Question

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?

Answer 1

Answer: Yes, because $42,000 is not contained in the 95% confidence interval, the null hypothesis would be rejected in favor of the alternative, and it could be concluded that the mean family income is significantly different from $42,000 at the α = 0.05 level

Explanation:

took the test

Answer 2

`43100 \leq \mu \leq 59710`

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

Null hypothesis:
`\mu =42000`

Alternative hypothesis:
`\mu \\eq 42000`

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

Explanation:

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

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

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

`43100 \leq \mu \leq 59710`

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

Null hypothesis:
`\mu =42000`

Alternative hypothesis:
`\mu \\eq 42000`

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