A recent survey of 8,000 high school students found that the mean price of a prom dress was [tex]$195[/tex] with a standard deviation of [tex]$12.00[/tex]. Alyssa thinks that her school is more fashion-conscious and that students spend more on prom dresses. She collected data from 20 people in her high school and found that the average price spent was [tex]$208.00[/tex].

Which of the following are the correct null hypothesis and alternate hypothesis?

[tex]
\begin{array}{l}
H_0: \mu = 195 ; H_{a}: \mu > 195 \\
H_0: \mu \neq 195 ; H_{a}: \mu = 208 \\
H_0: \mu = 195 ; H_{a}: \mu \neq 195 \\
H_0: \mu < 195 ; H_{a}: \mu \geq 208
\end{array}
[/tex]

Answer :

To determine the correct null hypothesis and alternate hypothesis from the options given, let's first understand the context:

- A survey reported the mean price of a prom dress to be [tex]$195.00.
- Alyssa believes the students at her high school spend differently, with her collected data from 20 people showing an average price of $[/tex]208.00.

Based on this information, Alyssa is questioning whether the mean price at her high school is different from the stated [tex]$195.00. Here's how we can set up the hypotheses:

1. Null Hypothesis (\(H_0\)): This is a statement of no effect or no difference. It suggests that any observed difference is due to random variation. In this case, the null hypothesis would state that the mean price of prom dresses at Alyssa's school is the same as the overall mean, which is $[/tex]195.00.

So, [tex]\(H_0: \mu = 195\)[/tex]

2. Alternate Hypothesis ([tex]\(H_a\)[/tex]): This reflects the claim that we want to test. For Alyssa, she thinks her school's average price is not equal to [tex]$195.00, indicating a difference.

Therefore, \(H_a: \mu \neq 195\)

By analyzing her claim and the options provided, the most fitting pair of hypotheses for Alyssa's study is:

- \(H_0: \mu = 195\)
- \(H_a: \mu \neq 195\)

This choice correctly represents her suspicion that the average price her classmates paid for prom dresses is different from the general average of $[/tex]195.00.