Question

You are conducting a study to see if the probability of a true negative on a test for a certain cancer is significantly less than 0.87. You use a significance level of α=0.01α=0.01.

H0:p=0.87H0:p=0.87
H1:p<0.87H1:p<0.87
You obtain a sample of size n=682n=682 in which there are 567 successes.
What is the test statistic for this sample? (Report answer accurate to three decimal places.)
test statistic =
What is the p-value for this sample? (Report answer accurate to four decimal places.)
p-value =
The p-value is...
a. less than (or equal to) αα
b. greater than αα
This test statistic leads to a decision to...
a. reject the null
b. accept the null
c. fail to reject the null
As such, the final conclusion is that...
There is sufficient evidence to warrant rejection of the claim that the probability of a true negative on a test for a certain cancer is less than 0.87.
There is not sufficient evidence to warrant rejection of the claim that the probability of a true negative on a test for a certain cancer is less than 0.87.
The sample data support the claim that the probability of a true negative on a test for a certain cancer is less than 0.87.
There is not sufficient sample evidence to support the claim that the probability of a true negative on a test for a certain cancer is less than 0.87.

Answer 1

The test statistic for this sample is -2.797 and the p-value is approximately 0.0028. The p-value being less than the significance level leads to a decision to reject the null hypothesis. Thus, there is sufficient evidence to reject the claim that the probability of a true negative on a test for a certain cancer is less than 0.87.

Step-by-step explanation:

The test statistic for this sample can be calculated as follows:

1. Calculate the sample proportion, p', by dividing the number of successes (567) by the sample size (682). p' = 567/682 = 0.8312.
2. Calculate the standard error (SE) using the formula SE = sqrt((p' * (1 - p')) / n). SE = sqrt((0.8312 * (1 - 0.8312)) / 682) = 0.0138.
3. Calculate the test statistic (Z) using the formula Z = (p' - p) / SE, where p is the null hypothesis value. Z = (0.8312 - 0.87) / 0.0138 = -2.797.

Therefore, the test statistic for this sample is -2.797.

The p-value for this sample can be obtained by finding the probability of obtaining a test statistic lower than -2.797 in a standard normal distribution. Using a standard normal distribution table or a statistical software, we find that the p-value is approximately 0.0028.

Since the p-value (0.0028) is less than the significance level (0.01), we reject the null hypothesis. The final conclusion is that there is sufficient evidence to warrant rejection of the claim that the probability of a true negative on a test for a certain cancer is less than 0.87.