Math-131 Test 3 Name that best completes the statement or answers the question.
ID: 2909188 • Letter: M
Question
Math-131 Test 3 Name that best completes the statement or answers the question. MULTIPLE CHOICE Choose the one alternative 1) Finine critical value zc that corresponds to a 94% confidence level. B) +1.645 Provide an appropriate response. 1) C) 2.33 D) +1.96 A) 1.88 ) Find the margin of error for the given values of c, o, and n e-aso, ?-115, n-120 2) A) 0.94 B) 1.05 C) 1.73 D) 0.16 3) In a random sample of 60 computers, the mean repair cost was $150. Assume the population standard deviation is $36. Construct 90% confidence interval for the population mean. 3) A) ($142, $158) B) ($141, $159) C) ($138, $162) D) ($537, 5654) 4) The standard IQ test has a mean of 96 and a standard deviation of 14. We want to be 99% certain 4) . that we are within 4 IQ points of the true mean. Determine the required sample size. A) 82 B) 178 C) 1 D) 10 5) Find the critical value, to forc=095 and n-16. 5) A) 2.120 B) 2947 C)2.131 D) 2.602 6) Find the value of E, the margin of error, for c 0.90, n-10 and s 3.1 B) 1.80 6) A) 057 C) 1.78 D) 1.36 7) Construct the indicated confidence interval for the population mean ? using the t-distribution. 7) c " 0.95, x 645, s A) (531.2, 612.9) 31, n = 16 C) 1.7, 365.8) D) (876.2, 981.5) 8) In a random sample of 28 families, the average weekly food expense was $95.60 with a standard 8) deviation of $22.50. Determine whether a normal distribution or a t-distribution should be used or whether neither of these can be used to construct a confidence interval. Assume weekly food expenses is normally shaped the distribution of A) Cannot use normal distribution or t-distribution. B) Use normal distribution. C) Use the t-distribution, For a sample of 20 whether a normal distribution or a t ton or a t-distribution should be used or whether neither of these can be deviation, a, is 15. Determine 9) used to construct a confidenice intervalL Assume that Qcres A) Use the t-distribution. B) Cannot use normal distribution or C) Use normal distribution.Explanation / Answer
Solution:-
1) option A) +/- 1.88
2) option C) 1.73
=> Z*sigma/sqrt(n) = 1.645*11.5/sqrt(120) = 1.7269
3) option A) ($142,$158)
=> X +/- Z*s/sqrt(n) = 150 +/- 1.645*36/sqrt(60) = (142.354 , 157.645)
4) option A) 82
=> n = (Z*sigma/E)^2 = (2.58*14/4)^2 = 81.54 = 82
5) option C) 2.131
6) option C) 1.78
=> M = 1.833*3.1/sqrt(10) = 1.7969 = 1.78
7) option B) (628.5,661.5)
=> x +/- t*s/sqrt(n) = 645 +/- 2.131*31/sqrt(16) = (628.485 , 661.515)
8) option C) use the t-distribution
9) option C) Use normal distribution