52-54 on ti 84 We know that 65% of all Americans prefer chocolate over vanilla i

Question

52-54 on ti 84 We know that 65% of all Americans prefer chocolate over vanilla ice cream. Suppose that 1000 people were randomly selected. The standard error of the sample proportion is a) 0.03567 b) 0.01508 c) 0.01798 d) 0.3785 The sampling Distribution of the sample proportion is a) Binomial (1000, 0.65) b) Normal(0.65, 0.01508) c) Normal(10000, 0.65) d) None of the above What is the probability that our sample will have more than 70% of people prefer chocolate ice cream? a) 0.9995 b) 0.0005 c) 0.70 d) none of the above We are doing an experiment where we record the number of heads when we get when we flip an unbiased coin many times. For what sample sizes below would the sampling distribution of the sample proportion be approximately normally distributed? a) 5 b) 28 c) 50 d) All of the above e) None of the above For a test with the null hypothesis Ho: p = 0.5 vs. the alternative Ha: p > 0.5, the null hypothesis was not rejected at level alpha=.05. Das wants to perform the same test at level alpha = .025. What will be his conclusion? a) Reject H_0. b) Fail to Reject H_0. c) No conclusion can be made. d) Reject Ha. The null hypothesis Ho: p = .5 against the alternative Ha: p >.5 was rejected at level alpha = 0.01. Nate wants to know what the test will result at level alpha = 0.10. What will be his conclusion? e) Reject H_0.

Explanation / Answer

52) The standard error can be calculated using the below formula:

sqrt[ P * ( 1 - P ) / n ] = sqrt[ 0.65 * 0.35 / 1000 ] = 0.0150831

So option b is correct.

53) Here option B is correct as the both the assumptions are satisifed.

1. Sample includes at least 10 cases for each case

i.e. 0.65 * 1000 = 650 > 10

and 0.35 * 1000 = 350 > 10

2. And the population size should be more than 20 times the size of sample which also seems fine here.

So option B is correct here.

54) To calculate the probability that our sample will have more than 70% of the people prefer chocolate ince cream, we first need to calculate the z score corresponding to proprtion of 70%

z = (p - P) /

= (0.7 - 0.65)/0.0150831 = 3.314968 = 3.32 approx

So P( Z> 3.32) = 1 – P(Z 3.32)

= 1 - 0.99955 = 0.00045 = 0.0005 approx

So option B is correct.

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