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 $120 \mathrm{~min} . ?$ 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 P-VALUE from Table A2 for testing HO? See two entry formats below.
Note: Two formats possible: 1) Enter LTO1 if P-value is less than 0.01 2) Enter XXX.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 134.784 is entered as $134.78,35.295$ is entered as $35.30, .2749$ is entered as $0.27,-.5650$ is entered as -0.57

Studdy Solution

STEP 1

What is this asking? We want to find out if UGA students study a different amount of time than the national average, and how confident we are in that difference. Watch out! Don't mix up one-tailed and two-tailed tests!
We're looking for a difference either way, not if UGA students study more or less.

STEP 2

1. Set up the hypothesis test
2. Calculate the test statistic
3. Find the p-value

STEP 3

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

STEP 4

Our alternative hypothesis ( $H_a$ ) is that UGA students study a different amount than the national average, so $\mu \ne 120$ minutes.
This makes it a two-tailed test!

STEP 5

Our significance level ( $\alpha$ ) is $\alpha = 0.05$ , meaning we're willing to accept a 5% chance of being wrong if the null hypothesis is actually true.

STEP 6

Since we know the population standard deviation ( $\sigma = 30$ ), we'll use the z-test: $z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ .
This tells us how far our sample mean is from the assumed population mean, in terms of standard deviations.

STEP 7

We have $\bar{x} = 110$ , $\mu = 120$ , $\sigma = 30$ , and $n = 39$ .
Let's plug those in: $z = \frac{110 - 120}{\frac{30}{\sqrt{39}}}$ .

STEP 8

First, let's simplify the denominator: $\frac{30}{\sqrt{39}} \approx \frac{30}{6.245} \approx 4.804$ .
Now, we can calculate z: $z \approx \frac{-10}{4.804} \approx -2.082$ .
So our test statistic is $z \approx -2.082$ .

STEP 9

Since this is a two-tailed test, we need to find the area in both tails.
We look up the absolute value of our z-score, $|z| = 2.08$ , in the z-table.

STEP 10

The table gives us the area to the left of the z-score.
For $z = 2.08$ , the area is approximately $0.9812$ .
Since we want the area to the right, we subtract from 1: $1 - 0.9812 = 0.0188$ .

STEP 11

Because it's a two-tailed test, we multiply the one-tail area by 2: $0.0188 \cdot 2 = 0.0376$ .
This is our p-value: $p \approx 0.0376$ .

STEP 12

The problem asks for the p-value rounded up to two decimal places.
So, our final p-value is 0.04.

STEP 13

The p-value is 0.04.

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Studdy Solution

STEP 1

STEP 2

1. Set up the hypothesis test
2. Calculate the test statistic
3. Find the p-value

STEP 3

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

STEP 4

Our alternative hypothesis ( $H_a$ ) is that UGA students study a different amount than the national average, so $\mu \ne 120$ minutes.
This makes it a two-tailed test!

STEP 5

Our significance level ( $\alpha$ ) is $\alpha = 0.05$ , meaning we're willing to accept a 5% chance of being wrong if the null hypothesis is actually true.

STEP 6

Since we know the population standard deviation ( $\sigma = 30$ ), we'll use the z-test: $z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ .
This tells us how far our sample mean is from the assumed population mean, in terms of standard deviations.

STEP 7

We have $\bar{x} = 110$ , $\mu = 120$ , $\sigma = 30$ , and $n = 39$ .
Let's plug those in: $z = \frac{110 - 120}{\frac{30}{\sqrt{39}}}$ .

STEP 8

First, let's simplify the denominator: $\frac{30}{\sqrt{39}} \approx \frac{30}{6.245} \approx 4.804$ .
Now, we can calculate z: $z \approx \frac{-10}{4.804} \approx -2.082$ .
So our test statistic is $z \approx -2.082$ .

STEP 9

Since this is a two-tailed test, we need to find the area in both tails.
We look up the absolute value of our z-score, $|z| = 2.08$ , in the z-table.

