Sample Exam 3 for Elementary Statistics | MATH 365, Exams of Statistics

Material Type: Exam; Professor: Porter; Class: Elementary Statistics; Subject: Mathematics; University: University of Kansas; Term: Unknown 1989;

Typology: Exams

Pre 2010

Uploaded on 03/10/2009

koofers-user-noi
koofers-user-noi 🇺🇸

10 documents

1 / 2

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Math 365 - Sample Exam 3
SHOW YOUR WORK on Problems 2(a,b,c), 3, 4(b,c) and 5. These problems require a solution
(with an explanation) as well as an answer. A correct, but unsubstantiated answer on these prob-
lems is worth only one point.
For Problems 1, 2(d), and 4(a,d), fill in the blanks; no partial credit given for these problems.
Do not use books or notes. BOX your answers. You will need your graphing calculator on some
of the problems, and you are encouraged to use your calculator to verify your work. Calculator
answers must be accurate to 3 decimal places.
1. (a) Start with a population with mean µand standard deviation σ. For large random sample of
size n, identify the approximate distribution of X.
(b) Given a random sample of size 42 with x= 23.4and s= 3.1, a 94% MOE for the true
mean is .
(c) Start with a population with proportion p. For large random sample of size n, identify the
approximate distribution of ˆp.
(d) Given a random sample of size 52 with ˆp=.4, a 94% MOE for the true proportion is
.
2. In a sample of 1,277 Americans (New York Times, November 10, 1985), 22% thought Russians
were more patriotic than Americans.
(a) Find a 96% confidence interval for the true proportion.
(b) For n= 1,277 and a 96% confidence level, what is the largest possible margin of error
(sampling error) for any surveyed question? (Hint: The margin of error in (a) is not the answer
here.)
(c) Before the sample was taken, the belief was that 18% thought Russians were more patriotic
than Americans. Test the claim that the proportion is actually higher than 18% with a significance
level of .04.
(d) The P-value of the test in (c) is .
3. Maxwell House wants to know the proportion of adult Americans who drink at least one cup
of coffee every day. How many adults must be surveyed to be 98% confident that the estimate is
within 2% of the true proportion?
4. A population of over 25,000 people follow a special diet for one year. At the end of one year
the weights (in kilograms) of the individuals in a sample of 30 people are listed below:
78 71 82 84 77 77 83 70 65 81
87 74 70 89 80 79 73 72 77 74
83 58 81 74 84 88 76 85 78 81
(a) x=and s=.
(b) Find a 92% confidence interval for µusing the results of (a).
1
pf2

Partial preview of the text

Download Sample Exam 3 for Elementary Statistics | MATH 365 and more Exams Statistics in PDF only on Docsity!

Math 365 - Sample Exam 3

SHOW YOUR WORK on Problems 2(a,b,c), 3, 4(b,c) and 5. These problems require a solution (with an explanation) as well as an answer. A correct, but unsubstantiated answer on these prob- lems is worth only one point. For Problems 1, 2(d), and 4(a,d), fill in the blanks; no partial credit given for these problems. Do not use books or notes. BOX your answers. You will need your graphing calculator on some of the problems, and you are encouraged to use your calculator to verify your work. Calculator answers must be accurate to 3 decimal places.

  1. (a) Start with a population with mean μ and standard deviation σ. For large random sample of size n, identify the approximate distribution of X.

(b) Given a random sample of size 42 with x = 23. 4 and s = 3. 1 , a 94% MOE for the true mean is.

(c) Start with a population with proportion p. For large random sample of size n, identify the approximate distribution of pˆ.

(d) Given a random sample of size 52 with pˆ =. 4 , a 94% MOE for the true proportion is .

  1. In a sample of 1,277 Americans (New York Times, November 10, 1985), 22% thought Russians were more patriotic than Americans.

(a) Find a 96% confidence interval for the true proportion. (b) For n = 1, 277 and a 96% confidence level, what is the largest possible margin of error (sampling error) for any surveyed question? (Hint: The margin of error in (a) is not the answer here.)

(c) Before the sample was taken, the belief was that 18% thought Russians were more patriotic than Americans. Test the claim that the proportion is actually higher than 18% with a significance level of .04.

(d) The P-value of the test in (c) is.

  1. Maxwell House wants to know the proportion of adult Americans who drink at least one cup of coffee every day. How many adults must be surveyed to be 98% confident that the estimate is within 2% of the true proportion?
  2. A population of over 25,000 people follow a special diet for one year. At the end of one year the weights (in kilograms) of the individuals in a sample of 30 people are listed below:

(a) x = and s =. (b) Find a 92% confidence interval for μ using the results of (a).

(c) The mean weight of the 25,000 people before the diet starts was μ 0 = 82 kilograms. Test the company claims at a level of significance of .03 that the mean weight after one year is less than 82 kilograms.

(d) The P-value for the test of (c) is.

  1. A machine to measure the “bounce” of a tennis ball is used on 45 randomly selected balls from a population of 10,000 yellow tennis balls. Experience has shown that the standard deviation σ of the “bounce” is 0. 30. If x = 1. 7 , find a 90% confidence interval for the true mean “bounce” of the population of yellow tennis balls.

Solutions

  1. (a) N(μ,

σ √ n

) normal distribution. (b) zα 2

σ √ n

≈ z. 03

≈. 8997. (c) N

p,

p(1 − p) n

normal distribution. (d) zα 2

pˆ(1 − pˆ) n

≈ z. 03

  1. (a). 22 ± z. 02

: (. 1962 , .2438) is a 96% confidence interval for p.

(b) 96% MOE = z. 02

(ˆp)(1 − pˆ) 1277

. But pˆ(1 − pˆ) ≤

and z. 02

(c) H 0 : p 0 =. 18 vs. H 1 : p >. 18. Z =

pˆ − p 0 √ (p 0 )(1 − p 0 ) n

≈ 3. 7206 , z. 04 ≈ 1. 751. As Z ≥ z. 04 ,

reject H 0 at .04 significance level.

(d) P (Z ≥ 3 .7206) ≈. 0000994.

  1. n =

zα 2 d

z. 01

. 02

≈ 3382. 4 , n = 3383.

  1. (a) x = 77. 7 , s ≈ 6. 899.

(b) 77. 7 ± Z. 04

: (75. 495 , 79 .905) is a 92% confidence interval for μ.

(c) H 0 : μ 0 = 82 vs. H 1 : μ < 82. Z =

x − μ 0 √^ s n

≈ − 3. 4138 , −z. 03 ≈ − 1. 881. As Z ≤ −z. 03 ,

reject H 0 at .03 significance level.

(d) P (Z ≤ − 3 .4138) ≈. 0003204

  1. x ± zα 2

σ √ n

= 1. 7 ± z. 05

: (1. 626 , 1 .774) is a 90% confidence interval for μ.