Question

In a​ study, researchers wanted to measure the effect of alcohol on the hippocampal​ region, the portion of the brain responsible for​ long-term memory​ storage, in adolescents. The researchers randomly selected 22 adolescents with alcohol use disorders to determine whether the hippocampal volumes in the alcoholic adolescents were less than the normal volume of 9.02 cm cubed. An analysis of the sample data revealed that the hippocampal volume is approximately normal with no outliers and x overbarequals8.02 cm cubed and sequals0.7 cm cubed. Conduct the appropriate test at the alphaequals0.01 level of significance. State the null and alternative hypotheses. Upper H 0​: mu equals 9.02 Upper H 1​: mu less than 9.02 ​(Type integers or decimals. Do not​ round.) Identify the​ t-statistic.

Answer 1

We reject H₀ we have evidence to claim that the hippocampal volumes in the alcoholic adolescents are less than the normal values.

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

Answer 2

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.