Final answer:
Using a one-sample proportion z-test, the calculated z-score for the test statistic is 2.335. Looking this up on a standard normal distribution table or using a software tool gives us a P-value of approximately 0.0096. This low P-value suggests that there is a statistically significant difference from the 45% claim, with a high level of confidence.
Step-by-step explanation:
To test the claim that the percentage of residents who favor annexation is over 45%, we use a one-sample proportion z-test. Since a sample of 1500 voters shows that 48% favor annexation, our sample proportion (p-hat) is 0.48. The null hypothesis for this test is that the true proportion of residents who favor annexation is 45% (p0 = 0.45).
The test statistic z is calculated using the formula:
z = (p-hat - p0) / sqrt((p0*(1-p0)) / n)
Plugging in the values given, we have:
z = (0.48 - 0.45) / sqrt((0.45*(1-0.45)) / 1500)
z = 0.030 / sqrt((0.45*0.55) / 1500)
z = 0.030 / sqrt(0.2475 / 1500)
z = 0.030 / sqrt(0.000165)
z = 0.030 / 0.01285
z = 2.335
To find the P-value, we look up the z-score in a standard normal distribution table or use a software tool. The P-value corresponds to the probability of observing a z-score as extreme as the one calculated, assuming the null hypothesis is true. Given the z-score of 2.335, the P-value is approximately 0.0096.
This P-value of 0.0096 indicates that, if the true proportion was actually 45%, there's a 0.96% chance of observing a sample proportion of 48% or more, simply due to random sampling error.