Question

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Provide an appropriate response. 1) Many people think that a national lobby's successful fight against gun control legislation is reflecting the will of a minority of Americans. A random sample of 4000 citizens yielded 2250 who are in favor of gun control legislation. Estimate the true proportion of all Americans who are in favor of gun control legislation using a 98% confidence interval. Express the answer in the form p E and round to the nearest ten-thousandth. 04375 00182 D6250.5727 B) 0.5625 0.0182 C) 0.4375 0.0182 A) 0.4375:0.5727 2) A senator wishes to estimate the proportion of United States voters who favor abolishing the Electoral College. How large a sample is needed in order to be 90% confident that the sample proportion will not differ from the true proportion by more than 5%? C) 9 D) 164 A) 542 B) 271 3) 3) The area under the graph of every Student's t-distribution is A) 1 B) Greater than the standard normal distribution C) Less than the standard normal distribution Area of the standard normal distribution D) Find the t-value. 4) Find the t-value such that the area left of the t-value is 0.2 with 4 degrees of freedom 4) A) 0.978 B)-0.941 C) 0.941 D)-2.999 rovide an appropriate response. 5) Construct a 95% confidence interval for the population mean, μ. Assume the population has a 5) normal distribution. A sample of 20 part-time workers had mean annual earnings of S3120 with a standard deviation of $677. Round to the nearest dollar. A) ($2657, $2891) B) ($2135, $2567 C) (S2803, $3437) D) ($1324, $1567) 6) Construct a 95% confidence interval for the population mean, μ. Assume the population has a 6) normal distribution. A sample of 25 randomly English majors has a mean test score of 81.5 with a standard deviation of 10.2. Round to the nearest hundredth A) (77.29, 85.71) B) (87.12, 98.32) C) (56.12, 78.34) D) (66.35, 69.89) 7) A local bank needs information concerning the savings account balances of its customers. A 7) random sample of 15 accounts was checked. The mean balance was $686.75 with a standard deviation of S25620. Find a 98% confidence interval for the true mean. Assume that the account balances are normally distributed. Round to the nearest cent. A) (5326.21, $437.90) C) ($513.17, 5860.33) B) ($238.23, $326.41) D) ($487.31, $563.80) 8) The prinipal at Riverside High School would like to estimate the mean length of time each day that it takes all the buses to arrive and unload the students. How large a sample is needed if the prírwipal would like to assert with 90% confidence that the eample mean eoll by-mo minutes. Assume that s14 minutes based on previous studies A) 10 B) 13 D) 12 9) It is claimed that the mean age of professors at a university is greater than 57 7 hypothests test is performed, how should you interpret a decision that rejects the null hypothesis A) There is not sufficient evidence to reject the claim 57.7 D) There is not sufficient evidence to support the claim u > 57.7 C) There is sufficient evidence to repot the claim μ > 57.7. D) There is sufficient evidence to support the claim μ > 577 10) Find the critical z value for a two-tailed test with a 0.01 10) A) 12.33 B) 1.96 C) 1.645 D) 12.575 11) A survey claims that 9 out of 10 doctors (Le, 90%) recommend brand Z for their patients who have 11) children. To test this claim against the alternative that the actual proportion of doctors who recommend brand Z is less than 90%, a random sample of doctors was taken. Suppose the test statistic is z -2.23. Can we conclude that Ho should be rejected at the a) a 0.10, b) a-0.05, and ca 0.01 level? A) a) noj b) no; c) no C) a) yes; b) yes c) yes B) a) no; b) no; c) yes D) a) yes, b) yes; c) no Find the critical value. 12) Determine the critical value for a right-tailed test of a population mean at the a 0.025 level of12) significance with 19 degrees of freedom. A) 3.174 B) 2.093 C)-2.093 D) 2.101 Provide an appropriate response. 13) Find the standardized test statistic t for a sample with n-12, x-192, 22, and α-0.01 if 13) Ho: H-18. Round your answer to three decimal places A) 2.132 B) 2.001 C) 1.890 D) 1.991 14) Find the standardized test statistic t for a sample with n- 12, x 19.5, s2.1, and a 0.01 if 14) HI : μ 20. Round your answer to three decimal places. C)-0.825 D) -0008 A) -0.381 B)-0.037 15) Find the standardized test statistic estimate, z, to test the hypothesis that p1> p2. Use a 0.01. The 15) sample statistics listed below are from independent samples 100, x1 38, and n2-140, x2 50 16) Construct a 98% confidence interval for P1-P2. The sample statistics listed below are from 250, and n2 Sample statistics:n1 A) 2.116 D) 0.638 B) 0.362 C)1.324 16) independent samples Sample statistics: n1 1000, x1 1200, x2-195 D) (0.047, 0.128) A) (1.516,3.021) C) (0.581, 1.819) B) (-0.621, 0.781) 17) Data sets A and B are dependent. Find d 17) A 1 31 41 6.0 3.0 3.1 B35 44 43 42 56 Assume that the paired data came from a population that is normally distributed A)-0.76 B) 0.58 D)-0.94 C) 0.89 18) Construct a 95% confidence interval for data sets A and B. Data sets A and B are dependent. 18) A 30 28 47 43 31 B 28 24 25 35 22 Assume that the paired data came from a population that is normally distributed B) (-0.113, 12.761) D) (-15.341, 15.431) A) (-0.696, 18.700) C) (-1.324, 8.981) u1 H2. Two samples are randomly selected and 19) 19) Find the test statistic, t, to test the hypothesis that come from populations that are normal. The sample statistics are given below. n2 30 n1-25 2x2-26 $2-1.9 1- 1.5 A) 4.361 D) 3.287 C) 2892 B) 1.986 20) made a 20) If we do not reject the null hypothesis when the null hypothesis is in error,then we have B) Type II error D) Type I error A) Correct decision C) Type B error

Answer 1

1)

Sample size = n = 4000

x = 2250

Sample proportion is

P 000-0.5625 n 4000

We have to construct 98% confidence interval for the population proportion.

Formula is

$(\hat{p}-E , \hat{p}+E)$

Here E is a margin of error.

$E =Zc*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

Zc = 2.32

0.5625(1-0.5625) 4000 E = 2.32 * 2.32 * V0.0000615 0.0182

So confidence interval is ( 0.5625 - 0.0182 , 0.5625 + 0.0182)

B)0.5625 0.0182

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