Sam puts $500 in the bank and does not put any more money in or take any out for elght years. He receives 6% Interest

Mathematics

2 answers:

kipiarov [429]3 years ago

4 0

Answer:

$793.44

Step-by-step explanation:

ra1l [238]3 years ago

4 0

793.44 the first answer

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3 years ago

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Suppose that we wish to assess whether more than 60 percent of all U.S. households in a particular income class bought life insu

Yakvenalex [24]

Answer:

a) The probability is P=0.3982.

b) No. There is no enough evidence to say that the proportion of the population is not 0.6.

Step-by-step explanation:

a) In this question, we have a sample of the population, of size n=1000. To know what is the probability of observing a sample proportion that is at least 0.64 we have to know the parameters of the sampling distribution.

The parameters of the sampling distribution of the proportion will be:

$\mu=\pi=0.6\\\\ \sigma=\sqrt{\frac{\pi(1-\pi)}{N} } =\sqrt{\frac{0.6(1-0.6)}{1000} }= 0.0155$

With these parameters, we can calculate the z-value for p=0.64 as:

$z=\frac{p-\pi}{\sigma}=\frac{0.64-0.6}{0.0155}= 0.258$

Then, we can calculate the probability of having a sample with p≥0.64:

$P(p\geq0.64)=P(z\geq0.258)=0.3982$

b) To know if the proportion of the population is greater than 0.6 the rigth thing to do is perform a hypothesis test, in which we test the following hypothesis:

$H_0: \pi\leq0.6\\\\H_1:\pi>0.6$

If the null hypothesis is rejected, we can conclude that there is evidence that the proportion of the population is greater than 0.6.

First, we assume a significance level of 0.05.

Second, we calculate the z value:

$z=\frac{p-\pi-0.5/N}{\sigma} =\frac{0.64-0.6-0.5/1000}{0.0155} =\frac{0.0395}{0.0155} =2.54$

The P-value for this z is P=0.4. The P-value is greater than the significance level, what means that there is no evidence to reject the null hypothesis.

In other words, a sample mean of 0.64 is a quite probable value even if the proportion of the population is 0.6.

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3 years ago

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kati45 [8]

we can use pythagorean identieis

remember that $sin^2(x)+cos^2(x)=1$