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 14 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. An analysis of the sample data revealed that the hippocampal volume is approximately normal with no outliers and X=8.19 cm' and s=0.8 cm Conduct the appropriate test at the a= 0.01 level of significance State the null and alternative hypotheses How H: (Type integers or decimals. Do not round.) Identify the t-statistic 6- Round to wo decimal places as needed.) Identity the P-value P values (Round to three decimal places as needed.) Make a conclusion regarding the hypothesis the null hypothesis. There suficient evidence to claim that the man hippocampal volume is cm? Enter your answer in each of the answer boxes. quest vey of 10 MacBook Pro $ % 1 2 3 4 5 & 7 6 8 9 Q W E R. tab Т. U А S D חד caps lock G Н. N Х shift C V B N M

Answer 1

Given that the researchers wants to measure the effect of alcohal on hippocampal region in brain.

They wanted to detrmine weather the hippocampal volumes in the alcohalic adolescents were less than $\mu$ = 9.02 cm³ , to test this tesy took a sample of n = 14 adolescents with alcohal disorder and found the mean volume of = 8.19 cm³ and sample standard deviation of s = 0.8 cm³

T

Also given that the distribution is normal with no outliers and the significance level as 0.01.

Based on the claim the hypotheses are:

Ho: p = 9.02 Ha: p< 9.02

Based on the claim it will be left-tailed test, but since the popualtion standard deviation is unknown hence t-distribution is applicable and degree of freedom is used which is calculated as df = n-1 = 14-1= 13.

Test statistic:

X - μo t= 8.19 - 9.02 0.8/14 = -3.88 8/ Vn

Rejection region:

Based on the significance level reject Ho if P-value is less than 0.01.

P-value:

The p-value is computed using the excel formula for t-distribution which is =T.DIST(-3.88, 13, TRUE), thus the P-value is computed as 0.001.

Conclusion:

Since the P-value is less than 0.01 hence we reject the null hypothesis and conclude that there is sufficient evidence to support the claim the mean is less than 9.02 cm³