Math / Data & Statistics

Question1. University and Community College: A Savannah at a four-year college claims that mean enrollment at four-year colleges is higher than at two-year colleges in the United States. Two surveys are conducted. Of the 35 two-year colleges surveyed, the mean enrollment was 5,068 with a standard deviation of 4,777 . Of the 35 four-year colleges surveyed, the mean enrollment was 5,466 with a standard deviation of 8,191. a. $H_{0}$ : b. $H_{a}$ : c. Test statistic: d. $P$ -value:

Studdy Solution

STEP 1

What is this asking? We want to check if four-year colleges really have more students on average than two-year colleges. Watch out! Don't mix up the numbers for the two types of colleges!
Also, remember we're dealing with average enrollments, not the actual headcount of any specific college.

STEP 2

1. Set up the hypotheses
2. Calculate the test statistic
3. Find the p-value
4. Interpret the results

STEP 3

We're assuming there's no difference in average enrollment.
So, our null hypothesis $H_0$ is that the mean enrollment at four-year colleges ( $\mu_4$ ) is the same as the mean enrollment at two-year colleges ( $\mu_2$ ).
$H_0: \mu_4 = \mu_2$

STEP 4

Our Savannah friend claims four-year colleges have higher enrollment.
This is our alternative hypothesis $H_a$ :
$H_a: \mu_4 > \mu_2$

STEP 5

We're comparing two sample means, and we don't know the true population standard deviations.
That screams t-test!
Specifically, it's a two-sample t-test.

STEP 6

Here's the formula for our t-statistic:
$t = \frac{(\bar{x}_4 - \bar{x}_2)}{\sqrt{\frac{s_4^2}{n_4} + \frac{s_2^2}{n_2}}}$ Where: * $\bar{x}_4$ and $\bar{x}_2$ are the sample means for four-year and two-year colleges, respectively. * $s_4$ and $s_2$ are the sample standard deviations. * $n_4$ and $n_2$ are the sample sizes.

STEP 7

Let's plug in our values: $\bar{x}_4 = \textbf{5466}$ , $\bar{x}_2 = \textbf{5068}$ , $s_4 = \textbf{8191}$ , $s_2 = \textbf{4777}$ , and $n_4 = n_2 = \textbf{35}$ .
$t = \frac{(5466 - 5068)}{\sqrt{\frac{8191^2}{35} + \frac{4777^2}{35}}}$

STEP 8

Time to crunch those numbers!
$t = \frac{398}{\sqrt{\frac{67092481}{35} + \frac{22820529}{35}}}$ $t = \frac{398}{\sqrt{1916928.03 + 652015.11}}$ $t = \frac{398}{\sqrt{2568943.14}}$ $t = \frac{398}{1602.79}$ $t \approx 0.248$

STEP 9

We need the degrees of freedom for our t-test.
A common approximation is:
$df = \min(n_4 - 1, n_2 - 1)$ In our case, $n_4 = n_2 = 35$ , so $df = \min(34, 34) = 34$ .

STEP 10

Using a t-table or calculator with our t-statistic ( $t \approx 0.248$ ) and degrees of freedom ( $df = 34$ ), we find a p-value greater than 0.25 (one-tailed test).

STEP 11

Our p-value is way bigger than a typical significance level (like 0.05).

STEP 12

This means we fail to reject the null hypothesis.
In simple terms, we don't have enough evidence to support Savannah's claim.

STEP 13

a. $H_0: \mu_4 = \mu_2$ b. $H_a: \mu_4 > \mu_2$ c. Test statistic: $t \approx 0.248$ d. $P$ -value: $> 0.25$

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Studdy Solution

STEP 1

What is this asking? We want to check if four-year colleges really have more students on average than two-year colleges. Watch out! Don't mix up the numbers for the two types of colleges!
Also, remember we're dealing with average enrollments, not the actual headcount of any specific college.

STEP 2

1. Set up the hypotheses
2. Calculate the test statistic
3. Find the p-value
4. Interpret the results

STEP 3

We're assuming there's no difference in average enrollment.
So, our null hypothesis $H_0$ is that the mean enrollment at four-year colleges ( $\mu_4$ ) is the same as the mean enrollment at two-year colleges ( $\mu_2$ ).
$H_0: \mu_4 = \mu_2$

STEP 4

Our Savannah friend claims four-year colleges have higher enrollment.
This is our alternative hypothesis $H_a$ :
$H_a: \mu_4 > \mu_2$

STEP 5

We're comparing two sample means, and we don't know the true population standard deviations.
That screams t-test!
Specifically, it's a two-sample t-test.

STEP 6

STEP 7

Let's plug in our values: $\bar{x}_4 = \textbf{5466}$ , $\bar{x}_2 = \textbf{5068}$ , $s_4 = \textbf{8191}$ , $s_2 = \textbf{4777}$ , and $n_4 = n_2 = \textbf{35}$ .
$t = \frac{(5466 - 5068)}{\sqrt{\frac{8191^2}{35} + \frac{4777^2}{35}}}$

STEP 8

Time to crunch those numbers!
$t = \frac{398}{\sqrt{\frac{67092481}{35} + \frac{22820529}{35}}}$ $t = \frac{398}{\sqrt{1916928.03 + 652015.11}}$ $t = \frac{398}{\sqrt{2568943.14}}$ $t = \frac{398}{1602.79}$ $t \approx 0.248$

STEP 9

We need the degrees of freedom for our t-test.
A common approximation is:
$df = \min(n_4 - 1, n_2 - 1)$ In our case, $n_4 = n_2 = 35$ , so $df = \min(34, 34) = 34$ .

