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

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(195381)/(√(26))=364621.22`

`443555+2.06(195381)/(√(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.

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)(s)/(√(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(195381)/(√(26))=364621.22`

`443555+2.06(195381)/(√(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.