Mathematics, 20.03.2020 07:47 borgesalfonso12

In a random sample of 50 Americans five years ago (Group 1), the average credit card debt was $5,779. In a random sample of 50 Americans in the present day (Group 2), the average credit card debt is $6,499, Let the population standard deviation be $1,152 five years ago, and let the current population standard deviation be $1,634. Using a 0.01 level of significance, test if there is a difference in credit card debt today versus five years ago. What is the p-value

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Answer from: rdeloney4

No, there is a no difference in credit card debt today versus five years ago.

P-value = 0.0054

Step-by-step explanation:

We are given that in a random sample of 50 Americans five years ago (Group 1), the average credit card debt was $5,779. In a random sample of 50 Americans in the present day (Group 2), the average credit card debt is $6,499, Let the population standard deviation be $1,152 five years ago, and let the current population standard deviation be $1,634.

Group 1 Group 2 Sample Size : 50 50

Average credit card debt : $5,779 $6,499

Population Standard Deviation : $1,152 $1,634

And we have to test if there is a difference in credit card debt today versus five years ago, using a significance level α of 0.01.

Firstly, we will specify our null and alternate hypothesis;

Let $\mu_1$ = Population average credit card debt five years ago

$\mu_2$ = Population average credit card debt in present day

So, Null Hypothesis, H_0 : $\mu_1-\mu_2$ = 0 {means that there is no difference in the credit card debt today versus five years ago}

Alternate Hypothesis, H_0 : $\mu_1-\mu_2\neq$ 0 {means that there is statistical difference in the credit card debt today versus five years ago}

The test statistics that will be used here is Two-sample z-test statistics;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{ \sqrt{\frac{\sigma^{2} _1}{n_1}+\frac{\sigma^{2}_2 }{n_2} } }$ ~ N(0,1)

where, $\bar X_1$ = Sample average credit card debt five years ago = $45.67

$\bar X_2$ = Sample average credit card debt in present day = $39.87

$\sigma_1$ = population standard deviation five years ago = $1,152

$\sigma_2$ = current population standard deviation = $1,634

n_1 = sample of Americans in group 1 = 50

n_2 = sample of Americans in group 2 = 50

So, test statistics = $\frac{(5,779-6,499)-(0)}{ \sqrt{\frac{1,152^{2}}{50}+\frac{1.634^{2} }{50} } }$

= -2.546

Now, at 0.01 significance level, z table gives critical value between -2.5758 and 2.5758. Since our test statistics lies in the given range of critical values so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that there is no difference in credit card debt today versus five years ago.

Also, P-value is given by = P(Z < -2.55) = 1 - P(Z $\leq$ 2.55)

= 1 - 0.99461 = 0.0054

Answer from: Quest

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step-by-step explanation:

Answer from: Quest

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In a random sample of 50 Americans five years ago (Group 1), the average credit card debt was $5,779...