Question

1) A) The student union wants to negotiate a subsidized bus pass with your school's administration. They have polled 270 students as to whether they support this proposal, which would add a small fee to their tuition since both the college and the students would share the cost. The union found that opinions were evenly split: 50% of students were in favor of the proposal, and the rest were opposed. The margin of error for their finding at the 90% level of confidence is ± _____ %. (NUMERIC ANSWER

B)Referring to the data in Example 1 in the section above, what if we wanted to increase our level of confidence in the Angela May election poll to 99%? Match the following components of the C.I. with their correct values.

Premise

ResponseDrag and drop to match

1)Critical value -----------> A) 0.55

2)Statistic---------> B) 0.035

3) Margin of error ---------->C)64.0%

4)Upper limit----------> D)2.58

5)Lower limit--------> E)46.0%

6)Standard error for proportions-------> F) ± 9.0%

C)Referring to Example 2 in the section above, suppose that a smaller sample size was used to conduct Nielsen ratings. What would the margin of error be (assume it is a ± value, in %) for a 95% confidence interval with a sample size of just 200 cases?

Answer 1

A) The sample size is n=270. The proportion of students that are in favor of the proposal is p=0.50. The margin of error is calculated using the following formula: Margin of error = -0.05 SE (P) P(1-p) | = 0.05 + n The critical value = is calculated at the following level of significance: 0.10 a=- = 0.05 The standard error is calculated as shown below: SE(P) = (1-2) Vn 10.50x0.50 270 The vobifahalise is obtained using -NORN = 0.0304 The critical value is obtained using -NORM.S.INV(0.05) function of the MS-Excel. The value obtained is 2005 = 1.6449.

The margin of error is calculated as shown below: Margin of error = -0.05 SE (P) = 1.6449x0.0304 = 0.05005 -0.0501 Thus, the value of the margin of error is Margin of error=0.0501. 1) The level of confidence is 99%, therefore the level of significance is a =0.01. The critical value is obtained using -NORM.S.INV(0.005) function of the MS-Excel. The value obtained is -o005 = 2.58. Thus, the critical value is -0.005 = 2.58).

The statistic under consideration is the sample proportion. The value of the statistic is p=0.50). 3) The margin of error is calculated as shown below: Margin of error = -0.005 SE (P) = 2.58 0.0304 = 0.0784 Thus, the value of the margin of error is Margin of error=0.0784. 4) The 99% confidence interval for the population proportion is calculated as shown below: [p--0005 SE (p), p + 0.005 SE (P)]=[0.50 – 2.58x0.0304, 0.50 +2.58x0.0304 = [0.4216,0.5784] The value of the upper limit is 0.5784).

5) The value of the lower limit is 0.4216. The standard error is calculated as shown below: p(1-P) SE( PVn 0.500.50 V 270 = 0.0304 Thus, the value of the standard error is SE (p) = 0.0304

C) The sample size in the question is reduced to n= 200. The level of significance is a = 0.05 The margin of error is calculated using the following formula: Margin of error = -0.025 SE (P) P(1-p) = 20.025 V The critical value = is calculated at the following level of significance: 0.05 a=- 2 = 0.025 The standard error from part A of the answer is SE(P) = 0.0304. The critical value is obtained using -NORM.S.INV(0.025) function of the MS-Excel. The value obtained is -0.025 = 1.96. The margin of error is calculated as shown below: Margin of error = -0.025 SE (P) = 1.96x0.0304 = 0.0596 Thus, the value of the margin of error is Margin of error=0.0596.