Question

Nine homes are chosen at random from real estate listings in two suburban neighborhoods, and the square footage of each home is noted in the following table Size of Homes in Two Subdivisions Subdivision Greenwood Pinewood Square Footage 2,450 2,348 2,300 2,418 2,754 2,425 2,735 2,521 2,799 2,466 2,883 2,421 2,736 2,783 2,339 2,649 2,896 3,975 Click here for the Excel Data File (a) Choose the appropriate hypothesis to test if there is a difference between the average sizes of homes in the two neighborhoods at the .10 significance level. Assume μι is the mean of home sizes in Greenwood and μ2 is the mean of home sizes in Pinewood Ob (b) Specify the decision rule with respect to the p-value Reject the null hypothesis if the p-value is less than (c) Find the test statistic fcalc. (A negative value should be indicated by a minus sign. Round your answer to 3 0.10. decimal places.) calc (d) Assume unequal variances to find the p-value. (Use the quick rule to determine degrees of freedom. Round your answer to 4 decimal places.) p-value (e) Make a decision. We do not reject the null hypothesis. (f) State your conclusion. We cannot conclude that there is a difference between the sizes of homes in the two neighborhoods

Answer 1

a) Here we need to test if there is difference between average sizes of two homes. Hence the correct hypothesis are

H0 : v/s H1 :

b) Here = 0.10

Decision Rule : Reject null hypothesis if the p value is less than 0.10.

c) From data we get : n1 =9, n2=9, $overline{x}_{1}$ =2527.78. $overline{x}_{2}$ =2794.22 , s₁ = 187.28,s₂ = 486.23

The test statistic is

= -1.534

d)

  

= 10

P value = 2 * p ( t < -1.534 )

= 0.1560

( We use excel formula to find p value ' =t.dist.2T( 1.534,10)' )

e) Here p value > 0.10.

Hence we do not reject null hypothesis.

f) Conclusion : We can not conclude that there is difference between sizes of home in the two neighborhoods.