Mathematics, 20.05.2020 12:58, summer5716

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 1010 Visa Gold cardholders resulted in a mean household income of $50,930$ 50,930 with a standard deviation of $9500$ 9500. A random survey of 66 MasterCard Gold cardholders resulted in a mean household income of $45,720$ 45,720 with a standard deviation of $9800$ 9800. 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.

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We conclude that there is no difference in the mean household income for credit cardholders of Visa Gold and of MasterCard Gold.

Step-by-step explanation:

We are given that a random survey of 10 Visa Gold cardholders resulted in a mean household income of $50,930 with a standard deviation of $9500.

A random survey of 66 MasterCard Gold cardholders resulted in a mean household income of $45,720 with a standard deviation of $9800.

Let $\mu_1$ = true mean household income for Visa Gold cardholders.

$\mu_2$ = true mean household income for MasterCard Gold cardholders.

So, Null Hypothesis, H_0 : $\mu_1=\mu_2$ {means that there is no difference in the mean household income for credit cardholders of Visa Gold and of MasterCard Gold}

Alternate Hypothesis, H_A : $\mu_1\neq \mu_2$ {means that there is a difference in the mean household income for credit cardholders of Visa Gold and of MasterCard Gold}

The test statistics that would be used here Two-sample t test statistics as we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ t__n__1-_n__2-2

where, $\bar X_1$ = sample mean household income for Visa Gold cardholders = $50,930

$\bar X_2$ = sample mean household income for MasterCard Gold cardholders = $45,720

s_1 = sample standard deviation for Visa Gold cardholders = $9,500

s_2 = sample standard deviation for MasterCard Gold cardholders = $9,800

n_1 = sample of Visa Gold cardholders = 10

n_2 = sample of MasterCard Gold cardholders = 6

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(10-1)\times 9,500^{2} +(6-1)\times 9.800^{2} }{10+6-2} }$ = 9,608.22

So, the test statistics = $\frac{(50,930-45,720)-(0)}{9,608.22 \times \sqrt{\frac{1}{10} +\frac{1}{6} } }$ ~ t_1_4

= 1.05

The value of t test statistics is 1.05.

Now, at 0.01 significance level the t table gives critical values of -2.977 and 2.977 at 14 degree of freedom for two-tailed test.

Since our test statistic lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that there is no difference in the mean household income for credit cardholders of Visa Gold and of MasterCard Gold.

answered: Guest

2x^2+5x+17

your welcome. : )

answered: Guest

the answer is 8/21

hope this

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