Question

A drug manufacturer claimed that the mean potency of one of its antibiotics was 80%. A random sample of 100 capsules were tested, giving a sample mean of 79.7%, with a standard deviation of 0.8%. To test if the data present sufficient evidence to refute the manufacturer's claim, let

a) 0.05
b) 0.01
c) 0.10
d)0.20

Answer 1

The p-value for the test at
`\[ alpha = 0.05} \]` is less than 0.05; therefore, we reject the null hypothesis.

Therefore, option a) 0.05 is correct.

Step-by-step explanation:

To assess whether the data refutes the manufacturer's claim, conduct a one-sample t-test. With a sample mean of 79.7%, a claimed mean of 80%, a sample size of 100, and a standard deviation of 0.8%, calculate the t-statistic.

Using the formula
`\(t = \frac{{\bar{x} - \mu}}{{s/√(n)}}\)`, where
`\(\bar{x}\)` is the sample mean,
`\(\mu\)` is the claimed mean, s is the standard deviation, and n is the sample size, we find the t-value. Then, find the p-value associated with the t-value.

With the calculated p-value, compare it to the significance levels
`\alpha\)` of 0.05, 0.01, 0.10, and 0.20. If the p-value is less than
`\(\alpha\)`, it suggests that there is sufficient evidence to reject the null hypothesis. In this case, the p-value at
`\(\alpha = 0.05\)` is less than 0.05, leading to the rejection of the null hypothesis.

Therefore, option a) 0.05 is correct.