

Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Material Type: Exam; Professor: Porter; Class: Elementary Statistics; Subject: Mathematics; University: University of Kansas; Term: Unknown 1989;
Typology: Exams
1 / 2
This page cannot be seen from the preview
Don't miss anything!


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.
(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 .
(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.
(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.
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
: (. 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.
zα 2 d
z. 01
. 02
≈ 3382. 4 , n = 3383.
(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
σ √ n
= 1. 7 ± z. 05
: (1. 626 , 1 .774) is a 90% confidence interval for μ.