Nine homes are chosen at random from real estate listings in two suburban neighb
ID: 3227377 • Letter: N
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. (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 mu_1 is the mean sizes in Greenwood and mu_2 is the mean of home sizes in Pinewood. a. H_0: mu_1 - mu_2 = 0 vs. H_1: mu_1 - mu_2 notequalto 0 b. H_0: mu_1 - mu_2 notequalto 0 vs. H_1; mu_1 - mu_2 = 0 a b (b) Specify the decision rule with respect to the p-value. Reject the null hypothesis if the p-value is 0.10 (c) Find the test statistic t_calc. (A negative value should be indicated by a minus sign. Round your answer to 3 decimal places.) t_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 the null hypothesis. (f) State your conclusion.Explanation / Answer
Given that,
mean(x)=2606
standard deviation , s.d1=177.7829
number(n1)=9
y(mean)=2592.4444
standard deviation, s.d2 =231.304
number(n2)=9
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.1
from standard normal table, two tailed t /2 =1.86
since our test is two-tailed
reject Ho, if to < -1.86 OR if to > 1.86
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =2606-2592.4444/sqrt((31606.75953/9)+(53501.54042/9))
to =0.139
| to | =0.139
critical value
the value of |t | with min (n1-1, n2-1) i.e 8 d.f is 1.86
we got |to| = 0.1394 & | t | = 1.86
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 0.1394 ) = 0.893
hence value of p0.1 < 0.893,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 0.139
critical value: -1.86 , 1.86
decision: do not reject Ho
p-value: 0.893
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