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3 years ago

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you deposited $5500 in a account that pays 3.6% annual interest. find the balance after 2 years if the interest is compound mont

hly

1 answer:

Sedaia [141]3 years ago

6 0

Answer:

$5,909.97

Step-by-step explanation:

We will use the compound interest formula for this problem as shown below:

$A=P(1+\frac{r}{n} )^{nt}$

<em>P = initial balance</em>

<em>r = interest rate (decimal)</em>

<em>n = number of times compounded annually</em>

<em>t = time</em>

<em />

First, change 3.6% into a decimal:

3.6% -> $\frac{3.6}{100}$ -> 0.036

Now plug in the values:

$A=5,500(1+\frac{0.036}{12})^{12(2)}$

$A=5,909.97$

After 2 years, the balance would be $5,909.97

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Answer:

We conclude that there is difference in the proportion of deaths between the 2 groups.

Step-by-step explanation:

We are given that among 2823 drivers not wearing seat belts, 31 died as a result of injuries, and among 7765 drivers wearing seat belts 16 were killed.

Let $p_1$ = <u><em>proportion of deaths when drivers were not wearing seat belts.</em></u>

$p_2$ = <u><em>proportion of deaths when drivers were wearing seat belts.</em></u>

So, Null Hypothesis, $H_0$ : $p_1=p_2$ {means that there is no difference in the proportion of deaths between the 2 groups}

Alternate Hypothesis, $H_A$ : $p_1\neq p_2$ {means that there is difference in the proportion of deaths between the 2 groups}

The test statistics that would be used here <u>Two-sample z test for proportions;</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 deaths when drivers were not wearing seat belts = $\frac{31}{2823}$ = 0.011

$\hat p_2$ = sample proportion of deaths when drivers were wearing seat belts = $\frac{16}{7765}$ = 0.002

$n_1$ = sample of drivers not wearing seat belts = 2823

$n_2$ = sample of drivers wearing seat belts = 7765

So, <u><em>the test statistics</em></u> = $\frac{(0.011-0.002)-(0)}{\sqrt{\frac{0.011(1-0.011)}{2823}+\frac{0.002(1-0.002)}{7765} } }$

= 4.438

The value of z test statistics is 4.438.

<u>Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.</u>

Since our test statistic doesn't 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 due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that there is difference in the proportion of deaths between the 2 groups.

Also, <u>Margin of error</u> (E) = $1.96 \times \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} }$

= $1.96 \times \sqrt{\frac{0.011(1-0.011)}{2823}+\frac{0.002(1-0.002)}{7765} }$

= <u>0.00397</u>

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