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2 years ago

Choose the proper rationale. "In the compound interest equation, time

Mathematics

1 answer:

k0ka [10]2 years ago

4 0

Answer:

The correct answer is:

c. Time is an exponent while the other units are factors.

Explanation:

In the compound interest formula, time is the exponent.

Since it is the exponent, this tells us how many times the base is multiplied by itself. As such, it has a larger effect on the answer than any other piece of the equation.

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2 years ago

An insurance company selected a random sample of 500 clients under 18 years of age and found that 180 of them had had an acciden

Butoxors [25]

Answer:

a) The pooled proportion is p=0.3.

b) P-value = 0.000078

c) Lower bound = 0.0556

d) Upper bound = 0.1644

Step-by-step explanation:

This is a hypothesis test for the difference between proportions.

The claim is that the accident proportions differ between the two age groups .

Then, the null and alternative hypothesis are:

$H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2\neq 0$

The significance level is 0.05.

The sample 1, of size n1=500 has a proportion of p1=0.36.

$p_1=X_1/n_1=180/500=0.36$

The sample 2, of size n2=600 has a proportion of p2=0.25.

$p_1=X_1/n_1=150/600=0.25$ .

The difference between proportions is (p1-p2)=0.11.

$p_d=p_1-p_2=0.36-0.25=0.11$

The pooled proportion, needed to calculate the standard error, is:

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{180+150}{500+600}=\dfrac{330}{1100}=0.3$

The standard error for the difference between proportions can now be calculated as:

The estimated standard error of the difference between means is computed using the formula:

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.3*0.7}{500}+\dfrac{0.3*0.7}{600}}\\\\\\s_{p1-p2}=\sqrt{0.00042+0.00035}=\sqrt{0.00077}=0.0277$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{0.11-0}{0.0277}=\dfrac{0.11}{0.0277}=3.964$

This test is a two-tailed test, so the P-value for this test is calculated as (using a z-table):

$P-value=2\cdot P(t>3.964)=0.000078$

As the P-value (0.000078) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the accident proportions differ between the two age groups.

If we want to calculate the bounds of a 95% confidence interval, we start by calculating the margin of error.

For a 95% CI, the critical value for z is z=1.96.

Then, the margin of error is:

$MOE=z \cdot s_{p1-p2}=1.96\cdot 0.0277=0.0544$

Then, the lower and upper bounds of the confidence interval are:

$LL=(p_1-p_2)-z\cdot s_{p1-p2} = 0.11-0.0544=0.05561\\\\UL=(p_1-p_2)+z\cdot s_{p1-p2}= 0.11+0.0544=0.16439$

The 95% confidence interval for the population mean is (0.0556, 0.1644).

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3 years ago