Final answer:
The sample data supports the conclusion that the memory booster is effective at improving memory performance.
Step-by-step explanation:
To test the effectiveness of the 'memory booster' herbal mix, the researcher obtained a sample of n = 16 people and had each person take the suggested dosage each day for 4 weeks. At the end of the four-week period, each individual took a standardized memory test.
The sample produced a mean of m = 26 with a sample variance of s² = 64. In the general population, the standardized test is known to have a mean of μ = 20.
We will conduct a hypothesis test to determine if the sample data supports the conclusion that the memory booster is effective at improving memory performance.
Step 1: Define the null and alternative hypotheses
The null hypothesis (H0) states that the mean memory test score for people who take the 'memory booster' is the same as the general population mean, μ = 20. The alternative hypothesis (Ha) states that the mean memory test score for people who take the 'memory booster' is greater than the general population mean, μ > 20.
H0: μ = 20
Ha: μ > 20
Step 2: Choose a significance level
We will use a significance level of α = 0.05, which means we are willing to accept a 5% chance of rejecting the null hypothesis when it is actually true.
Step 3: Conduct the hypothesis test
We will use a one-sample t-test to compare the sample mean of m = 26 to the population mean μ = 20. We have a small sample size (n = 16), so the t-test is appropriate.
Using the sample mean, sample variance, and sample size, we can calculate the t-statistic:
t = (m - μ) / (s / sqrt(n))
Where:
m is the sample mean: 26
μ is the population mean: 20
s is the sample standard deviation: sqrt(64) = 8
n is the sample size: 16
t = (26 - 20) / (8 / sqrt(16))
t = 6 / (8 / 4)
t = 6 / 2
t = 3
Using a t-table or statistical software, we can find the critical t-value for a one-tailed test with a significance level of α = 0.05 and degrees of freedom (df) = n - 1 = 16 - 1 = 15. The critical t-value is approximately 1.753.
Since the calculated t-value of 3 is greater than the critical t-value of 1.753, we reject the null hypothesis.
Step 4: Interpret the results
The sample data supports the conclusion that the 'memory booster' is effective at improving memory performance. The mean memory test score for people who take the 'memory booster' is significantly higher than the general population mean.