Many food products contain small quantities of substances th

Many food products contain small quantities of substances that would give an undesireable taste or smell if they are present in large amounts. An exaple is the \"off-odors\" caused by sulfur compounds in wine. Oenologists (wine experts) have determined the odor threshold, the lowest concentration of a compound that the human nose can detect. For example, the odor threshold for dimethyl sulfide (DMS) is given in the oenology literature as 25 micrograms per liter of wine (µg/l). Untrained noses may be less sensitive, however. Here are the DMS odor thresholds for 10 beginning students of oenology. 30 31 35 20 32 37 43 33 23 26 Assume (this is not realistic) that the standard deviation of the odor threshold for untrained noses is known to be = 7 µg/l. A normal quantile plot confirms that there are no systematic departures from normality.

(b) Give a 95% confidence interval for the mean DMS odor threshold among all beginning oenology students. (Round your answers to three decimal places.) - Wrong answers 25.997 , 36.003

(c) Are you convinced that the mean odor threshold for beginning students is higher than the published threshold, 25 µg/l? Carry out a significance test to justify your answer. (Use = 0.05. Round your value for z to two decimal places and round your P-value to four decimal places.)

Solution

a)
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=31
Standard deviation( sd )=6.766
Sample Size(n)=10
Confidence Interval = [ 31 ± t a/2 ( 6.766/ Sqrt ( 10) ) ]
= [ 31 - 2.262 * (2.14) , 31 + 2.262 * (2.14) ]
= [ 26.16,35.84 ]

b)
Set Up Hypothesis
Null, H0: U=25
Alternate, H1: U>25
Test Statistic
Population Mean(U)=25
Sample X(Mean)=31
Standard Deviation(S.D)=6.766
Number (n)=10
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =31-25/(6.766/Sqrt(9))
to =2.804
| to | =2.804
Critical Value
The Value of |t | with n-1 = 9 d.f is 1.833
We got |to| =2.804 & | t | =1.833
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value :Right Tail - Ha : ( P > 2.8043 ) = 0.01028
Hence Value of P0.05 > 0.01028,Here we Reject Ho

convinced that the mean odor threshold for beginning students is higher than the published threshold