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2 years ago

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Nancy borrowed $150 from her parents to buy a video game system. She has agreed to pay them back $25 each week Which equation re

presents the amount of money
Nancy still owes her parents, P. after w weeks?

1 answer:

Zinaida [17]2 years ago

7 0

Answer:

P = 150 - 25w

Step-by-step explanation:

Amount borrowed = $150

Repayment amount = $25 each week

P = amount Nancy owes

w = number of weeks

Amount Nancy owes = amount borrowed - repayment plan per week

P = 150 - 25w

During the first week of repayment: when w = 1

P = 150 - 25w

P = 150 - 25(1)

= 150 - 25

= 125

P = $125

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The average salary of all residents of a city is thought to be about $39,000. A research team surveys a random sample of 200 res

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Answer:

$z=\frac{40000-39000}{\frac{12000}{\sqrt{200}}}=1.179$

$p_v =2*P(z>1.179)=2*[1-P(Z$

Step-by-step explanation:

Data given and notation

$\bar X=4000$ represent the average score for the sample

$s=12000$ represent the sample standard deviation

$n=200$ sample size

$\mu_o =39000$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to apply a two tailed test.

What are H0 and Ha for this study?

Null hypothesis: $\mu = 39000$

Alternative hypothesis : $\mu \neq 39000$

Compute the test statistic

The sample size is large enough to assume the distribution for the statisitc normal. The statistic for this case is given by:

$z=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

We can replace in formula (1) the info given like this:

$z=\frac{40000-39000}{\frac{12000}{\sqrt{200}}}=1.179$

Give the appropriate conclusion for the test

Since is a two tailed test the p value would be:

$p_v =2*P(z>1.179)=2*[1-P(Z$

Conclusion

If we compare the p value and a significance level given for example $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, then the true population mean for the salary not differs significantly from the value of 39000.

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Step-by-step explanation:

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Suppose a simple random sample of size n is drawn from a large population with mean mu and standard deviation sigma. The samplin

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Answer:

$X \sim N(\mu,\sigma)$

Where $\mu$ the mean and $\sigma$ the deviation

Since the distribution for X is normal then we can conclude that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\mu_{\bar X} = \mu$

$\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}$

Step-by-step explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(\mu,\sigma)$

Where $\mu$ the mean and $\sigma$ the deviation

Since the distribution for X is normal then we can conclude that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\mu_{\bar X} = \mu$

$\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}$

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