The mean family income for a random sample of 600 suburban households in Loganville shows that a 95 percent confidence interval is ($43,100, $59,710). Alma is conducting a test of the null hypothesis H0: µ = 42,000 against the alternative hypothesis Ha: µ ≠ 42,000 at the α = 0.05 level of significance. Does Alma have enough information to conduct a test of the null hypothesis against the alternative?

And for this case we want to test the following hypothesis:

Null hypothesis:

Alternative hypothesis:

For this case since the lower value of the confidence interval is higher than 42000 we have enough evidence to reject the null hypothesis at the 55 of significance and we can conclude that the true mean is significantly different from 42000

Step-by-step explanation:

The confidence interval for the mean is given by the following formula:

  (1)

And for this case the 95% confidence interval is already calculated as:

And for this case we want to test the following hypothesis:

Null hypothesis:

Alternative hypothesis:

For this case since the lower value of the confidence interval is higher than 42000 we have enough evidence to reject the null hypothesis at the 55 of significance and we can conclude that the true mean is significantly different from 42000

a 10 percent increase

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl