Answer:
Yes, there is evidence that more than 50% of likely voters will likely vote for the incumbent.
Step-by-step explanation:
We are given that in order to estimate the proportion of all likely voters who will likely vote for the incumbent in the upcoming city’s mayoral race, a random sample of 267 likely voters is taken, finding that 65% state they will likely vote for the incumbent.
The polling agency wishes to test whether there is evidence that more than 50% of likely voters will likely vote for the incumbent.
<em>Let p = proportion of voters who will likely vote for the incumbent</em>
SO, <u>Null Hypothesis</u>,
: p
50% {means that less than or equal to 50% of likely voters will likely vote for the incumbent}
<u>Alternate Hypothesis</u>,
: p > 50% {means that more than 50% of likely voters will likely vote for the incumbent}
The test statistics that will be used here is <u>One-sample z proportion</u> <u>statistics</u>;
T.S. =
~ N(0,1)
where,
= sample proportion of voters who will likely vote for the incumbent in a sample of 267 voters = 65% or 0.65
n = sample of voters = 267
So, <em><u>test statistics</u></em> = ![\frac{0.65-0.50}{{\sqrt{\frac{0.65(1-0.65)}{267} } } } }](https://tex.z-dn.net/?f=%5Cfrac%7B0.65-0.50%7D%7B%7B%5Csqrt%7B%5Cfrac%7B0.65%281-0.65%29%7D%7B267%7D%20%7D%20%7D%20%7D%20%7D)
= 5.139
<em>Since in the question we are not given the level of significance so we assume it to be 5%. Now at 0.05 significance level, the z table gives critical value of 1.6449 for right-tailed test. Since our test statistics is more than the critical value of z so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.</em>
Therefore, we conclude that the more than 50% of likely voters will likely vote for the incumbent. The strength of the evidence is 95%.