Question

Suppose you are an expert on the fashion industry and wish to gather information to compare the amount earned per month by models featuring Liz Claiborne attire with those of Calvin Klein. The following is the amount ($000) earned per month by a sample of Claiborne models:

$5.0 4.5 3.4 3.4 6.0 3.3 4.5 4.6 3.5 5.2
4.8 4.4 4.6 3.6 5.0

The following is the amount ($000) earned by a sample of Calvin Klein models:

$3.1 3.7 3.6 4.0 3.8 3.8 5.9 4.9 3.6 3.6
2.3 4.0

Is it reasonable to conclude that Claiborne models earn more? Use the .05 significance level and assume the population standard deviations are not the same.

Answer 1

Answer:

$t=\frac{(4.39-3.81)-0}{\sqrt{\frac{0.795^2}{15}+\frac{0.907^2}{12}}}}=1.743$  

P value  

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

$p_v =P(t_{25}>1.743)=0.0468$  

Comparing the p value with the significance level $\alpha=0.05$ we see that $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis.  

Step-by-step explanation:

Data given and notation  

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

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

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

$n_1 = 15$ represent the sampe size for Claiborne

$n_2 =12$ represent the sample size for Calvin Klein

$\bar X_1 =4.39$ represent the sample mean for Claiborne

$\bar X_2 =3.81$ represent the sample mean for Calvin Klein

$s_1 = 0.795$ represent the sample deviation for Claiborne

$s^2_1 =0.631$ represent the sample variance for Claiborne

$s_2 = 0.907$ represent the sample deviation for Calvin Klein

$s^2_2 = 0.823$ represent the sample variance for Calvin Klein

$\alpha=0.05$ represent the significance level provided

Confidence =0.95 or 95%

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the mean for Claiborne is higher than the mean for Calvin Klein, the system of hypothesis would be:  

Null hypothesis:$\mu_{1}-\mu_{2}\leq 0$  

Alternative hypothesis:$\mu_{1} - \mu_{2}> 0$  

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:  

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)  

And the degrees of freedom are given by $df=n_1 +n_2 -2=15+12-2=25$  

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

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

$t=\frac{(4.39-3.81)-0}{\sqrt{\frac{0.795^2}{15}+\frac{0.907^2}{12}}}}=1.743$  

P value  

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

$p_v =P(t_{25}>1.743)=0.0468$  

Comparing the p value with the significance level $\alpha=0.05$ we see that $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis.