In the population, the average IQ is μ = 100 with a standard deviation of σ = 15. A team of scientists wants to test a new medication to see if it has a positive effect on intelligence. A sample of n = 10 participants who have taken the medication has a mean of M = 110. Did the medication significantly increase intelligence, using a one tail test with an alpha = 0.01?

Z = (score-mean)/SEm

SEm = SD/√n

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability of the Z score. Is it ≤ .01?

To determine if the medication significantly increased intelligence, we can conduct a hypothesis test. The first step is to establish the null and alternative hypotheses:

Null Hypothesis (H₀): The medication did not significantly increase intelligence.
Alternative Hypothesis (H₁): The medication significantly increased intelligence.

Next, we need to determine the critical value and the test statistic. Since we are conducting a one-tail test (to see if there is a significant increase), we need to find the critical value that corresponds to an alpha level of 0.01 in the right tail of the distribution.

To find the critical value, we can use a standard normal distribution table or a statistical software. A critical value of 2.33 corresponds to an alpha level of 0.01 in the right tail of the distribution.

The test statistic can be calculated by using the formula:

t = (M - μ) / (σ / √n)

where M is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.

In this case, M = 110, μ = 100, σ = 15, and n = 10. Plugging in these values, we get:

t = (110 - 100) / (15 / √10)
t = 10 / 4.743
t ≈ 2.11

Once we have the test statistic, we can compare it to the critical value. If the test statistic is greater than the critical value, we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.

In this case, the test statistic (2.11) is less than the critical value (2.33), so we fail to reject the null hypothesis. This means that there is not enough evidence to conclude that the medication significantly increased intelligence.

In conclusion, based on the given sample of participants, the medication did not have a significant effect on intelligence at the α = 0.01 level.