Question

In Hamilton County, Ohio the mean number of days needed to sell a home is days (Cincinnati Multiple Listing Service, April, 2012). Data for the sale of homes in a nearby county showed a sample mean of days with a sample standard deviation of days. Conduct a hypotheses test to determine whether the mean number of days until a home is sold is different than the Hamilton county mean of days in the nearby county. Round your answer to four decimal places. -value = Use for the level of significance, and state your conclusion.

Answer 1

Answer:

$t=\frac{80-86}{\frac{20}{\sqrt{40}}}=-1.8974$    

$p_v =2*P(t_{39}$

If we compare the p value with a significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, so there is not enough evidence to conclude that the mean for the data is significantly different from 86 days.    

Step-by-step explanation:

Assuming the following problem: "In Hamilton County, Ohio the mean number of days needed to sell a home is 86 days (Cincinnati Multiple Listing Service, April, 2012). Data for the sale 40 of homes in a nearby county showed a sample mean of 80 days with a sample standard deviation of 20 days. Conduct a hypotheses test to determine whether the mean number of days until a home is sold is different than the Hamilton county mean of 86 days in the nearby county. "

1) Data given and notation    

$\bar X=80$ represent the sample mean

$s=20$ represent the sample standard deviation

$n=40$ sample size    

$\mu_o =86$ 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)

2) State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to determine if the mean is different from 86, the system of hypothesis would be:    

Null hypothesis:$\mu = 86$    

Alternative hypothesis:$\mu \neq 86$    

We don't know the population deviation, so for this case we can use the t test to compare the actual mean to the reference value, and the statistic is given by:    

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

t-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".

3) Calculate the statistic    

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

$t=\frac{80-86}{\frac{20}{\sqrt{40}}}=-1.8974$    

4) Calculate the P-value    

First we need to find the degrees of freedom given by:

$df=n-1=40-1=39$

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

$p_v =2*P(t_{39}$

In Excel we can use the following formula to find the p value "=2*T.DIST(-1.8974,39,TRUE)"  

5) Conclusion    

If we compare the p value with a significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, so there is not enough evidence to conclude that the mean for the data is significantly different from 86 days.