A random sample of n = 12 individuals is selected from a population with
µ = 70 and a treatment is administered to each individual in the sample. After treatment the sample mean is found to be M = 74.5 with S = 5.20. Based on the sample data, the treatment has a significant effect as per the deduced test statistic. Following are the statements for Null and the Alternate Hypotheses,
H₀: The treatment does not have a significant effect
Hₐ: The treatment has a significant effect
According to the given information,
Size of the sample, n = 12
Population Mean, μ = 70
Sample Mean, M = 74.5
Standard Deviation, S = 5.20
The Null hypothesis is given by,
H₀: The treatment does not have a significant effect (μ = 70)
The Alternate hypothesis is given by,
Hₐ: The treatment has a significant effect (μ ≠ 70)
Since the population standard deviation is unknown, we will use one-sample t-test here.
Test statistic, TS = (M-μ) / (s/√n)
Substituting the given values of M, μ, S, and n, we get,
TS = (74.5-70) / (5.20/√12)
TS = 4.5 / 1.5
TS ≈ 3
The t table now provides critical values of -2.131 and 2.131 at 15 degrees of freedom for a two-tailed test using α = 0.05 (significance level).
Since our test statistic, TS does not fall in the range of the critical values, it falls in the rejection region. Thus, we can reject the null hypothesis and say that the treatment has a significant effect.
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