A random sample of 200 Vancouverites are asked whether they support the Trans Mo

Question

A random sample of 200 Vancouverites are asked whether they support the Trans Mountain pipeline expansion. Their responses are recorded as 1 (“yes”) or 0 (“no”) in the file pipeline(GIven Below).

a) Provide a 95% Wald confidence interval for the probability that a Vancouverite supports the pipeline.

b) Provide a 95% Agresti-Coull confidence interval for the probability that a Vancouverite supports the pipeline.

c) Which of your confidence intervals above (a or b) do you expect to be closer to valid and why?

Pipeline FIle:

response
1
1
1
0
0
0
0
0
1
0
0
0
0
1
0
1
0
1
0
1
0
0
1
0
0
0
0
1
0
1
1
0
0
0
0
0
1
0
0
1
0
1
0
0
1
1
1
0
1
1
0
0
0
0
1
0
1
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
1
1
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
1
0
0
0
1
0
1
0
1
0
1
0
1
0
0
1
0
0
1
0
0
1
1
0
1
0
1
1
1
1
0
0
0
1
0
0
1
0
0
0
1
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
1
0
0
0
1
0
0
1
0
1
0
0
0
0
0
1
0
1
0
0
0
1
1
0
0
0
0
0
1
0
0
1
0
1
0
0
0
0
1
0
0
0
0
1
1
1

Explanation / Answer

Solution

Back-up Theory

1. Probability that a Vancouverite supports the pipeline

    = proportion, p, of Vancouverites population who support the pipeline.

2. p is estimated by the sample proportion,

pcap = Number of Vancouverites in the sample who support the pipeline/Total sample size

= (Number of Vancouverites in the sample whose responses were 1) / 200

= 65/200 [65 is obtained using Excel Function ‘COUNTIF’]

= 0.325.

Part (a)

100(1 - )% Wald Confidence Interval for p = pcap ± Z/2SE(pcap)

= pcap ± Z/2{pcap(1 - pcap)/n}, where Z/2 = upper (/2) percent point of N(0, 1) and n = sample size.

So, 95% Wald Confidence Interval for p = 0.325 ± 1.96{0.325(0.675)/200}

= 0.325 ± (1.96 x 0.0331) = 0.325 ± 0.0649

Lower Bound = 0.2601 and Upper Bound = 0.3899 ANSWER

Part (b)

100(1 - )% Agresti-Couli Confidence Interval for p = N/D, where

N = {pcap + (Z/22/2n) ± Z/2[{pcap(1 - pcap)/n} + (Z/22/4n2)]

D = 1 + (Z/22/n).

Substituting the values,

N = {0.325 + (1.962/400)} ± 1.96{0.0010969 + (1.962/160000)}

    = 0.337604 ± 1.960.00112091

    = 0.337604 ± (1.96 x 0.0335)

    = 0.337604 ± 0.0656

D = 1 + 0.019208 = 1.019208

Lower Bound = 0.2669 and Upper Bound = 0.3956 ANSWER

Part (c)

In general, Agresti-Couli Confidence Interval is preferred when the sample size is not large enough. But,when n is large as in the present case, both are equally reliable.

This is also reflected in the values obtained – the values are pretty close.

DONE

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