Respuesta :
Answer:
Step-by-step explanation:
We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean
For the null hypothesis,
µ = 9.02
For the alternative hypothesis,
µ < 8.02
This is a left tailed test due to the inequality symbol.
Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.
Since n = 22
Degrees of freedom, df = n - 1 = 22 - 1 = 21
t = (x - µ)/(s/√n)
Where
x = sample mean = 8.02
µ = population mean = 9.02
s = samples standard deviation = 0.7
t = (8.02 - 9.02)/(0.7/√22) = - 6.701
We would determine the p value using the t test calculator. It becomes
p = 0.00001
Since alpha, 0.01 > than the p value, 0.00001, then we would reject the null hypothesis. Therefore, At a 1% level of significance, the sample data showed significant evidence that the hippocampal volumes in the alcoholic adolescents were less than the normal volume.
Answer:
We reject H₀ we have evidence to claim that the hippocampal volumes in the alcoholic adolescents are less than the normal values.
Step-by-step explanation:
Assume Normal Distribution
Population mean μ₀ = 9,02
Standard deviation Unknown
Sample:
Size 22 then n = 22 and degree of fredom df = 22 - 1 df = 21
Sample mean X = 8,02
sample s = 0,7
Confidence interval:
significance level α = 0,01 confidence interval 99%
1.-Hypoyhesis test:
We will find out if the hippocampal volume in alcoholic adolescents is less than the normal value; therefore the test is a one tail test (left)
Null Hypothesis: H₀ μ = μ₀
Alternative Hypothesis Hₐ μ < μ₀
2.-From t-student tables, we find for α = 0,01 and df = 21
t(c) = - 2,831
3.-We calculate t(s) as:
t(s) = [μ - μ₀ ] / s/√n ⇒ t(s) = (8,02 - 9,02)* √22 / 0,7
t(s) = (-1)* 4,6904 / 0,7
t(s) = - 6,70059
4.-We compare t(s) and T(c)
t(s) = - 6,70059 t(c) = - 2,831
t(s) < t(c)
T(s) fall in the rejection region, so we reject H₀