Question

In one​ community, a random sample of 26 foreclosed homes sold for an average of ​$443 comma 555 with a standard deviation of ​$195 comma 381. ​a) What assumptions and conditions must be checked before finding a confidence interval for the​ mean? How would you check​ them? ​b) Find a 95​% confidence interval for the mean value per home. ​c) Interpret this interval and explain what 95​% confidence means. ​d) Suppose​ nationally, the average foreclosed home sold for ​$300 comma 000. Do you think the average sale price in the sampled community differs significantly from the national​ average? Explain.c

Answer 1

Answer:

i) a) Normality : We assume that the data follows approximately a normal distribution

ii) Random sample: The data comes from a random sample

iii) The sample size represent <10% of the population size

We assume that all the conditions are satisfied for this case.

b) $443555-2.06\frac{195381}{\sqrt{26}}=364621.22$    

$443555+2.06\frac{195381}{\sqrt{26}}=522488.775$    

So on this case the 95% confidence interval would be given by (364621.22;522488.775)    

c) We are confident at 95% that the true mean of foreclosed homes sold's are between (364621.22;522488.775)    

d) Since the lower value for the 95% confidence interval is higher than 300000 we can conclude that yes differes significantly and the true mean is different from 300000 at 5% of significance.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

$\bar X=443555$ represent the sample mean

$\mu$ population mean (variable of interest)

s=195381 represent the sample standard deviation

n=26 represent the sample size  

Part a

We need some conditions:

a) Normality : We assume that the data follows approximately a normal distribution

b) Random sample: The data comes from a random sample

c) The sample size represent <10% of the population size

We assume that all the conditions are satisfied for this case.

Part b

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=26-1=25$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,25)".And we see that $t_{\alpha/2}=2.06$

Now we have everything in order to replace into formula (1):

$443555-2.06\frac{195381}{\sqrt{26}}=364621.22$    

$443555+2.06\frac{195381}{\sqrt{26}}=522488.775$    

So on this case the 95% confidence interval would be given by (364621.22;522488.775)    

Part c

We are confident at 95% that the true mean of foreclosed homes sold's are between (364621.22;522488.775)    

Part d

Since the lower value for the 95% confidence interval is higher than 300000 we can conclude that yes differes significantly and the true mean is different from 300000 at 5% of significance.