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?
A No, because the value of σ is not known
B No, because it is not known whether the data are Normally distributed
C No, because the entire data set is needed to do this test
D Yes, because $42,000 is not contained in the 95% confidence interval, the null hypothesis would not be rejected, and it could be concluded that the mean family income is not significantly different from $42,000 at the α = 0.05 level
E 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
It is given that the 95% confidence interval is $43,100 to $59,710 and the population mean value is $42,000. It is clear that the null value or the population mean is not included in the confidence interval, which means that the confidence interval is significantly difference from the population mean.
Therefore, we can say that at 0.05 significance level, mean family income is significantly different from population mean family income of $42,000 because confidence interval lower limit is above the population mean
so, option E is correct answer
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