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WEEK 4 PARTICIPATION EXERCISES
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Post the exercises, any 2 of the 5 total, from different categories listed below. Work offline and use the equation editor to present both exercises in the same “Start a New Tread.”
Be as clear and brief as possible. No need to re-state the question for each exercise. Do your own work and only ask the Instructor for help.
List of 5 exercises below are: 7.1.63,8.1.95,8.2.104,8.3.119,8.3.133.
Title of the Thread could be “7.3.84 & 8.3.119” or similar.
I. Central Limit Theory: Exercise 7.1.63 below.
According to the Internal Revenue Service, the average length of time for an individual to complete (keep records for, learn, prepare, copy, assemble, and send) IRS Form 1040 is 10.53 hours (without any attached schedules). The distribution is unknown. Let us assume that the standard deviation is two hours. Suppose we randomly sample 36 taxpayers.
a) Would you be surprised if the 36 taxpayers finished their Form 1040s in an average of more than 12 hours? Explain why or why not in complete sentences.
b. Would you be surprised if one taxpayer finished his or her Form 1040 in more than 12 hours? In a complete sentence, explain why.
Click on the link below for example 7.1.62 which has a mean of 250 feet for fly balls hit to the outfield. Also, the std dev is 50 feet with a sample size of 49. Also, see below for the solution to Example 1.
Example 1: Given a sample mean of 90 and a std dev of 15, samples of size n=25 are drawn randomly from the population. Find the probability that the sample mean is between 85 and 92.
Recall that for each value of the sample mean there is a corresponding value of Z. GIven μ,σ,n {“version”:”1.1″,”math”:”μ,σ,n”} the value of Z1 correspoding to the sample mean = 85 is,
Z1=85−901525 √ =−53{“version”:”1.1″,”math”:”<math xmlns=”http://www.w3.org/1998/Math/MathML”><mi… displaystyle=”true”><mfrac><mn>15</mn><msqrt><mn>25</mn></msqrt></mfrac></mstyle></mfrac><mo>=</mo><mfrac><mrow><mo>-</mo><mn>5</mn></mrow><mn>3</mn></mfrac></math>”}
For the sample mean = 92,
Z2=92−901525 √ =23{“version”:”1.1″,”math”:”<math xmlns=”http://www.w3.org/1998/Math/MathML”><mi… displaystyle=”true”><mfrac><mn>15</mn><msqrt><mn>25</mn></msqrt></mfrac></mstyle></mfrac><mo>=</mo><mfrac><mn>2</mn><mn>3</mn></mfrac></math>”}
So, the question restated is find,
P(Z1<Z<Z2) = P(-5/3<Z<2/3).
Using the Z tables (0 to Z)
P((-5/3<Z<0) = P(0<Z<5/3) = .4515 by symmetry of normal curve. Also, P(0<Z<2/3) = .2454. The total area under the curve is the probability. So, P(85<X ¯ ¯ ¯ <92) = .4515+.2454 = .6969 {“version”:”1.1″,”math”:”X¯<92) = .4515+.2454 = .6969″}
