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2 years ago

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Jim received a $2000 loan from his bank. The loan accrues 3% interest every 3 months. How much will Jim owe the bank after 4 yea

rs? Round to the nearest cent

1 answer:

pashok25 [27]2 years ago

4 0

Answer:

$2253.98

Step-by-step explanation:

Jim received a $2000 loan from his bank. The loan accrues 3% interest every 3 months.

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

P=2000

r= 3%=0.03 and t= 4 years

interest every 3 months so n= 4

$A=2000(1+\frac{.03}{4} )^{4 \cdot 4}$

$A=2000(1+\frac{.03}{4} )^{16}\\A=2000(1.0075)^{16}\\\\A=2253.98$

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a) $\hat p_1 -\hat p_2= 0.2-0.35= -0.15$

b) $ME= 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =0.169$

c) $(0.2-0.35) - 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =-0.319$

$(0.2-0.35) + 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =0.0185$

And the 99% confidence interval would be given (-0.319;0.0185).

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$p_1$ represent the real population proportion 1

$\hat p_1=0.2$ represent the estimated proportion 1

$n_1=60$ is the sample size required 1

$p_2$ represent the real population proportion for 2

$\hat p_2 =0.35$ represent the estimated proportion 2

$n_2=100$ is the sample size required for Brand B

$z$ represent the critical value for the margin of error

Solution to the problem

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_1 -\hat p_2) \pm z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_2)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}$

Part a

The best estimate is given by:

$\hat p_1 -\hat p_2= 0.2-0.35= -0.15$

Part b

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

The margin of error is given by:

$ME= 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =0.169$

Part c

And replacing into the confidence interval formula we got:

$(0.2-0.35) - 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =-0.319$

$(0.2-0.35) + 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =0.0185$

And the 99% confidence interval would be given (-0.319;0.0185).

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3 years ago