STEP 10

Using a t-table or calculator with our t-statistic ( $t \approx 0.248$ ) and degrees of freedom ( $df = 34$ ), we find a p-value greater than 0.25 (one-tailed test).

STEP 11

Our p-value is way bigger than a typical significance level (like 0.05).

STEP 12

This means we fail to reject the null hypothesis.
In simple terms, we don't have enough evidence to support Savannah's claim.

STEP 13

a. $H_0: \mu_4 = \mu_2$ b. $H_a: \mu_4 > \mu_2$ c. Test statistic: $t \approx 0.248$ d. $P$ -value: $> 0.25$

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Studdy Solution

STEP 1

What is this asking? We want to check if four-year colleges really have more students on average than two-year colleges. Watch out! Don't mix up the numbers for the two types of colleges!Also, remember we're dealing with *average* enrollments, not the actual headcount of any specific college.

STEP 2

1. Set up the hypotheses2. Calculate the test statistic3. Find the p-value4. Interpret the results

STEP 3

We're assuming there's *no* difference in average enrollment.So, our **null hypothesis** H0H_0H0​ is that the mean enrollment at four-year colleges (μ4\mu_4μ4​) is the same as the mean enrollment at two-year colleges (μ2\mu_2μ2​).H0:μ4=μ2H_0: \mu_4 = \mu_2H0​:μ4​=μ2​

STEP 4

Our Savannah friend *claims* four-year colleges have *higher* enrollment.This is our **alternative hypothesis** HaH_aHa​:Ha:μ4>μ2H_a: \mu_4 > \mu_2Ha​:μ4​>μ2​

STEP 5

We're comparing two *sample* means, and we don't know the true population standard deviations.That screams *t-test*!Specifically, it's a two-sample t-test.

STEP 6

STEP 7

STEP 8

STEP 9

We need the **degrees of freedom** for our t-test.A common approximation is:df=min⁡(n4−1,n2−1)df = \min(n_4 - 1, n_2 - 1)df=min(n4​−1,n2​−1)In our case, n4=n2=35n_4 = n_2 = 35n4​=n2​=35, so df=min⁡(34,34)=34df = \min(34, 34) = 34df=min(34,34)=34.

STEP 10

Using a t-table or calculator with our **t-statistic** (t≈0.248t \approx 0.248t≈0.248) and **degrees of freedom** (df=34df = 34df=34), we find a **p-value** greater than 0.25 (one-tailed test).

STEP 11

Our **p-value** is *way* bigger than a typical significance level (like 0.05).

STEP 12

This means we *fail to reject* the null hypothesis.In simple terms, we don't have enough evidence to support Savannah's claim.

STEP 13

a. H0:μ4=μ2H_0: \mu_4 = \mu_2H0​:μ4​=μ2​ b. Ha:μ4>μ2H_a: \mu_4 > \mu_2Ha​:μ4​>μ2​ c. Test statistic: t≈0.248t \approx 0.248t≈0.248 d. PPP-value: >0.25> 0.25>0.25

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? We want to check if four-year colleges really have more students on average than two-year colleges. Watch out! Don't mix up the numbers for the two types of colleges!
Also, remember we're dealing with average enrollments, not the actual headcount of any specific college.

1. Set up the hypotheses
2. Calculate the test statistic
3. Find the p-value
4. Interpret the results

We're assuming there's no difference in average enrollment.
So, our null hypothesis $H_0$ is that the mean enrollment at four-year colleges ( $\mu_4$ ) is the same as the mean enrollment at two-year colleges ( $\mu_2$ ).
$H_0: \mu_4 = \mu_2$

Our Savannah friend claims four-year colleges have higher enrollment.
This is our alternative hypothesis $H_a$ :
$H_a: \mu_4 > \mu_2$

We're comparing two sample means, and we don't know the true population standard deviations.
That screams t-test!
Specifically, it's a two-sample t-test.

We need the degrees of freedom for our t-test.
A common approximation is:
$df = \min(n_4 - 1, n_2 - 1)$ In our case, $n_4 = n_2 = 35$ , so $df = \min(34, 34) = 34$ .

Using a t-table or calculator with our t-statistic ( $t \approx 0.248$ ) and degrees of freedom ( $df = 34$ ), we find a p-value greater than 0.25 (one-tailed test).

Our p-value is way bigger than a typical significance level (like 0.05).

This means we fail to reject the null hypothesis.
In simple terms, we don't have enough evidence to support Savannah's claim.

a. $H_0: \mu_4 = \mu_2$ b. $H_a: \mu_4 > \mu_2$ c. Test statistic: $t \approx 0.248$ d. $P$ -value: $> 0.25$