral value so it converges and if the limits become infinite or the limit does not exist of the function then we can not find the value of the integral and it diverges. so first we have to check the limit of the function that exists or not so u=\int_{1}^{\propto}f(x)dx u=\int_{1}^{\propto}x^{2}dx f(x)=x^{2} now taking the limit of the function consider k is limiting value then k=\lim_{x \Rightarrow \propto}f(x) put the f(x)=x^{2} k=\lim_{x \Rightarrow \propto}f(x) k=\lim_{x \Rightarrow \propto}x^{2} now put the limits k=\propto^{2}=\propto k=\propto or k=\lim_{x \Rightarrow \propto}x^{2}=\propto now the value of the kis infinite it means that we can not find the value of the given integral so the integral is diverges. now we move to the next step for the result Step 3 Result Result: from the above analysis we concluded that the given integral is diverges since its functional limiting value is infinite so itis very easy to say about the given integral is diverges. Question #20 In the given equation as follows, use Taylor’s Theorem to obtain an upper bound for the error of the approximation. Then calculate the exact value of the error|:see the equation as attached here:\cos (0.3)\approx1-\frac{(0.3)^{2}}{2!}+\frac{(0.3)^{4}}{4!} Answer: Step 1 Approximation is a result that is near, not exactly, but almost accurate. To find the simpler value of f(x), which has approximately the same value as F(x) over a specified interval. the analysis is then carried out on the more accessible f(x), knowing the difference will be small. Upper bound on the size |R_{n}(x)| lof the error since T_{0}(x)=f(a) then R_{0}(x)=f(x)-f(a) . T_{1}(x)=f(a)+f`(a)(x-a) is the tangent line to f at a the remainder R_{1}(x) is the difference between f(x) and the tangent line approximation at f. Step 2 The formula for upper bound error estimate in Taylor's series |R_{n}(x)|\le \frac{\max|f^{n+1}(c)|}{(n+1)!}|x-a|^{n+1} in the given problem n=4max|f^{n+1}(c)| is the maximum value off f^{n+1}(c) between c and a, where a=0,x=0.3 and f(x)=\cos (x) to find the fifth derivative of f(x). f(x)=\cos (x) f`(x)=-\sin (x) f``(x)=-\cos (x) f```(x)=\sin (x) f^{4}(x)=\cos (x) f^{5}(x)=-\sin (x) |R_{n}(x)|\le \frac{\max|f^{n+1}(c)|}{(n+1)!}|x-a|^{n+1} |R_{4}(x)|\le \frac{|-\sin(0.3)|}{5!}|0.3-0|^{5} |R_{4}(x)|\le \frac{\sin(0.3)}{5!}(0.3)^{5}\approx6\times10^{-6} Upper bound for the error of the approximation is 6\times10^{-6} . Step 3 To find the exact value of the error, subtract the given Taylor theorem by \cos (0.3) . |1-\frac{(0.3)^{2}}{2!}+\frac{(0.3)^{4}}{4!}-\cos(0.3)|\approx1\times10^{-6} . Therefore, the exact error is 1\times10^{-6} .