Math Snap

STEP 1

What is this asking?
We want to find out if UGA students study for a different amount of time than the national average, and we'll use a hypothesis test to do it!
Watch out!
Don't mix up the UGA sample mean with the national average!
Also, make sure to use the correct z-score for a two-tailed test.

STEP 2

1. Set up the hypotheses
2. Calculate the test statistic
3. Find the p-value
4. Make a decision

STEP 3

Our null hypothesis $H_0$ is that the UGA average study time is the same as the national average, which is $\mu = 120$ minutes.

STEP 4

Our alternative hypothesis $H_a$ is that the UGA average study time is different from the national average, so $\mu \ne 120$ minutes.
This is a two-tailed test because we're looking for any difference, not just greater than or less than.

STEP 5

We'll use the z-statistic formula because we know the population standard deviation:
$$z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$$ where $\bar{x}$ is the UGA sample mean ($110$), $\mu$ is the national average ($120$), $\sigma$ is the population standard deviation ($30$), and $n$ is the sample size ($39$).

STEP 6

Let's plug in the values:
$$z = \frac{110 - 120}{\frac{30}{\sqrt{39}}}$$ $$z = \frac{-10}{\frac{30}{6.245}}$$$$z = \frac{-10}{4.803}$$$$z \approx -2.082$$So, our test statistic is $z \approx -2.082$.

STEP 7

Since this is a two-tailed test, we need to find the area in both tails of the standard normal distribution.
We're looking for the probability of getting a z-score more extreme than $-2.082$ or $2.082$.

STEP 8

Using a z-table, we find that the area to the left of $z = -2.08$ is approximately $0.0188$.
Since it's a two-tailed test, we multiply this value by $2$ to get the p-value: $2 \cdot 0.0188 = 0.0376$.

STEP 9

Our p-value ($0.0376$) is less than our significance level ($0.05$).

STEP 10

This means we reject the null hypothesis.
There's enough evidence to suggest that the UGA average study time is different from the national average.

The correct answer is 2.
The UGA study times are different from the national average.