Question

Inorganic phosphorous is a naturally occurring element in all plants and animals, with concentrations increasing progressively up the food chain (fruit < vegetables < cereals < nuts < corpse). Geochemical surveys take soil samples to determine phosphorous content (in ppm, parts per million). A high phosphorous content may or may not indicate an ancient burial site, food storage site, or even a garbage dump. The Hill of Tara is a very important archaeological site in Ireland. It is by legend the seat of Ireland's ancient high kings†. Independent random samples from two regions in Tara gave the following phosphorous measurements (ppm). Assume the population distributions of phosphorous are mound-shaped and symmetric for these two regions. Region I: x1; n1 = 12540 810 790 790 340 800890 860 820 640 970 720Region II: x2; n2 = 16750 870 700 810 965 350 895 850635 955 710 890 520 650 280 993(a) Use a calculator with mean and standard deviation keys to calculate x1, s1, x2, and s2. (Round your answers to one decimal place.) x1, s1, x2, s2 "(b) Let ?1 be the population mean for x1 and let ?2 be the population mean for x2. Find a 95% confidence interval for ?1 ?2. (Use 1 decimal place.)lower limit upper limit (c) Explain what the confidence interval means in the context of this problem. Does the interval consist of numbers that are all positive? all negative? of different signs? At the 95% level of confidence, is one region more interesting than the other from a geochemical perspective?Because the interval contains only negative numbers, we can say that region II is more interesting than region I.We can not make any conclusions using this confidence interval. Because the interval contains only positive numbers, we can say that region I is more interesting than region II.Because the interval contains both positive and negative numbers, we can not say that one region is more interesting than the other.(d) Which distribution (standard normal or Student's t) did you use? Why?Student's t was used because ?1 and ?2 are unknown.Standard normal was used because ?1 and ?2 are unknown. Standard normal was used because ?1 and ?2 are known.Student's t was used because ?1 and ?2 are known.

Answer 1

Answer:

a) $s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X_1 =747.5$ represent the sample mean 1

$\bar X_2 =738.938$ represent the sample mean 2

$s_1 =170.407$ sample standard deviation for sample 1

$s_2 =212.146$ sample standard deviation for sample 2

b) The 95% confidence interval would be given by $-140.5 \leq \mu_1 -\mu_2 \leq 157.6$  

c) The interval contains values of different signs

II.Because the interval contains both positive and negative numbers, we can not say that one region is more interesting than the other.

d) We use the T distribution

Student's t was used because $\sigma_1$ and $\sigma_2$ are known.

Step-by-step explanation:

Data given and previous concepts

A 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".  

Data given

Region I: 540 810 790 790 340 800 890 860 820 640 970 720

Region II: 750 870 700 810 965 350 895 850 635 955 710 890 520 650 280 993

Part a

We can calculate the sample mean and deviation with the following formulas:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X_1 =747.5$ represent the sample mean 1

$\bar X_2 =738.938$ represent the sample mean 2

n1=12 represent the sample 1 size  

n2=16 represent the sample 2 size  

$s_1 =170.407$ sample standard deviation for sample 1

$s_2 =212.146$ sample standard deviation for sample 2

$\mu_1 -\mu_2$ parameter of interest.

Part b

The confidence interval for the difference of means is given by the following formula:  

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}$ (1)  

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =747.5-738.938=8.563$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:  

$df=n_1 +n_2 -1=12+16-2=26$  

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: "=-T.INV(0.025,26)".And we see that $t_{\alpha/2}=2.06$  

The standard error is given by the following formula:

$SE=\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}$

And replacing we have:

$SE=\sqrt{\frac{170.407^2}{12}+\frac{212.146^2}{16}}=72.338$

Confidence interval

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

$8.563-2.06\sqrt{\frac{170.407^2}{12}+\frac{212.146^2}{16}}=-140.453$  

$8.563+2.06\sqrt{\frac{170.407^2}{12}+\frac{212.146^2}{16}}=157.579$  

So on this case the 95% confidence interval would be given by $-140.5 \leq \mu_1 -\mu_2 \leq 157.6$  

Part c

The interval contains values of different signs

II.Because the interval contains both positive and negative numbers, we can not say that one region is more interesting than the other

Part d

We use the T distribution

Student's t was used because $\sigma_1$ and $\sigma_2$ are known.