Question

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.

Answer 1

Answer:

a) $29.4-1.96\frac{7}{\sqrt{10}}=25.06$  

$29.4+1.96\frac{7}{\sqrt{10}}=33.74$  

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

b) $z=\frac{29,4-25}{\frac{7}{\sqrt{10}}}=1.988$  

$p_v =P(z>1.988)=0.0234$    

If we compare the p value and the significance level given $\alpha=Step-by-step explanation:Previous conceptsA confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval". The margin of error is the range of values below and above the sample statistic in a confidence interval. Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean". Part a Data given: 30 30 42 35 22 33 31 29 19 23 We can calculate the sample mean with the following formula:[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}= 29.4$ the sample mean

$\mu$ population mean (variable of interest)  

$\sigma=7$ 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:  

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)  

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, 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 $z_{\alpha/2}=1.96$  

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

$29.4-1.96\frac{7}{\sqrt{10}}=25.06$  

$29.4+1.96\frac{7}{\sqrt{10}}=33.74$  

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:  $\mu \leq 25$  

Alternative hypothesis :$\mu>25$  

Compute the test statistic

The statistic for this case is given by:  

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (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:  

$z=\frac{29,4-25}{\frac{7}{\sqrt{10}}}=1.988$  

Give the appropriate conclusion for the test

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

$p_v =P(z>1.988)=0.0234$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.