II. Confidence Intervals: Exercises 8.1.95 and 8.2.104.
μ= x ±Z α 2 / ⋅σn √ n>30 (Large Sample Size− Normal Z)μ= x ±t α 2 / ⋅sn √ n<30 (Small Sample Size−t Distribution) {“version”:”1.1″,”math”:”<math xmlns=”http://www.w3.org/1998/Math/MathML”><mi… notation=”top”><mi>x</mi></menclose><mo>±</mo><msub><mi>Z</mi><mfrac bevelled=”true”><mi>α</mi><mn>2</mn></mfrac></msub><mo>·</mo><mfrac><mi>σ</mi><msqrt><mi>n</mi></msqrt></mfrac><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mi>n</mi><mo>></mo><mn>30</mn><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo>(</mo><mi>L</mi><mi>a</mi><mi>r</mi><mi>g</mi><mi>e</mi><mo> </mo><mi>S</mi><mi>a</mi><mi>m</mi><mi>p</mi><mi>l</mi><mi>e</mi><mo> </mo><mi>S</mi><mi>i</mi><mi>z</mi><mi>e</mi><mo>-</mo><mo> </mo><mi>N</mi><mi>o</mi><mi>r</mi><mi>m</mi><mi>a</mi><mi>l</mi><mo> </mo><mi>Z</mi><mo>)</mo><mspace linebreak=”newline”></mspace><mi>μ</mi><mo>=</mo><menclose notation=”top”><mi>x</mi></menclose><mo>±</mo><msub><mi>t</mi><mfrac bevelled=”true”><mi>α</mi><mn>2</mn></mfrac></msub><mo>·</mo><mfrac><mi>s</mi><msqrt><mi>n</mi></msqrt></mfrac><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mi>n</mi><mo><</mo><mn>30</mn><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo>(</mo><mi>S</mi><mi>m</mi><mi>a</mi><mi>l</mi><mi>l</mi><mo> </mo><mi>S</mi><mi>a</mi><mi>m</mi><mi>p</mi><mi>l</mi><mi>e</mi><mo> </mo><mi>S</mi><mi>i</mi><mi>z</mi><mi>e</mi><mo>-</mo><mi>t</mi><mo> </mo><mi>D</mi><mi>i</mi><mi>s</mi><mi>t</mi><mi>r</mi><mi>i</mi><mi>b</mi><mi>u</mi><mi>t</mi><mi>i</mi><mi>o</mi><mi>n</mi><mo>)</mo></math>”}
Exercise 8.1.95
Among various ethnic groups, the standard deviation of heights is known to be approximately three inches. We wish to construct a 95% confidence interval for the mean height of male Swedes. Forty-eight male Swedes are surveyed. The sample mean is 71 inches. The sample standard deviation is 2.8 inches. Calculate the error bound. What will happen to the level of confidence obtained if 1,000 male Swedes are surveyed instead of 48? Why?
Exercise 8.2.104 (using t-distribution)
In six packages of “The Flintstones® Real Fruit Snacks” there were five Bam-Bam snack pieces. The total number of snack pieces in the six bags was 68. We wish to calculate a 96% confidence interval for the population proportion of Bam-Bam snack pieces. Construct a 96% confidence interval for the population proportion of Bam-Bam snack pieces per bag and calculate the error bound.
Do you think that six packages of fruit snacks yield enough data to give accurate results? Why or why not?
Exercise 8.3.119: (The estimated interval for the true proportion, p.)
p=p ˆ ±Z α 2 / ⋅p ˆ q ˆ n − − − √ p ˆ −Sampled proportion {“version”:”1.1″,”math”:”<math xmlns=”http://www.w3.org/1998/Math/MathML”><mi… bevelled=”true”><mi>α</mi><mn>2</mn></mfrac></msub><mo>·</mo><msqrt><mfrac><mrow><mover><mi>p</mi><mo>^</mo></mover><mover><mi>q</mi><mo>^</mo></mover></mrow><mi>n</mi></mfrac></msqrt><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mover><mrow><mi>p</mi><mo> </mo></mrow><mo>^</mo></mover><mo>-</mo><mi>S</mi><mi>a</mi><mi>m</mi><mi>p</mi><mi>l</mi><mi>e</mi><mi>d</mi><mo> </mo><mi>p</mi><mi>r</mi><mi>o</mi><mi>p</mi><mi>o</mi><mi>r</mi><mi>t</mi><mi>i</mi><mi>o</mi><mi>n</mi></math>”}
According to a recent survey of 1,200 people, 61% feel that the president is doing an acceptable job. We are interested in the population proportion of people who feel the president is doing an acceptable job. Construct a 90% confidence interval for the population proportion of people who feel the president is doing an acceptable job and calculate the error bound.
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