A researcher is testing the effect of a new cold and flu medication on mental alertness. A sample of n=9 college students is obtained and each student is given the normal dose of the medicine. Thirty minutes later, each students performance is measured on a video game that requires careful attention and quick decision making. The scores for the nine students are as follows: 6,8,10,6,7,13,5,5,3.

a. Assuming that scores for students in the regular population average u=10, are the data sufficient to conclude that the medication has a significant effect on mental performance? Test at the .05 level of significance.

Try a one-sample t-test since your sample size is small. You will need to calculate the mean and standard deviation. Use a t-table to find the critical value at .05 level of significance for a one-tailed test. Degrees of freedom is (n - 1). Sample size (n) is 9.

To determine whether the medication has a significant effect on mental performance, we can perform a hypothesis test.

Step 1: State the null and alternative hypotheses:
Null hypothesis (H0): The medication does not have a significant effect on mental performance. (u = 10)
Alternative hypothesis (Ha): The medication has a significant effect on mental performance. (u ≠ 10)

Step 2: Select the appropriate test statistic:
Since we have a sample size less than 30 and the population standard deviation is unknown, we'll use a t-test.

Step 3: Determine the significance level:
The significance level, also known as alpha (α), is given as 0.05.

Step 4: Calculate the test statistic:
To calculate the test statistic, we use the formula:
t = (sample mean - population mean)/(sample standard deviation / sqrt(sample size))

The sample mean is (6+8+10+6+7+13+5+5+3)/9 = 7.78

To calculate the sample standard deviation, first calculate the squared differences from the mean for each score:
(6-7.78)^2=3.1684, (8-7.78)^2=0.0476, (10-7.78)^2=4.6924, (6-7.78)^2=3.1684, (7-7.78)^2=0.6084, (13-7.78)^2=27.2624, (5-7.78)^2=8.1584, (5-7.78)^2=8.1584, (3-7.78)^2=17.7024
Then calculate the sample variance by summing up these squared differences: 3.1684+0.0476+4.6924+3.1684+0.6084+27.2624+8.1584+8.1584+17.7024=72.9572
Finally, calculate the sample standard deviation: sqrt(72.9572/8)=sqrt(9.11965)=3.0182

Now plug in these values to calculate the test statistic:
t = (7.78 - 10)/(3.0182 / sqrt(9))
t = -2.22

Step 5: Determine the critical value:
Since the alternative hypothesis is two-tailed, we need to find the critical values for a t-distribution with 8 degrees of freedom (n-1). Looking up the critical value in a t-table or using software, we find the critical values to be approximately -2.306 and 2.306 at a significance level of 0.05.

Step 6: Make a decision:
If the test statistic falls within the critical region (outside the accepted range), we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.

In this case, -2.22 falls within the range between -2.306 and 2.306. Therefore, we fail to reject the null hypothesis.

Step 7: State the conclusion:
Based on the data, there is not sufficient evidence to conclude that the medication has a significant effect on mental performance at the 0.05 level of significance.

To determine if the medication has a significant effect on mental performance, we need to conduct a hypothesis test.

First, let's state our null hypothesis (H0) and alternative hypothesis (Ha):

H0: The medication has no significant effect on mental performance (u = 10)
Ha: The medication has a significant effect on mental performance (u ≠ 10)

Next, we calculate the sample mean (x̄) and sample standard deviation (s) of the scores:

x̄ = (6 + 8 + 10 + 6 + 7 + 13 + 5 + 5 + 3) / 9 = 7.8
s = √[((6 - 7.8)² + (8 - 7.8)² + (10 - 7.8)² + (6 - 7.8)² + (7 - 7.8)² + (13 - 7.8)² + (5 - 7.8)² + (5 - 7.8)² + (3 - 7.8)²) / (9 - 1)] ≈ 2.93

Now, we can calculate the t-score using the formula:

t = (x̄ - u) / (s / √n)
= (7.8 - 10) / (2.93 / √9)
= -2.2

With a sample size of 9 (n = 9), we have 9 - 1 = 8 degrees of freedom.

Now, we need to compare the t-score to the critical value from the t-distribution at the 0.05 significance level. The critical value can be obtained from a t-table or using statistical software. For a two-tailed test, the critical value for a significance level of 0.05 with 8 degrees of freedom is approximately ±2.306.

Since our calculated t-score (-2.2) is within the acceptance range of -2.306 to 2.306, we fail to reject the null hypothesis.

Therefore, based on the given data, we do not have sufficient evidence to conclude that the medication has a significant effect on mental performance.