Question

An advertising executive claims that there is a difference in the mean household income for credit cardholders of Visa Gold and of MasterCard Gold. A random survey of 11 Visa Gold cardholders resulted in a mean household income of $66,970 with a standard deviation of $9500. A random survey of 17 MasterCard Gold cardholders resulted in a mean household income of $59,060 with a standard deviation of $10,000. Is there enough evidence to support the executive's claim? Let μ1 be the true mean household income for Visa Gold cardholders and μ2 be the true mean household income for MasterCard Gold cardholders. Use a significance level of α=0.1 for the test. Assume that the population variances are not equal and that the two populations are normally distributed.
State the null and alternative hypotheses for the test.

Answer 1

Answer:

Null hypothesis:$\mu_{Visa}=\mu_{Mastercard}$

Alternative hypothesis:$\mu_{Visa} \neq \mu_{Mastercard}$

$t=\frac{66970-59060}{\sqrt{\frac{9500^2}{11}+\frac{10000^2}{17}}}}=2.108$  

$p_v =2*P(t_{26}>2.108)=0.0448$

Comparing the p value with the significance level given $\alpha=0.1$ we see that $p_v$ so we can conclude that we can reject the null hypothesis, and a would be a significant difference between the  in the mean household income for credit cardholders of Visa Gold and of MasterCard Gold at 10% of significance .

Step-by-step explanation:

Data given and notation

$\bar X_{Visa}=66970$ represent the mean for Visa

$\bar X_{Mastercard}=59060$ represent the mean for the sample Mastercard

$s_{Visa}=9500$ represent the population standard deviation for Visa

$s_{Mastercard}=10000$ represent the population standard deviation for Mastercard

$n_{Visa}=11$ sample size for the group Visa

$n_{Mastercard}=17$ sample size for the group Mastercard

t would represent the statistic (variable of interest)

$\alpha=0.1$ significance level provided

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the means for the two groups are different, the system of hypothesis would be:

Null hypothesis:$\mu_{Visa}=\mu_{Mastercard}$

Alternative hypothesis:$\mu_{Visa} \neq \mu_{Mastercard}$

Since we don't know the population deviations for each group, for this case is better apply a t test to compare means, and the statistic is given by:

$z=\frac{\bar X_{Visa}-\bar X_{Masterdcard}}{\sqrt{\frac{s^2_{Visa}}{n_{Visa}}+\frac{s^2_{Mastercard}}{n_{Mastercard}}}}$ (1)

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.

Calculate the value of the test statistic for this hypothesis testing.

Since we have all the values we can replace in formula (1) like this:

$t=\frac{66970-59060}{\sqrt{\frac{9500^2}{11}+\frac{10000^2}{17}}}}=2.108$  

What is the p-value for this hypothesis test?

First we need to calculate the degrees of freedom given by:

$df= n_{Visa}+n_{Mastercard}-2 = 11+17-2= 26$

Since is a bilateral test the p value would be:

$p_v =2*P(t_{26}>2.108)=0.0448$

Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given $\alpha=0.1$ we see that $p_v$ so we can conclude that we can reject the null hypothesis, and a would be a significant difference between the  in the mean household income for credit cardholders of Visa Gold and of MasterCard Gold at 10% of significance .