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3 years ago

1. Calculate the simple interest over an amount of N$ 2500 that is invested for 3 years and 6 months at a rate of 7% pa.

Mathematics

1 answer:

Umnica [9.8K]3 years ago

3 0

Answer:

N$ 612.5

Step by step explaination:

Given,

Principal or'P'=N$2500

Time or'n'=3 years & 6 months or 3.5 years

Rate of profit or 'r'=7% or 7/100

Profit or 'I'=?

_____________________________________

We know,

I=P*n*r

I=2500*3.5*7/100

I=612.5

So,simple interest is N$ 612.5.

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Answer:

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1,58,18,400

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2 years ago

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3 years ago

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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 .

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

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.

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

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

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).

5 0

3 years ago

1. Calculate the simple interest over an amount of N$ 2500 that is invested for 3 years and 6 months at a rate of 7% pa.​

1. Calculate the simple interest over an amount of N$ 2500 that is invested for 3 years and 6 months at a rate of 7% pa.