Question

core: 1 of 5 pts 4 of 6 (6 complete) HW Score: 56%, 11.2 of 20 pts *9.3.17-T Question Help Buppose the average size of a new house built in a certain county in 2006 was 2,276 square feet. A random sample of 30 new homes built in this county was selected n 2010. The average square footage was 2,195, with a sample standard deviation of 222 square feet. a. Using a 0.05, does this sample provide enough evidence to conclude that the average house size of a new home in the county has changed since 2006? b. Use technology to determine the p-value for this test O op CIUT TO C. Ho = 2,276 and H u #2,276 OD. Ho: 22,276 and Hu<2,276 Determine the critical value(s). The critical value(s) is(are) -2.045, 2.045 (Round to three decimal places as needed. Use a comma to separate answers as needed.) Determine the test statistic, = (Round to two decimal places as needed.)

Answer 1

8)

a) Hypothesis :
Η :μ = 2276
Vs
H, :μ + 2276

Critical values are -2.145 , 2.045

From t-table,

In-1,0/2 = t30-1,0.05/2 = +2.045

The test statistic $t_{\bar{X}}$ is ,

X - HO tx = s/n

2195 - 2276 222/30 -=-2.00

Decision : Here ,

tx= 2.00 <tn-1,0/2 = 2.045

Therefore , fail to reject the null hypothesis.

Conclusion : Hence , there is sufficient evidence to conclude that the average size of new house built in a certain county in 2006 was 2176 square feet.

b. p-value = ; From t table "=TDIST(2,29,2)"

P(tn-1 > tx) = P(t30-1 > 2.00) = 0.0549

9 ) Given : Sample size=n=10

Significance level=

a = 0.10

Therefore , the critical value for one (lower tail) tailed test is ,

; From t-table

tn-1,0 = t10–1,0.10 = -1.383

10 ) In hypothesis testing , if the null hypothesis is rejected then the alternative hypothesis is true.