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
You can try a proportional one-sample z-test for this one since this problem is using proportions.
Using a formula for a proportional one-sample z-test with your data included, we have:
z = .16 - .10 -->test value (13/81 is .16) minus population value (.10) divided by
√[(.10)(.90)/81] --> .90 represents 1-.10 and 81 is sample size.
Finish the calculation. Use a z-table to determine the p-value for the z-test statistic.
To find the p-value for this situation, we need to perform a hypothesis test.
First, we need to set up our null and alternative hypotheses.
Null hypothesis (H0): The proportion of crushed buns is equal to or less than 10% (or 0.10).
Alternative hypothesis (Ha): The proportion of crushed buns is greater than 10% (or 0.10).
Next, we need to calculate the test statistic and p-value. In this case, we'll be using a one-sample proportion z-test.
The test statistic (z) is calculated using the formula:
z = (p̂ - p) / sqrt(p * (1 - p) / n)
Where:
- p̂ is the sample proportion of crushed buns
- p is the hypothesized proportion of crushed buns under the null hypothesis (0.10)
- n is the sample size (81)
In this case, p̂ = 13/81 ≈ 0.1605
Now, let's calculate the test statistic:
z = (0.1605 - 0.10) / sqrt(0.10 * (1 - 0.10) / 81)
≈ 0.1605 - 0.10 / 0.030973
≈ 0.0605 / 0.030973
≈ 1.951
To find the p-value, we need to determine the probability of observing a test statistic as extreme as or more extreme than the one calculated (1.951) under the null hypothesis.
We then use a standard normal distribution table or calculator to find the area under the curve that corresponds to the test statistic. In this case, we're looking for the area to the right of 1.951.
The p-value for this situation is the probability of getting a z-value greater than 1.951. From the standard normal distribution table, we find that the area to the right of 1.951 is approximately 0.0278.
Therefore, the p-value for this situation is approximately 0.0278, which is not one of the listed options.