Question

Test the null hypothesis that there is no difference between the estimates of the two garages. Be sure to specify the null and alternative hypotheses, the test statistic with degrees of freedom, and the P-value. What do you conclude using the 0.05 significance level

Answer 1

Answer:

t = [ (x1' - x2' ) - d ] / SE . p < 0.05 suggests means are equal, p > 0.05 suggests means are unequal.

Step-by-step explanation:

Null Hypothesis [H0] : There is no difference between average estimates of two garages, M1 = M2

Alternate Hypothesis [H1] : There is difference between average estimates of two garages, M1  ≠ M2

Hypothesis can be tested by t test

t = [ (x1' - x2' ) - d ] / SE , where :-

x1' & x2' are sample means, SE = sqrt[ (s1^2/n1) + (s2^2 /n2) ] , d is the hypothesized difference between population means, and SE is the standard error ie 0 here

If the p value corresponding to calculated t value is < p value as per significance level (0.05) : We reject null hypothesis & state that M1  ≠ M2

If the p value corresponding to calculated t value >  p value as per significance level (0.05) : We accept null hypothesis & state that M1 = M2