Question

5. If the sample size had been 2,500 registered voters, and the results stated 47% would vote for the Republican governor, and 45% said they would vote for the Democratic challenger,

what would the margin of error have been?
Could the director still say that the election was a toss-up?

Answer 1

If we replace for the proportion estimated for the republican governor we got:

`ME=1.96\sqrt{(0.47 (1-0.47))/(2500)}=0.0196`

And the confidence interval given by:

`0.47-0.0196= 0.45`

`0.47+0.0196= 0.49`

If we replace for the proportion estimated for the democratic chalenger we got:

`ME=1.96\sqrt{(0.45 (1-0.45))/(2500)}=0.0195`

And the confidence interval given by:

`0.45-0.0195= 0.43`

`0.45+0.0195= 0.47`

As we can see both confidence intervals not contain the value 0.5 for the true proportion so then the director cannot said that the election was a toss up at 5% of significance.

Explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

`p \sim N(p,\sqrt{(p(1-p))/(n)})`

Solution to the problem

We assume for this case a confidence of 95%. In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by
`\alpha=1-0.95=0.05` and
`\alpha/2 =0.025`. And the critical value would be given by:

`z_(\alpha/2)=-1.96, z_(1-\alpha/2)=1.96`

The margin of error for the proportion interval is given by this formula:

`ME=z_(\alpha/2)\sqrt{(\hat p (1-\hat p))/(n)}` (a)

If we replace for the proportion estimated for the republican governor we got:

`ME=1.96\sqrt{(0.47 (1-0.47))/(2500)}=0.0196`

And the confidence interval given by:

`0.47-0.0196= 0.45`

`0.47+0.0196= 0.49`

If we replace for the proportion estimated for the democratic chalenger we got:

`ME=1.96\sqrt{(0.45 (1-0.45))/(2500)}=0.0195`

And the confidence interval given by:

`0.45-0.0195= 0.43`

`0.45+0.0195= 0.47`

As we can see both confidence intervals not contain the value 0.5 for the true proportion so then the director cannot said that the election was a toss up at 5% of significance.