Math / Data & Statistics

QuestionA study time survey is conducted of 39 UGA students and a mean x-bar 110 minutes is obtained. Test whether the UGA study times differ from the national average of 120 min ? Assume study times have a normal distribution with a known standard deviation of $\sigma=30$ minutes. Use the $5 \%$ significance level.
1 What is the correct interpretation of the hypothesis testing result at the $5 \%$ level of significance? 1) = The UGA study times were greater than the national average. 2) = The UGA study times are different from the national average. 3) = There was no difference between the UGA and national study times.
Note: Enter integer 1, 2, 3 No decimal. $A$

Studdy Solution

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.

STEP 11

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

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Studdy Solution

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

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.

STEP 11

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

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Studdy Solution

STEP 1

STEP 2

1. Set up the hypotheses2. Calculate the test statistic3. Find the p-value4. Make a decision

STEP 3

Our **null hypothesis** H0H_0H0​ is that the UGA average study time is the same as the national average, which is μ=120\mu = 120μ=120 minutes.

STEP 4

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

STEP 5

STEP 6

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 −2.082-2.082−2.082 or 2.0822.0822.082.

STEP 8

Using a z-table, we find that the area to the left of z=−2.08z = -2.08z=−2.08 is approximately 0.01880.01880.0188.Since it's a two-tailed test, we multiply this value by 222 to get the **p-value**: 2⋅0.0188=0.03762 \cdot 0.0188 = 0.03762⋅0.0188=0.0376.

STEP 9

Our **p-value** (0.03760.03760.0376) is less than our **significance level** (0.050.050.05).

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.

STEP 11

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

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

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

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.

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 $2.082$ .

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: $2 \cdot 0.0188 = 0.0376$ .

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

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.