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gayaneshka [121]
3 years ago
11

To the nearest cent, what is the finance charge, if the charge is computed on average daily balance at a periodic monthly rate o

f 3.2%?

Mathematics
1 answer:
Amanda [17]3 years ago
8 0

Answer:

Finance charge is $100.34

Step-by-step explanation:

Given : The table attached in the question and the charge is computed on average daily balance at a periodic monthly rate of 3.2%.

To find : What is the finance charge?

Solution :

We have given the number of days and sum of balances.

So, we find the finance charge of each dates one by one.

Finance rate = 3.2%=0.032

Using the formula,

F=\frac{\text{No. of days}}{\text{total days}}\times \text{sum of balance}\times \text{finance rate}

Refer the table,

First finance charge is,

Number of days=5, Total number of days = 31 , Finance rate = 0.032 , sum of balance=$1375

F_1=\frac{5}{31}\times 1375\times 0.032

F_1=7.096

Similarly, we put all the values from the table and we get,

F_2=\frac{1}{31}\times 385\times 0.032

F_2=0.397

F_3=\frac{9}{31}\times 3465\times 0.032

F_3=32.190

F_4=\frac{1}{31}\times 260\times 0.032

F_4=0.268

F_5=\frac{15}{31}\times 3900\times 0.032

F_5=60.387

Now, the total finance charge is

F=F_1+F_2+F_3+F_4+F_5

F=7.096+0.397+32.190+0.268+60.387

F=\$100.34

$1=100 cent

$100.34=10034 cent.

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sergiy2304 [10]

Answer:

(a) 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

(b) We conclude that there is a statistical difference in these two proportions measured initially and then one year later.

Step-by-step explanation:

We are given that a random sample of 1,250 drivers found that 98.9% were in compliance. A year after the implementation, compliance was again measured to see if compliance was the same (or not) as previously measured.

A different random sample of 1,100 drivers found 96.9% compliance."

<em />

<em>Let </em>p_1<em> = proportion of drivers that were in compliance initially</em>

p_2<em> = proportion of drivers that were in compliance one year later</em>

(a) <u>Null Hypothesis</u>, H_0 : p_1-p_2=0  or  p_1= p_2      {means that there is not any statistical difference in these two proportions measured initially and then one year later}

<u>Alternate Hypothesis</u>, H_A : p_1-p_2\neq 0  or  p_1\neq p_2     {means that there is a statistical difference in these two proportions measured initially and then one year later}

The test statistics that will be used here is <u>Two-sample z proportion statistics</u>;

                     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 drivers in compliance initially = 98.9%

\hat p_2 = sample proportion of drivers in compliance one year later = 96.9%

n_1 = sample of drivers initially = 1,250

n_2 = sample of drivers one year later = 1,100

(b) So, <u><em>the test statistics</em></u>  =  \frac{(0.989-0.969)-(0)}{\sqrt{ \frac{0.989(1-0.989)}{1,250} + \frac{0.969(1-0.969)}{1,100}} }  

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<u>Now, P-value of the test statistics is given by;</u>

         P-value = P(Z > 3.33) = 1 - P(Z \leq 3.33)

                                            = 1 - 0.99957 = <u>0.00043</u>

Since in the question we are not given with 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.

<em>Since our test statistics does not lies 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 due to which </em><u><em>we reject our null hypothesis.</em></u>

Therefore, we conclude that there is a statistical difference in these two proportions measured initially and then one year later.

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