Mathematics, 15.04.2020 22:58 deena7

The average price of homes sold in the U. S. in 2016 was $240,000. A sample of 150 homes sold in Chattanooga in 2016 showed an average price of $236,000. It is known that the standard deviation of the population (σ) is $36,000. We are interested in determining whether the average price of homes sold in Chattanooga is significantly less than the national average.

Use a significance level of α equal to 0.05.

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Answer from: donald1255

$z=\frac{236000-240000}{\frac{36000}{\sqrt{150}}}=-1.36$

p_v =P(z

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is lower than 240000 at 5% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=236000$ represent the sample mean

$\sigma=36000$ represent the population standard deviation for the sample

n=150 sample size

$\mu_o =240000$ represent the value that we want to test

$\alpha=0.05$ 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 conduct a hypothesis in order to check if the average price of homes sold in Chattanooga is significantly less than the national average, the system of hypothesis would be:

Null hypothesis: $\mu \geq 240000$

Alternative hypothesis: $\mu < 240000$

If we analyze the size for the sample is > 30 and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (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".

Calculate the statistic

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

$z=\frac{236000-240000}{\frac{36000}{\sqrt{150}}}=-1.36$

P-value

Since is a one sided test the p value would be:

p_v =P(z

Conclusion

Answer from: ayeeeee98

There is no enough evidence that the average price of homes sold in Chattanooga is significantly less than the national average.

Step-by-step explanation:

We have to perform a hypothesis test on the mean. The claim is that the average price of homes sold in Chattanooga is significantly less than the national average.

The null and alternative hypothesis are:

$H_0: \mu=240,000\\\\H_a: \mu$

The significance level is α=0.05.

The sample mean is $236,000.

The sample size is n=150.

The population standard deviation is $36,000.

Then, the statistic z is

$z=\frac{\bar x-\mu}{\sigma/\sqrt{n}}=\frac{236,000-240,000}{36,000/\sqrt{150}}= \frac{-4,000}{2939}=-1.36$

The P-value for z=-1.36 is

P(z

The P-value is bigger than the significance level, so the effect is not significant.

The null hypothesis failed to be rejected.

There is no enough evidence that the average price of homes sold in Chattanooga is significantly less than the national average.

Answer from: Quest

step-by-step explanation:

Answer from: Quest

Your maths is very weak try yourself this would improve your skills don't depend on others

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