Multiply 613 by the reciprocal of −716.

The Nellie Mae organization conducts an extensive annual study of credit card usage by college students. For their 2004 study, t

Masja [62]

Answer:

1. Null Hypothesis, $H_0$ : $p_1-p_2=0$ or $p_1=p_2$

Alternate Hypothesis, $H_A$ : $p_1-p_2\neq 0$ or $p_1\neq p_2$

2. Test statistics = 4.63

P-value = 0.00001

3. We conclude that the proportion of undergraduate students who held a credit card differed between these two years.

Step-by-step explanation:

We are given that the Nellie Mae organization conducts an extensive annual study of credit card usage by college students.

For their 2004 study, they analyzed credit bureau data for a random sample of 1,413 undergraduate students between the ages of 18 and 24. They found that 76% of the students sampled held a credit card. Three years earlier they had found that 83% of undergraduates sampled held a credit card.

Let $p_1$ = population proportion of undergraduate students who held a credit card in year 2001

$p_2$ = population proportion of undergraduate students who held a credit card in year 2004

1. Null Hypothesis, $H_0$ : $p_1-p_2=0$ or $p_1=p_2$ {means that the proportion of undergraduate students who held a credit card does not differed between these two years}

Alternate Hypothesis, $H_A$ : $p_1-p_2\neq 0$ or $p_1\neq p_2$ {means that the proportion of undergraduate students who held a credit card differed between these two years}

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

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of undergraduate students who held a credit card in 2001 = 83%

$\hat p_2$ = sample proportion of undergraduate students who held a credit card in 2004 = 76%

$n_1$ = sample of students surveyed in 2001 = 1,413

$n_2$ = sample of students surveyed in 2004 = 1,413

So, test statistics = $\frac{(0.83-0.76)-(0)}{\sqrt{\frac{0.83(1-0.83)}{1,413} +\frac{0.76(1-0.76)}{1,413}} }$

= 4.63

2. Hence, the value of test statistics is 4.63.

Also, P-value is given by the following formula;

P-value = P(Z > 4.63) = 1 - P(Z $\leq$ 4.63)

= 1 - 0.99999 = 0.00001

3. Since in the question we are not given the level of significance so we assume it to be 5%. Now at 5% significance level, the z table gives critical values between -1.96 and 1.96 for two-tailed test. Since our test statistics is does not lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the proportion of undergraduate students who held a credit card differed between these two years.

6 0

3 years ago

Can someone help me please A) tanL = B) sinN= C) cosN=

timofeeve [1]

Answer:

a.) tanL= 0.533

b.) sinN= 0.882

c.) cosN= 0.471

Step-by-step explanation:

Triangle ABC is similar to Triangle LMN, which means that their corresponding angles are congruents.

Using SOH-CAH-TOA we have,

tanL~tanA= 8/15 = 0.533

sinN~sinC= 15/17 = 0.882

cosN~cosC= 8/17 = 0.471

8 0

3 years ago

What is the variable

krek1111 [17]

A quantity that may change

5 0

3 years ago

Which number is not in the solution set of x-6>10?

givi [52]

Answer:

options?

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Lean wants to giftwrap a present she got for her little brother. how many square inches of giftwrap will be needed to cover a bo

Natalka [10]

Check the picture below.

so.. notice, the box for the gift is really just a rectangular prism, which is really just 6 rectangles stacked up to each other at the edges.

the front and back are two 5x3 rectangles

the left and right are just two 3x7 rectangles

and the top and bottom are just two 5x7 rectangles

so, just get the area of each, sum them up, and that's the area of the rectangular prism and thus of the box, and that's how many in² she will need.

4 0

3 years ago