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. |Enter the ALTERNATIVE HYPOTHESIS, H1 number.
Note: Enter INGEGER 1, 2 or 3, No decimal for selection: 1) H 1 : mean $>120$ 2) H1: mean < 120 3) H1: mean Not Equal To 120

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 we'll use a hypothesis test to do it! Watch out! Don't mix up the null hypothesis and the alternative hypothesis!
Also, make sure to use the correct test statistic since we know the population standard deviation.

STEP 2

1. State the hypotheses
2. Calculate the test statistic
3. Determine the p-value
4. Make a decision

STEP 3

We're testing if the UGA average study time is different from the national average.
This means it could be greater or lesser, so we'll use a two-tailed test.

STEP 4

Our null hypothesis $\text{H}_0$ is that the UGA average study time is the same as the national average: $\mu = 120$ .

STEP 5

Our alternative hypothesis $\text{H}_1$ is that the UGA average study time is not equal to the national average: $\mu \neq 120$ .

STEP 6

Since we know the population standard deviation $\sigma = 30$ , we'll use a z-test.
The formula for the z-test statistic is: $z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ where $\bar{x}$ is the sample mean, $\mu$ is the population mean we're testing against, $\sigma$ is the population standard deviation, and $n$ is the sample size.

STEP 7

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

STEP 8

Now, let's calculate the test statistic: $z = \frac{-10}{\frac{30}{6.245}} = \frac{-10}{4.803} \approx -2.082$ So, our test statistic is approximately $z = -2.082$ .

STEP 9

Since this is a two-tailed test, we need to find the area in both tails of the standard normal distribution that is beyond our test statistic.
Because the normal distribution is symmetric, we can find the area in one tail and multiply it by two.

STEP 10

Looking up a z-score of $-2.08$ in a z-table (or using a calculator), we find a one-tailed p-value of approximately $0.0188$ .

STEP 11

Now, we multiply by two to get the two-tailed p-value: $2 \cdot 0.0188 = 0.0376$ .

STEP 12

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

STEP 13

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

STEP 14

The alternative hypothesis is that the mean is not equal to 120, which corresponds to option 3.

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Studdy Solution

STEP 1

STEP 2

1. State the hypotheses
2. Calculate the test statistic
3. Determine the p-value
4. Make a decision

STEP 3

We're testing if the UGA average study time is different from the national average.
This means it could be greater or lesser, so we'll use a two-tailed test.

STEP 4

Our null hypothesis $\text{H}_0$ is that the UGA average study time is the same as the national average: $\mu = 120$ .

STEP 5

Our alternative hypothesis $\text{H}_1$ is that the UGA average study time is not equal to the national average: $\mu \neq 120$ .

STEP 6

STEP 7

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

STEP 8

Now, let's calculate the test statistic: $z = \frac{-10}{\frac{30}{6.245}} = \frac{-10}{4.803} \approx -2.082$ So, our test statistic is approximately $z = -2.082$ .

STEP 9

Since this is a two-tailed test, we need to find the area in both tails of the standard normal distribution that is beyond our test statistic.
Because the normal distribution is symmetric, we can find the area in one tail and multiply it by two.

STEP 10

Looking up a z-score of $-2.08$ in a z-table (or using a calculator), we find a one-tailed p-value of approximately $0.0188$ .

STEP 11

Now, we multiply by two to get the two-tailed p-value: $2 \cdot 0.0188 = 0.0376$ .

STEP 12

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

STEP 13

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

STEP 14

The alternative hypothesis is that the mean is not equal to 120, which corresponds to option 3.

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Studdy Solution

STEP 1

STEP 2

1. State the hypotheses2. Calculate the test statistic3. Determine the p-value4. Make a decision

STEP 3

We're testing if the UGA average study time is *different* from the national average.This means it could be *greater* or *lesser*, so we'll use a **two-tailed test**.

STEP 4

Our **null hypothesis** H0\text{H}_0H0​ is that the UGA average study time is the same as the national average: μ=120\mu = 120μ=120.

STEP 5

Our **alternative hypothesis** H1\text{H}_1H1​ is that the UGA average study time is *not* equal to the national average: μ≠120\mu \neq 120μ=120.

STEP 6

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 values into our formula: z=110−1203039 z = \frac{110 - 120}{\frac{30}{\sqrt{39}}} z=39​30​110−120​

STEP 8

Now, let's calculate the **test statistic**: z=−10306.245=−104.803≈−2.082 z = \frac{-10}{\frac{30}{6.245}} = \frac{-10}{4.803} \approx -2.082 z=6.24530​−10​=4.803−10​≈−2.082 So, our **test statistic** is approximately z=−2.082z = -2.082z=−2.082.

STEP 9

Since this is a **two-tailed test**, we need to find the area in *both* tails of the standard normal distribution that is beyond our test statistic.Because the normal distribution is symmetric, we can find the area in one tail and multiply it by two.

STEP 10

Looking up a *z*-score of −2.08-2.08−2.08 in a *z*-table (or using a calculator), we find a **one-tailed p-value** of approximately 0.01880.01880.0188.

STEP 11

Now, we **multiply by two** to get the **two-tailed p-value**: 2⋅0.0188=0.03762 \cdot 0.0188 = 0.03762⋅0.0188=0.0376.

STEP 12

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

STEP 13

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

STEP 14

The alternative hypothesis is that the mean is *not equal to* 120, which corresponds to option 3.

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1. State the hypotheses
2. Calculate the test statistic
3. Determine the p-value
4. Make a decision

We're testing if the UGA average study time is different from the national average.
This means it could be greater or lesser, so we'll use a two-tailed test.

Our null hypothesis $\text{H}_0$ is that the UGA average study time is the same as the national average: $\mu = 120$ .

Our alternative hypothesis $\text{H}_1$ is that the UGA average study time is not equal to the national average: $\mu \neq 120$ .

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

Now, let's calculate the test statistic: $z = \frac{-10}{\frac{30}{6.245}} = \frac{-10}{4.803} \approx -2.082$ So, our test statistic is approximately $z = -2.082$ .

Since this is a two-tailed test, we need to find the area in both tails of the standard normal distribution that is beyond our test statistic.
Because the normal distribution is symmetric, we can find the area in one tail and multiply it by two.

Looking up a z-score of $-2.08$ in a z-table (or using a calculator), we find a one-tailed p-value of approximately $0.0188$ .

Now, we multiply by two to get the two-tailed 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 average study time for UGA students is different from the national average.

The alternative hypothesis is that the mean is not equal to 120, which corresponds to option 3.