A local McDonald's manager will return a shipment of hamburger buns if more than 10% of the buns are crushed. A random sample of 81 buns finds 13 crushed buns. A 5% significance test is conducted to determine if the shipment should be accepted. The p value for this situation is:
0.0348
0.0500
0.0700
0.0436
0.0218
0.0348
To find the p-value for this situation, we need to perform a hypothesis test.
First, let's set up the hypotheses:
Null hypothesis (H0): The proportion of crushed buns is equal to or less than 10%.
Alternative hypothesis (Ha): The proportion of crushed buns is greater than 10%.
Next, we need to calculate the test statistic. In this case, we are dealing with proportions, so we can use the z-test for proportions.
The test statistic can be calculated using the formula:
z = (p̂ - p) / sqrt(p(1-p) / n)
Where:
p̂ is the sample proportion of crushed buns (13/81)
p is the hypothesized proportion (0.10)
n is the sample size (81)
Calculating the test statistic:
z = (13/81 - 0.10) / sqrt(0.10(1-0.10)/81)
z = (0.1605 - 0.10) / sqrt(0.09/81)
z = 0.0605 / sqrt(0.0011)
z ≈ 2.829
Now, we need to find the p-value associated with this test statistic. Since the alternative hypothesis is that the proportion is greater than 10%, we need to find the area to the right of the test statistic on the standard normal distribution.
Using a table or a statistical software, we can find the p-value associated with a z-score of 2.829. The p-value turns out to be approximately 0.00218.
However, we need to be careful when interpreting the p-value. Since the alternative hypothesis is one-sided (greater than), we need to double the p-value to obtain the final result.
So the p-value for this situation is approximately 0.00436, which is closest to 0.0436 (answer choice D).