Question

2. (Hypothesis testing) Suppose you collect a random sample of 2,610 homes sold in Stockton, CA, between the years 1996 and 1998. For each of the 2,610 sold homes within this sample, you obtain data on the home size, sale price, number of bedrooms, etc. You then summarize this data and variables in the table of descriptive statistics displayed below. Please use the results in that table of descriptive statistics to answer the following questions related to statistical inference. (a) Your friend claims that the population mean for house prices in Stockton (CA) between the years 1996 and 1998 was equal to $124,500. Based on the results displayed in the table of descriptive statistics, and using a two-sided alternative hypothesis, do you statistically “reject” or “fail to reject” your friend’s null hypothesis at the 5% significance level? How about at the 1% significance level? Please use the “t-stat” method to answer this question. (b) Additionally, please calculate the “p-value” of this hypothesis test. Please make sure to explain and show your work. (unable to post table)

Answer 1

Sol:

NUll hypothesis:

Ho: =124,500

Alterntative Hypothesis:

Ha:  124,500

alpha=0.05

test statistic:

t=sample mean-population mean/samplesd/sqrt(n)

=sample mean-124,500/sample std deviation/sqrt(2610)

provide data so that sample mean ,samploe std deviation can be calculated .hence t statistic can be calcualted

After calculating t

p value can be calculated fot t and degrees of freedom=n-1=2610-1=2609

alpha=0.05

alpha/2=0.05/2=0.025

p<0.05

Reject Null Hypothesis.

Accept Alternative Hypothesis.

We can statistically conclude at 5% level of signifcance that he population mean for house prices in Stockton (CA) between the years 1996 and 1998 was different from $124,500

if p>0.05

Fail to reject Null hypothesis

Accept Null hypothesis

We can statistically conclude at 5% level of signifcance that he population mean for house prices in Stockton (CA) between the years 1996 and 1998 was equal from $124,500

similarly at 1% level

alpha=0.01

alpha/2=0.01/2=0.005

df=n-1=2610-1=2609

For 2609 df and alpha=0.005 calculate p value

if p<0.01

Reject Null hypothesis

if p>0.01

Fail to reject null hypoothesis.