Listed in the data table are amounts of strontium-90 (in millibecquerels, or mBq, per gram of calcium) in a simple random sample of baby teeth obtained from residents in two cities. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Use a 0.05 significance level to test the claim that the mean amount of strontium- 90 from city \#1 residents is greater than the mean amount from city \#2 residents.
Click the icon to view the data table of strontium-90 amounts.
What are the null and alternative hypotheses? Assume that population 1 consists of amounts from city \#1 levels and population 2 consists of amounts from city \#2.
A.
\[
\begin{array}{l}
H_{0}: \mu_{1}=\mu_{2} \\
H_{1}: \mu_{1} \neq \mu_{2}
\end{array}
\]
C.
\[
\begin{array}{l}
H_{0}: \mu_{1}=\mu_{2} \\
H_{1}: \mu_{1}> \mu_{2}
\end{array}
\]
B.
\[
\begin{array}{l}
H_{0}: \mu_{1} \neq \mu_{2} \\
H_{1}: \mu_{1}> \mu_{2}
\end{array}
\]
D.
\[
\begin{array}{l}
H_{0}: \mu_{1} \leq \mu_{2} \\
H_{1}: \mu_{1}> \mu_{2}
\end{array}
\]
Answer
So, the final answer is D. \[ \boxed{D} \]
Step 1 ：The null hypothesis (H0) is a statement of no effect, difference, or relationship. It is the hypothesis that the researcher is trying to disprove. The alternative hypothesis (H1) is a statement that contradicts the null hypothesis. It is the hypothesis that the researcher believes to be true and is trying to prove.
Step 2 ：In this case, we are testing the claim that the mean amount of strontium-90 from city #1 residents is greater than the mean amount from city #2 residents. Therefore, the null hypothesis should be that the means are equal or that the mean of city #1 is less than or equal to the mean of city #2. The alternative hypothesis should be that the mean of city #1 is greater than the mean of city #2.
Step 3 ：The correct hypotheses are: \[ H_{0}: \mu_{1} \leq \mu_{2} \] \[ H_{1}: \mu_{1}>\mu_{2} \]
Step 4 ：So, the final answer is D. \[ \boxed{D} \]
