Sulfur compounds cause "off-odors" in wine, so winemakers want to know the odor threshold, the lowest concentration of a compound that the human nose can detect. the odor threshold for dimethyl sulfide (dms) in trained wine tasters is about 25 μg/l (micrograms per liter). the untrained noses of consumers may have a higher threshold, however.
here are the dms odor thresholds for 10 untrained students:

30 30 42 35 22 33 31 29 19 23

assume that the standard deviation of the odor threshold for untrained noses is known to be σ = 7 μg/l.
(a) give a 95% ci for the mean dms odor threshold among all students.
(b) are you convinced that the mean odor threshold for students is higher than the published threshold, 25 μg/l? carry out a significance test with an α = 0.05 significance level to justify your answer.

a)  

So on this case the 95% confidence interval would be given by (25.06;33.74)  

b)  

If we compare the p value and the significance level given the sample mean

population mean (variable of interest)  

represent the population standard deviation  

n=10 represent the sample size  

95% confidence interval  

The confidence interval for the mean is given by the following formula:  

(1)  

Since the Confidence is 0.95 or 95%, the value of and , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that  

Now we have everything in order to replace into formula (1):  

So on this case the 95% confidence interval would be given by (25.06;33.74)  

Part b

What are H0 and Ha for this study?  

Null hypothesis:    

Alternative hypothesis :  

Compute the test statistic

The statistic for this case is given by:  

(1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

We can replace in formula (1) the info given like this:  

Give the appropriate conclusion for the test

Since is a one side right tailed test the p value would be:  

Conclusion  

If we compare the p value and the significance level given we see that so we can conclude that we have enough evidence to reject the null hypothesis.    

a) 0

step-by-step explanation:

i^413 x i^0 = 0 as any number multiplied by a zero is

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