Question

⦁   A nutrition store in the mall is selling “Memory Booster,” which is a concoction of herbs and minerals that is intended to improve memory performance, but there is no good reason to think it couldn't possibly do the opposite. To test the effectiveness of the herbal mix, a researcher obtains a sample of 8 participants and asks each person to take the suggested dosage each day for 4 weeks. At the end of the 4-week period, each individual takes a standardized memory test. In the general population, the standardized test is known to have a mean of μ = 7

Answer 1

Solution:

Solution:

The formulas for mean and standard deviation are given as below:

Mean = M = ∑X/n

Variance = S^2 = ∑(X - M)^2 / (n – 1)

Standard deviation = S = Sqrt[∑(X - M)^2 / (n – 1)]

SS = ∑(X - M)^2

The calculation table is given as below:

Sample scores	X - M	(X - M)^2
8	0	0
9	1	1
6	-2	4
8	0	0
9	1	1
8	0	0
7	-1	1
9	1	1
∑X = 64		SS = 8
n = 8		S^2 = 1.142857143
M = 8		S = 1.069044968

Now, we have to use one sample t test for the population mean.

The null and alternative hypotheses are given as below:

Null hypothesis: H0: The average standardized test score is 7.

Alternative hypothesis: Ha: The average standardized test score is greater than 7.

H0: µ = 7 versus Ha: µ > 7

This is an upper tailed or right tailed test.

The test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

µ = 7

Xbar = 8

S = 1.069044968

n = 8

df = n – 1 = 7

We assume α = 0.05

Critical value = 1.8946

(by using t-table or excel)

t = (8 – 7)/[ 1.069044968/sqrt(8)]

t = 2.6458

P-value = 0.0166

(by using t-table)

P-value < α = 0.05

So, we reject the null hypothesis.

There is sufficient evidence to conclude that the average standardized test score is greater than 7.

There is sufficient evidence to conclude that the herbal mix is effective.

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