To determine the correct null hypothesis ([tex]\(H_0\)[/tex]) and alternative hypothesis ([tex]\(H_a\)[/tex]), we need to focus on the claims made in the problem:
1. Alyssa believes that the students in her school spend more than [tex]$195.00 on prom dresses.
2. This indicates that we are dealing with a right-tailed test (one-tailed test) where we want to test if the mean price of a prom dress in Alyssa's school is greater than $[/tex]195.00.
Given this setup, the correct hypotheses would be:
- Null Hypothesis ([tex]\(H_0\)[/tex]): The mean price of a prom dress at Alyssa's school is equal to [tex]$195.00.
- Alternative Hypothesis (\(H_a\)): The mean price of a prom dress at Alyssa's school is greater than $[/tex]195.00.
In mathematical terms, these hypotheses are expressed as:
- [tex]\(H_0: \mu = 195\)[/tex]
- [tex]\(H_a: \mu > 195\)[/tex]
So, the correct choice among the given options is:
- [tex]\(H_0: \mu = 195 ; H_2, \mu > 195\)[/tex]
Next, let's outline the steps for the hypothesis testing:
1. Calculate the Standard Error of the Mean (SEM):
[tex]\[
\text{Standard Error} = \frac{\text{Population Standard Deviation}}{\sqrt{\text{Sample Size}}}
\][/tex]
Using the given values:
[tex]\[
\text{Standard Error} = \frac{12.00}{\sqrt{20}} \approx 2.683
\][/tex]
2. Calculate the Z-score:
[tex]\[
Z = \frac{\text{Sample Mean} - \text{Population Mean}}{\text{Standard Error}}
\][/tex]
Using the given values:
[tex]\[
Z = \frac{208.00 - 195.00}{2.683} \approx 4.845
\][/tex]
3. Based on the Z-score and a standard normal distribution table, a Z-score of approximately 4.845 is quite extreme and lies far in the right tail. This suggests that the observed sample mean is significantly higher than the population mean of [tex]$195.00, supporting Alyssa's claim.
Thus, the findings would lead us to reject the null hypothesis (\(H_0\)) in favor of the alternative hypothesis (\(H_a\)), indicating that students at Alyssa's school do spend more than $[/tex]195.00 on prom dresses.
