Question

Question 2

Suppose you want to determine if average mercury levels in Albacore tuna are less than 400 parts per billion (ppb). In a random sample of 43 specimens of Albacore tuna, the average mercury level was found to be 378 ppb and the standard deviation of mercury level was found to be 172 ppb. You then run an appropriate test of significance with null hypothesis that average mercury levels in Albacore tuna is equal to 400 ppb and alternative hypothesis that average mercury levels in Albacore tuna is less than 400 ppb..

Your test of significance produces a p-value of 0.20.

Which of the following statements best describes what that p-value means?

(A) If average mercury levels in Albacore tuna is 400 ppb, the probability you would get an observed statistic of 378 ppb or less in a random sample of 43 specimens of Albacore tuna is 0.20.

(B) The probability that average mercury levels in Albacore tuna is less than 400 ppb is 0.20.

(C) Your observed statistic of 378 ppb is approximately 0.20 standard deviations above the mean of the null distribution.

(D) If average mercury levels in Albacore tuna is less than 400 ppb, the probability you would get an observed statistic of 378 ppb or less in a random sample of 43 specimens of Albacore tuna is 0.20.

Answer 1

Answer is

(A) If the average mercury levels in Albacore tuna is 400 ppb , the probability you would get an observed statistic of 378 ppb or less in a random sample of 43 specimens of Albacore tuna is 0.20 .

Note :

P value is the probability of observed sample value or more extreme (sample mean is 378 ppb or less) when the null hypothesis is true (the average mercury levels in Albacore tuna is 400 ppb)

The null and alternative hypothesis is

H0: = 400

Ha: < 400

P value is calculated on the basis of the test statistic

Test statistic that is used in this scenario is

where ,

378(sample mean)

S=172 ( sample standard deviation)

n= 43 (sample size)

is the hypothesised mean = 400

the research hypothesis is Ha: < 400

So this is a one tailed test ( left tailed)

P value = P( t < to )  

where t o=- 0.84 is the calculated value of test statistic

P value = P( t < -0.84 ) =0.20

Note : We find p value using t table or excel "=T.DIST.RT(0.84,42)"