In a random sample of 380 waiters in New York, 285 admitted to lying when filing
ID: 3294306 • Letter: I
Question
In a random sample of 380 waiters in New York, 285 admitted to lying when filing their income taxes in 2012. a. Use your calculator to construct the 90% confidence interval for the proportion of all waiters in New York who lied when filing their income taxes in 2012. Test Used: 90%-Confidence Interval: b. What is the value of the margin of error of your confidence interval? c. Write a sentence to explain the meaning of the confidence interval you constructed in part a. d. Fill in the blanks to complete the interpretation of what we mean when we say that our confidence level is 90% in the context of this problem. If we were to construct 500 confidence intervals using the same population, the same _____, and the same _____, we would expect about _____ of this intervals to contain the true _____ of all waiters in New York who _____ when they filed their income taxes in 2012. A politician running for a senate seat stated that 70% of all married couples with children in which both parents work would choose to have one parent stay home if money was not a factor. In a survey of 645 married couples with children in which both parents work, 428 indicated that they would choose to have one parent stay home if money was not a factor. Does this data provide sufficient evidence to conclude that the true proportion in question is less than 70%? Use the appropriate test in your calculator. a. What is the population of interest? b. State the null and alternative hypotheses using proper notation. H_0: H_a: c. Test Used: d. Give the test statistic and the p-value: Test statistic = _____ p-value = _____ e. Decision: f. Conclusion:Explanation / Answer
Q.6 Number of waiters = 380
Lyers while filling income tax = 285
proportion p^ = 285/ 380 = 0.75 ; q^ = 0.25
(A) Test Used = Z Test
90% Confidence Interval = p^ +- Z95% sqrt [p^ q^ /N] = 0.75 +- 1.645 * sqrt [0.75 * 0.25/ 380]
= 0.75 +- 1.645 * 0.0222
= (0.7135, 0.7865)
(B) Margin of error = 0.75 - 0.7135 = 0.0365
(C) Meaning of confidence interval here is that It is 95 % confident that the proportions of waiters who lie about their income while filing income tax retuern is in between 71.35% to 78.65%.
(D) If we were to construct 500 confidence interval using the same population, with same sample size and the same confidence interval, we would expect about 475 (95% ) of this intervals to contain the true proprtion of all waiters in New York who lying when they filled their income taxes in 2012.
Questilon 7
(a) Here population of interst is " all married couple with children in which both parents work".
(b) Sample size = N = 645
Postivie success = 428
Proportion of all married couple with Children in which both parents work and willing to sacrifice the job if money is not the matter. p^ = 428/645 = 0.664
H0 : p = 0.70
Ha : p < 0.70
standard error of the sample se0 = sqrt [p * (1-p)/N] = sqrt [0.70 * 0.30/ 645] = 0.018
(c) Test used : We will use Z test here.
(d) Test Statistic :
Z = (p^ - p)/ se0 = (0.664 - 0.70)/ 0.018 = -2
p - value = (z < -2) = 0.0228
(e) Decision : We shall reject the null hypothesis
(f) Conculusion : We shall conclude that there is sufficnet evidence that the true proportions in question is less than 70%.
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