STEP 10

The table gives us the area to the left of the z-score.
For $z = 2.08$ , the area is approximately $0.9812$ .
Since we want the area to the right, we subtract from 1: $1 - 0.9812 = 0.0188$ .

STEP 11

Because it's a two-tailed test, we multiply the one-tail area by 2: $0.0188 \cdot 2 = 0.0376$ .
This is our p-value: $p \approx 0.0376$ .

STEP 12

The problem asks for the p-value rounded up to two decimal places.
So, our final p-value is 0.04.

STEP 13

The p-value is 0.04.

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Studdy Solution

STEP 1

STEP 2

1. Set up the hypothesis test2. Calculate the test statistic3. Find the p-value

STEP 3

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

STEP 4

Our **alternative hypothesis** (HaH_aHa​) is that UGA students study a *different* amount than the national average, so μ≠120\mu \ne 120μ=120 minutes.This makes it a **two-tailed test**!

STEP 5

Our **significance level** (α\alphaα) is α=0.05\alpha = 0.05α=0.05, meaning we're willing to accept a 5% chance of being wrong if the null hypothesis is actually true.

STEP 6

Since we know the population standard deviation (σ=30\sigma = 30σ=30), we'll use the *z*-test: z=xˉ−μσnz = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}z=n​σ​xˉ−μ​.This tells us how far our sample mean is from the assumed population mean, in terms of standard deviations.

STEP 7

We have xˉ=110\bar{x} = 110xˉ=110, μ=120\mu = 120μ=120, σ=30\sigma = 30σ=30, and n=39n = 39n=39.Let's plug those in: z=110−1203039z = \frac{110 - 120}{\frac{30}{\sqrt{39}}}z=39​30​110−120​.

STEP 8

STEP 9

Since this is a two-tailed test, we need to find the area in *both* tails.We look up the absolute value of our *z*-score, ∣z∣=2.08|z| = 2.08∣z∣=2.08, in the *z*-table.

STEP 10

The table gives us the area to the *left* of the *z*-score.For z=2.08z = 2.08z=2.08, the area is approximately 0.98120.98120.9812.Since we want the area to the *right*, we subtract from 1: 1−0.9812=0.01881 - 0.9812 = 0.01881−0.9812=0.0188.

STEP 11

Because it's a two-tailed test, we multiply the one-tail area by 2: 0.0188⋅2=0.03760.0188 \cdot 2 = 0.03760.0188⋅2=0.0376.This is our **p-value**: p≈0.0376p \approx 0.0376p≈0.0376.

STEP 12

The problem asks for the p-value rounded up to two decimal places.So, our final p-value is **0.04**.

STEP 13

The p-value is **0.04**.

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1. Set up the hypothesis test
2. Calculate the test statistic
3. Find the p-value

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

Our alternative hypothesis ( $H_a$ ) is that UGA students study a different amount than the national average, so $\mu \ne 120$ minutes.
This makes it a two-tailed test!

Our significance level ( $\alpha$ ) is $\alpha = 0.05$ , meaning we're willing to accept a 5% chance of being wrong if the null hypothesis is actually true.

Since we know the population standard deviation ( $\sigma = 30$ ), we'll use the z-test: $z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ .
This tells us how far our sample mean is from the assumed population mean, in terms of standard deviations.

We have $\bar{x} = 110$ , $\mu = 120$ , $\sigma = 30$ , and $n = 39$ .
Let's plug those in: $z = \frac{110 - 120}{\frac{30}{\sqrt{39}}}$ .

Since this is a two-tailed test, we need to find the area in both tails.
We look up the absolute value of our z-score, $|z| = 2.08$ , in the z-table.

The table gives us the area to the left of the z-score.
For $z = 2.08$ , the area is approximately $0.9812$ .
Since we want the area to the right, we subtract from 1: $1 - 0.9812 = 0.0188$ .

Because it's a two-tailed test, we multiply the one-tail area by 2: $0.0188 \cdot 2 = 0.0376$ .
This is our p-value: $p \approx 0.0376$ .

The problem asks for the p-value rounded up to two decimal places.
So, our final p-value is 0.04.

The p-value is 0.04.