Question

Prices of diamonds are determined by what is known as the 4 Cs: cut, clarity, color, and carat weight. The prices of diamonds go up as the carat weight increases, but the increase is not smooth. For example, the di?erence between the size of a 0.99 carat diamond and a 1 carat diamond is undetectable to the naked human eye, but the price of a 1 carat diamond tends to be much higher than the price of a 0.99 diamond. In this question we use two random samples of diamonds, 0.99 carats and 1 carat, each sample of size 23, and compare the average prices of the diamonds. In order to be able to compare equivalent units, we first divide the price for each diamond by 100 times its weight in carats. That is, for a 0.99 carat diamond, we divide the price by 99. For a 1 carat diamond, we divide the price by 100. The distributions and some sample statistics are shown below.43

Conduct a hypothesis test to evaluate if there is a di?er- ence between the average standardized prices of 0.99 and 1 carat diamonds. Make sure to state your hypotheses clearly, check relevant conditions, and interpret your re- sults in context of the data.

Answer 1

Concepts and reason

The two-sample independent t-test use when the two samples are come from different populations and independent to each other. The two populations, one with mean ${\mu _1}$ and the other with mean ${\mu _2}$ , and it is needed to make inference about the difference ${\mu _1} - {\mu _2}$

Assumptions:

1)Simple random sample: The data is collected from a representative, randomly selected portion of the total population.

2)Normality: The samples taken from the population is normal.

Fundamentals

The formula for the test statistic is,

$t = \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right)}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }}$

Here, ${\bar x_1}$ = sample average of the first sample.

${\bar x_2}$ = sample average of the second sample.

= sample size of the first sample.

= sample size of the second sample.

= sample variance of the first sample.

= sample variance of the second sample.

The formula for degrees of freedom is,

$\begin{array}{c}\\df = \left( {{n_1} - 1} \right) + \left( {{n_2} - 1} \right)\\\\ = {n_1} + {n_2} - 2\\\end{array}$

Setup the null and alternative hypotheses:

Null hypothesis:

There is no significance difference between the average standardized prices of 0.99 and 1 carat diamonds.

${H_0}:{\mu _1} = {\mu _2}$

Alternative hypothesis:

There is a significance difference between the average standardized prices of 0.99 and 1 carat diamonds.

${H_1}:{\mu _1} \ne {\mu _2}$

The summary data is as shown below:

The test statistic foe testing the difference between the two means is,

$\begin{array}{c}\\t\left( {{\rm{observed}}} \right) = \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right)}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }}\\\\ = \frac{{\left( {44.51 - 56.8} \right)}}{{\sqrt {\frac{{{{\left( {13.3} \right)}^2}}}{{23}} + \frac{{{{\left( {16.1} \right)}^2}}}{{23}}} }}\\\\ = - \frac{{12.29}}{{4.3544}}\\\\ = - 2.82\\\end{array}$

The objective of the study is to test the significance difference between the average standardized prices of 0.99 and 1 carat diamonds.

Let the specified level of significanceis 0.05.

The degrees of freedom using the following formula,

$\begin{array}{c}\\df = {n_1} + {n_2} - 2\\\\ = 23 + 23 - 2\\\\ = 44\\\end{array}$

The P-value is as shown below:

$\begin{array}{c}\\P - value = tdist\left( {\left| t \right|,df,tail} \right)\\\\ = tdist\left( {2.82,44,2} \right)\\\\ = 0.0072\\\end{array}$

Here, it can be observed that the P-value is less than the level of the significance 0.05.

Hence, there is enough evidence to reject the null hypothesis.

Therefore, there is a sufficient evidence to support the claim that there is a significance difference between the average standardized prices of 0.99 and 1 carat diamonds.

Ans:

Null hypothesis ${H_0}:{\mu _1} = {\mu _2}$

Alternative hypothesis ${H_1}:{\mu _1} \ne {\mu _2}$

The value of the test statistic to test the significance difference between the two means is -2.82.

The P-value of the test statistic is 0.0072.