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 99 Visa Gold cardholders resulted in a mean household income of $56,930$ 56,930 with a standard deviation of $10,700$ 10,700. A random survey of 1414 MasterCard Gold cardholders resulted in a mean household income of $50,790$ 50,790 with a standard deviation of $10,600$ 10,600. Is there enough evidence to support the executive's claim? Let μ1μ1 be the true mean household income for Visa Gold cardholders and μ2μ2 be the true mean household income for MasterCard Gold cardholders. Use a significance level of α=0.01α=0.01 for the test. Assume that the population variances are not equal and that the two populations are normally distributed.

Answer 1

`t=\frac{(56930-50790)-0}{\sqrt{(10700^2)/(9)+(10600^2)/(14)}}}=1.348`

`p_v =2*P(t_((21))>1.348)=0.192`

If we compare the p value and the significance level given
`\alpha=0.01` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the we don't have significant difference between the two means at 1% of significance.

Explanation:

Data given and notation

`\bar X_(1)=56930` represent the mean for the sample 1

`\bar X_(2)=50790` represent the mean for the sample 2

`s_(1)=10700` represent the sample standard deviation for the sample 1

`s_(B)=10600` represent the sample standard deviation for the sample 2

`n_(1)=9` sample size selected 1

`n_(B)=14` sample size selected B

`\alpha=0.01` represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

`p_v` represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

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

Null hypothesis:
`\mu_(1)-\mu_(2) = 0`

Alternative hypothesis:
`\mu_(1)-\mu_(2) \\eq 0`

We don't know the population deviations, 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))-12}{\sqrt{(s^2_(1))/(n_(1))+(s^2_(2))/(n_(2))}}` (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 statistic

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

`t=\frac{(56930-50790)-0}{\sqrt{(10700^2)/(9)+(10600^2)/(14)}}}=1.348`

P-value

The first step is calculate the degrees of freedom, on this case:

`df=n_(1)+n_(2)-2=9+14-2=21`

Since is a two sided test the p value would be:

`p_v =2*P(t_((21))>1.348)=0.192`

Conclusion

If we compare the p value and the significance level given
`\alpha=0.01` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the we don't have significant difference between the two means at 1% of significance.