A politician claims that she will receive 66% of the vote in an upcoming election. The results of a properly designed random sample of 145 voters showed that 80 of those sampled will vote for her. Is it likely that her assertion is correct at the 0.01 level of significance?
Find the p-value. (Give your answer correct to four decimal places.)
p-value is the probability of obtaining the observed sample result, in this case the "properly designed random sample."
p(vote)= 80/145
= .55
= 55%
.55 DOES NOT equal to 66% or .66
To find the p-value, we need to perform a hypothesis test.
Step 1: State the hypotheses.
The null hypothesis (H0): The politician will receive 66% of the vote.
The alternative hypothesis (H1): The politician will not receive 66% of the vote.
Step 2: Set the level of significance.
The significance level (α) is given as 0.01. This means we are willing to accept a 1% chance of making a Type I error (rejecting the null hypothesis when it is actually true).
Step 3: Calculate the test statistic.
We can use the z-test statistic for proportions to compare the sample proportion to the claimed proportion.
The formula for the z-test statistic is:
z = (p̂ - p₀) / sqrt(p₀(1-p₀)/n)
Where:
p̂ is the sample proportion (80/145 ≈ 0.5517)
p₀ is the claimed proportion (0.66)
n is the sample size (145)
Plugging in the values, we get:
z = (0.5517 - 0.66) / sqrt(0.66(1-0.66)/145)
Step 4: Find the p-value.
The p-value is the probability of observing a test statistic as extreme as the one obtained, assuming that the null hypothesis is true.
We can find the p-value by looking up the z-score in the standard normal distribution table or by using a calculator. In this case, the test statistic is negative. Therefore, we want to find P(Z ≤ z).
For a two-tailed test, since we are interested in both tails, we need to find P(Z ≤ z) and P(Z ≥ z), and multiply the result by 2.
Using a calculator or a table, we find that P(Z ≤ z) ≈ 0.9483.
Since the test is two-tailed, the p-value is 2 * P(Z ≤ z) ≈ 2 * 0.9483 = 1.8966.
Step 5: Make a decision.
Finally, we compare the p-value to the significance level.
Since the p-value (1.8966) is greater than the significance level (0.01), we fail to reject the null hypothesis. This means that there is not enough evidence to conclude that the politician's claim of receiving 66% of the vote is incorrect at the 0.01 level of significance.
Therefore, it is likely that her assertion is correct at the 0.01 level of significance.