Q 8.24RP. - Elementary Statistics

Question

Diamond Pricing. In a Singapore edition of Business Times. diamond pricing was explored. The price of a diamond is based on the diamond's weight, color, and clarity. A simple random sample of $18$ one-half-carat diamonds had the following prices, in dollars.

a. Apply the $t-$interval procedure to these data to find a $90\%$ confidence interval for the mean price of all one-half-carat diamonds. Interpret your result. (Note: $\overline{x}=\backslash (1964.7$ and $s=\backslash )206.5$)

b. Obtain a normal probability plot, a boxplot, and a histogram. and a stem-and-leaf diagram of the data.

c. Based on your graphs from part (b), is it reasonable to apply the $t-$interval procedure as you did in part (a)? Explain your answer.

Step-by-Step Solution

Answer

Part (a) We can be $90\%$ confident that the mean price of all one-half carat diamonds, $\mu$ lies somewhere between $\$1880.1$ and $\$2049.4$

Part (b) 

Leaf Unit $=10$

Stem Leaf

$$\begin{gathered} 1 14 4 \\ 1 15 \\ 2 16 7 \\ 3 17 1 \\ 5 18 27 \\ \left(6\right) 19 448899 \\ 7 20 377 \\ 4 21 04 \\ 2 22 3 \\ 1 23 1 \end{gathered}$$

Part (c) No.

Part (a) Step 1: Given information

1676	1442	1995	1718	1826	2071	1947	1983	2146
1995	1876	2032	1988	2071	2234	2108	1941	2316

Part (a) Step 2: Explanation

$t-$When using the interval process on the supplied data, we wish to get the $90\%$ confidence interval of the population mean $\mu$, therefore $(1880.1,2049.4)$ is the $90\%$ confidence interval of the population mean $\mu$

[Using MINITAB]

i.e., we may be $90\%$ certain that the average price of all half-carat diamonds, $\mu$, is between $\$1880.1$ and $$ 2049.4$.

Part (b) Step 1: Explanation

Now, draw the probability plot for the given data.  

Now, construct the box plot for the given data.  

Now, construct the histogram for the given data.  

Part (b) Step 2: Calculation

Create the stem-and-leaf for the supplied data collection now.

Stem-and-Leaf Display: PRICE

Leaf Unit $=10$

Stem Leaf

$$\begin{gathered} 1 14 4 \\ 1 15 \\ 2 16 7 \\ 3 17 1 \\ 5 18 27 \\ \left(6\right) 19 448899 \\ 7 20 377 \\ 4 21 04 \\ 2 22 3 \\ 1 23 1 \end{gathered}$$

Part (c) Step 1: Explanation

No, the $t-$interval technique is not appropriate for the data. Because the sample is of a reasonable size and the graphical representations reveal that the data contains an outlier (observation $1442$).