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.
I What is the value of the TEST STATISTIC?
Note: 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 $A$

Studdy Solution

STEP 1

What is this asking? We're checking if UGA students study differently than the national average of 120 minutes, using a sample of 39 students who average 110 minutes. Watch out! Don't forget to use the known standard deviation of 30 minutes and the significance level of 5%!

STEP 2

1. Define the hypotheses
2. Calculate the test statistic
3. Determine the significance

STEP 3

Alright, let's start by setting up what we're trying to prove or disprove!
We have two hypotheses here:
- Null Hypothesis ( $H_0$ ): The mean study time for UGA students is equal to the national average, which is 120 minutes.
So, $H_0: \mu = 120$ .
- Alternative Hypothesis ( $H_a$ ): The mean study time for UGA students is different from the national average.
So, $H_a: \mu \neq 120$ .

STEP 4

Now, let's calculate the test statistic!
This is where we use the sample mean, the national average, and the standard deviation.
The formula for the test statistic $z$ is:
$z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ where: - $\bar{x} = 110$ (the sample mean), - $\mu = 120$ (the national average), - $\sigma = 30$ (the standard deviation), - $n = 39$ (the sample size).

STEP 5

Let's plug in those numbers and do the math:
$z = \frac{110 - 120}{\frac{30}{\sqrt{39}}}$

STEP 6

First, calculate the denominator:
$\frac{30}{\sqrt{39}} \approx \frac{30}{6.244} \approx 4.805$

STEP 7

Now, calculate the numerator:
$110 - 120 = -10$

STEP 8

Finally, divide the numerator by the denominator to find $z$ :
$z = \frac{-10}{4.805} \approx -2.081$

STEP 9

We're using a 5% significance level, which corresponds to a critical value of approximately $\pm 1.96$ for a two-tailed test.
Since our calculated $z$ value of $-2.081$ is outside this range, it suggests that the study times are significantly different from the national average!

STEP 10

The value of the test statistic is approximately $-2.08$ .

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Studdy Solution

STEP 1

What is this asking? We're checking if UGA students study differently than the national average of 120 minutes, using a sample of 39 students who average 110 minutes. Watch out! Don't forget to use the known standard deviation of 30 minutes and the significance level of 5%!

STEP 2

1. Define the hypotheses
2. Calculate the test statistic
3. Determine the significance

STEP 3

STEP 4

STEP 5

Let's plug in those numbers and do the math:
$z = \frac{110 - 120}{\frac{30}{\sqrt{39}}}$

STEP 6

First, calculate the denominator:
$\frac{30}{\sqrt{39}} \approx \frac{30}{6.244} \approx 4.805$

STEP 7

Now, calculate the numerator:
$110 - 120 = -10$

STEP 8

Finally, divide the numerator by the denominator to find $z$ :
$z = \frac{-10}{4.805} \approx -2.081$

STEP 9

We're using a 5% significance level, which corresponds to a critical value of approximately $\pm 1.96$ for a two-tailed test.
Since our calculated $z$ value of $-2.081$ is outside this range, it suggests that the study times are significantly different from the national average!

STEP 10

The value of the test statistic is approximately $-2.08$ .

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Studdy Solution

STEP 1

What is this asking? We're checking if UGA students study differently than the national average of 120 minutes, using a sample of 39 students who average 110 minutes. Watch out! Don't forget to use the known standard deviation of 30 minutes and the significance level of 5%!

STEP 2

1. Define the hypotheses2. Calculate the test statistic3. Determine the significance

STEP 3

STEP 4

STEP 5

Let's plug in those numbers and do the math:z=110−1203039z = \frac{110 - 120}{\frac{30}{\sqrt{39}}}z=39​30​110−120​

STEP 6

First, calculate the denominator:3039≈306.244≈4.805\frac{30}{\sqrt{39}} \approx \frac{30}{6.244} \approx 4.80539​30​≈6.24430​≈4.805

STEP 7

Now, calculate the numerator:110−120=−10110 - 120 = -10110−120=−10

STEP 8

Finally, divide the numerator by the denominator to find zzz:z=−104.805≈−2.081z = \frac{-10}{4.805} \approx -2.081z=4.805−10​≈−2.081

STEP 9

STEP 10

The value of the test statistic is approximately −2.08-2.08−2.08.

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1. Define the hypotheses
2. Calculate the test statistic
3. Determine the significance

Let's plug in those numbers and do the math:
$z = \frac{110 - 120}{\frac{30}{\sqrt{39}}}$

First, calculate the denominator:
$\frac{30}{\sqrt{39}} \approx \frac{30}{6.244} \approx 4.805$

Now, calculate the numerator:
$110 - 120 = -10$

Finally, divide the numerator by the denominator to find $z$ :
$z = \frac{-10}{4.805} \approx -2.081$

The value of the test statistic is approximately $-2.08$ .