The average price of homes sold in the U. S. in 2012 was $240,000. A sample of 144 homes sold in Chattanooga in 2012 showed an average price of $246,000. It is known that the standard deviation of the population (σ) is $36,000. At the 5% level of significance. We are interested in determining whether or not the average price of homes sold in Chattanooga is significantly more than the national average.

a. State the null and alternative hypotheses to be tested. b. Compute the test statistic. c. The null hypothesis is to be tested at the 5% level of significance. Determine the critical value(s) for this test. d. What do you conclude?

H0 : Average price of homes sold in US = 24000 ; H1 : Average price of homes sold in US ≠ 24000

t  calculated value = 2 , t critical (tabulated) value = 1.96

calculated t > critical t . Null Hypothesis is rejected, It is concluded that 'Average price of homes sold in US ≠ 24000 '

Explanation:

Null Hypothesis : Average price of homes sold in US = 24000

Alternate Hypothesis : Average price of homes sold in US ≠ 24000

t = (x' - u) / (s / √n)

x' = sample mean = 246000 (given)

u = population mean = 240000 (given)

s = standard deviation = 36000

n = no. of observations = 144

t = (246000 - 240000) / (36000/√144)

6000/ (36000/12000) = 6000/3000

t = 2

Critical value for a two tailed test at 5% significance level, 0.025 in t distribution = 1.96

Since calculated value, 2 > tabulated or critical value at significance level, 1.96. So, we reject the null hypothesis. This implies that 'Average price of homes sold in US ≠ 24000'

n

explanation:

i would say iii (3) because in order to compare a price you have to use as a reference other near stores because it's its

hope i explained myself