Problem 2: The Michigan Democratic primary took place on March 8th of this year.

Question

Problem 2: The Michigan Democratic primary took place on March 8th of this year. For the remainder of the problem let PH = the proportion of Michigan democrats who prefer Hillary Clinton to Bernie Sanders 1. On March 3rd, Monmouth released a poll from a sample of 81 Michigan democrats wherein 62% preferred Hillary Clinton to Bernie Sanders (ie 50 of 81 preferred Hillary Clinton). Based on this poll, what would be your best guess of pH? [1pt] 2. Suppose, for this part of the problem, that 58% of Michigan democrats preferred Hillary Clinton to Bernie Sanders (i.e., pH = 0.58). State the sampling distribution of pH for a sample of size 81 based on this information. [1pt] 3. What is the probability of Monmouth observing a sample of size 81 where 61% or more of those polled preferred Hillary Clinton if pH 0.58 (as we assumed in the previous question)?[2pt] 4. Based on your answer to the previous question, what would be your conclusion about the hypotheses Ho : pH = 0.58 Ha : pH >0.58 based on the Monmouth poll (assuming a significance level of = 0.05)? [1pt]

Explanation / Answer

Q2 Part (1)

Monmouth survey revealed that in a sample of 81, 50 preferred Hilary Clinton. => sample proportion is 0.62.

Since sample proportion is the best estimate of the population proportion,

our best guess for pH is 0.62. ANSWER

Q2 Part (2)

Sampling distribution of sample proportion (pcap) is approximately Normal with mean = p and variance {p(1 - p)/n}, where n = sample size, p = population proportion and pcap, the sample proportion is the estimate of p.

As seen in Part (1) sample proportion is the best estimate of pH and so pHcap = pcap.

So, given pH = 0.58, sampling distribution of pHcap is approximately Normal with mean = 0.58 and variance {0.58(0.42/81} i.e., N(0.58, 0.0030) ANSWER

[The above is as per the convention N(µ, 2). In terms of N(µ, ) convention, it would be N(0.58, 0.0554)

[additional input: a thumb rule to see if the above Normal approximation is valid, the condition is that both np and np(1 - p) must be greater than 10]

Q2 Part (3)

If pcap = proportion of voters preferring Hilary Clinton in the sample survey, then as seen in Part (2), pacp is approximately Normal with mean = p = 0.58 and standard deviation 0.0554.

And what we want is the probability that proportion of voters preferring Hilary Clinton is 61% or more. i.e.,P(pcap 0.61), which by approximation, = P[Z {(0.61 – 0.58)/0.0554]

= P(Z 0.5415) = 0.2941 ANSWER [probability is obtained using Excel Function]

Q2 Part (4)

Under the null hypothesis, H0 : pH = 0.58 Vs HA : pH > 0.58, and given level of significance is 0.05, we should have P(pcap 0.61) 0.05.

Since as per Part (3), P(pcap 0.61) = 0.2941 which is much greater than 0.05, the null hypothesis is accepted. i.e., we conclude that there is sufficient evidence to suggest that actual proportion of Michigan Democrats preferring Hilary Clinton is 58%. ANSWER

Q2 Part (5)

Here we want the probability that proportion of voters preferring Hilary Clinton is 61% or more given pH = 0.49. i.e.,P(pcap 0.61), which by approximation, = P[Z {(0.61 – 0.49)/0.0556] [0.0556 = sqrt{(0.49 x 0.51)/81]

= P(Z 2.1583) = 0.0145 ANSWER [probability is obtained using Excel Function]

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