Answer:
If we replace for the proportion estimated for the republican governor we got:
And the confidence interval given by:
If we replace for the proportion estimated for the democratic chalenger we got:
And the confidence interval given by:
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.
Step-by-step 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
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 and
. And the critical value would be given by:
The margin of error for the proportion interval is given by this formula:
(a)
If we replace for the proportion estimated for the republican governor we got:
And the confidence interval given by:
If we replace for the proportion estimated for the democratic chalenger we got:
And the confidence interval given by:
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.
