Questions: A sample of n=16 people is selected from a population with a mean of μ=80. A treatment is administered to the sample and after treatment, the sample mean is found to be M=86 with a standard deviation of s=8. Does the sample provide sufficient evidence to conclude the treatment has a significant effect? Use a two-tailed test with α=.05.

A sample of n=16 people is selected from a population with a mean of μ=80.

A treatment is administered to the sample and after treatment, the sample mean is found to be M=86 with a standard deviation of s=8.

Does the sample provide sufficient evidence to conclude the treatment has a significant effect?

Use a two-tailed test with α=.05.

Transcript text: A sample of $n=16$ people is selected from a population with a mean of $\mu=80$. A treatment is administered to the sample and after treatment, the sample mean is found to be $M=86$ with a standard deviation of $s=8$. Does the sample provide sufficient evidence to conclude the treatment has a significant effect? Use a two-tailed test with $\alpha=.05$.

Solution

Solution Steps

Step 1: Calculate the Standard Error

The standard error $ SE $ is calculated using the formula:

\[ SE = \frac{s}{\sqrt{n}} = \frac{8}{\sqrt{16}} = 2.0 \]

Step 2: Calculate the Test Statistic

The test statistic $ t $ is calculated using the formula:

\[ t = \frac{\bar{x} - \mu_0}{SE} = \frac{86 - 80}{2.0} = 3.0 \]

Step 3: Calculate the P-value

For a two-tailed test, the p-value $ P $ is calculated as:

\[ P = 2 \times (1 - T(|z|)) = 0.009 \]

Step 4: Conclusion

Since the p-value $ P = 0.009 $ is less than the significance level $ \alpha = 0.05 $, we reject the null hypothesis. This indicates that the sample provides sufficient evidence to conclude that the treatment has a significant effect.