Math / Data & Statistics

Question5 pts 1 Details
The numbers of online applications from simple random samples of college applications for 2004 and for the 2008 were taken. In 2004, out of 658 applications, 290 of them were completed online. In 2008, out of 236 applications, 118 of them were completed online. Test the claim that the proportion of online applications in 2004 was equal to the proportion of online applications in 2008 at the .025 significance level.
Claim: Select an answer which corresponds to Select an answerv Opposite: Select an answer which corresponds to Select an answer
The test is: Select an answer The test statistic is: $z=$ $\square$ (to 2 decimals)
The critical value is: $z=$ $\square$ (to 2 decimals)
Based on this we: Select an answer $\square$ Conclusion There Select an answer appear to be enough evidence to support the claim that the proportion of online applications in 2004 was equal to the proportion of online applications in 2008.

Studdy Solution

STEP 1

What is this asking? Are the percentages of online college applications the same in 2004 and 2008? Watch out! Don't mix up the years or the number of online vs. total applications!

STEP 2

1. Set up the hypothesis test
2. Calculate the pooled proportion
3. Calculate the test statistic
4. Find the critical value
5. Make a decision

STEP 3

We're testing if the proportions are equal, so our null hypothesis $H_0$ is $p_{2004} = p_{2008}$ .
The alternative hypothesis $H_1$ is $p_{2004} \ne p_{2008}$ .
This is a two-tailed test because we're looking for any difference, not just if one is greater than the other.

STEP 4

The significance level is $\alpha = 0.025$ .
This means there's a 2.5% chance we'll reject the null hypothesis even if it's true.
It's like setting the sensitivity on our difference-detector!

STEP 5

The pooled proportion is like a weighted average of the two proportions, pretending they're actually the same.
It's a useful tool for this type of test!
The formula is: $\hat{p} = \frac{x_{2004} + x_{2008}}{n_{2004} + n_{2008}}$ Where $x$ represents the number of online applications and $n$ represents the total number of applications.

STEP 6

Let's plug in the numbers: $x_{2004} = \textbf{290}$ , $n_{2004} = \textbf{658}$ , $x_{2008} = \textbf{118}$ , and $n_{2008} = \textbf{236}$ . $\hat{p} = \frac{290 + 118}{658 + 236} = \frac{408}{894} \approx \textbf{0.456}$ So, our pooled proportion is approximately $\textbf{0.456}$ .

STEP 7

The test statistic, often called "z", tells us how far apart our observed proportions are from what we'd expect if the null hypothesis were true.
Here's the formula: $z = \frac{\hat{p}_{2004} - \hat{p}_{2008}}{\sqrt{\hat{p}(1 - \hat{p})(\frac{1}{n_{2004}} + \frac{1}{n_{2008}})}}$

STEP 8

First, let's calculate the individual proportions: $\hat{p}_{2004} = \frac{290}{658} \approx \textbf{0.441}$ and $\hat{p}_{2008} = \frac{118}{236} \approx \textbf{0.500}$ .

STEP 9

Now, plug everything into the z-score formula: $z = \frac{0.441 - 0.500}{\sqrt{0.456(1 - 0.456)(\frac{1}{658} + \frac{1}{236})}} \approx \frac{-0.059}{\sqrt{0.456 \cdot 0.544 \cdot 0.0057}} \approx \frac{-0.059}{0.038} \approx \textbf{-1.55}$ Our test statistic is approximately $\textbf{-1.55}$ .

STEP 10

The critical value is like a cutoff point.
If our test statistic is beyond this value, it's so unlikely that we reject the null hypothesis.
Since this is a two-tailed test with $\alpha = 0.025$ , we'll look up the critical value for $\frac{\alpha}{2} = 0.0125$ in a z-table (or use a calculator).

STEP 11

The critical value for a two-tailed test at $\alpha = 0.025$ is approximately $\textbf{±2.24}$ .

STEP 12

Our test statistic ( $-1.55$ ) is between the critical values ( $-2.24$ and $2.24$ ).
This means it's not extreme enough to reject the null hypothesis.

STEP 13

The test statistic is $z \approx \textbf{-1.55}$ .
The critical value is $z \approx \textbf{±2.24}$ .
We fail to reject the null hypothesis.
There does not appear to be enough evidence to reject the claim that the proportion of online applications in 2004 was equal to the proportion of online applications in 2008.

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Studdy Solution

STEP 1

What is this asking? Are the percentages of online college applications the same in 2004 and 2008? Watch out! Don't mix up the years or the number of online vs. total applications!

STEP 2

1. Set up the hypothesis test
2. Calculate the pooled proportion
3. Calculate the test statistic
4. Find the critical value
5. Make a decision

STEP 3

We're testing if the proportions are equal, so our null hypothesis $H_0$ is $p_{2004} = p_{2008}$ .
The alternative hypothesis $H_1$ is $p_{2004} \ne p_{2008}$ .
This is a two-tailed test because we're looking for any difference, not just if one is greater than the other.

STEP 4

The significance level is $\alpha = 0.025$ .
This means there's a 2.5% chance we'll reject the null hypothesis even if it's true.
It's like setting the sensitivity on our difference-detector!

STEP 5

STEP 6

Let's plug in the numbers: $x_{2004} = \textbf{290}$ , $n_{2004} = \textbf{658}$ , $x_{2008} = \textbf{118}$ , and $n_{2008} = \textbf{236}$ . $\hat{p} = \frac{290 + 118}{658 + 236} = \frac{408}{894} \approx \textbf{0.456}$ So, our pooled proportion is approximately $\textbf{0.456}$ .

STEP 7

The test statistic, often called "z", tells us how far apart our observed proportions are from what we'd expect if the null hypothesis were true.
Here's the formula: $z = \frac{\hat{p}_{2004} - \hat{p}_{2008}}{\sqrt{\hat{p}(1 - \hat{p})(\frac{1}{n_{2004}} + \frac{1}{n_{2008}})}}$

STEP 8

First, let's calculate the individual proportions: $\hat{p}_{2004} = \frac{290}{658} \approx \textbf{0.441}$ and $\hat{p}_{2008} = \frac{118}{236} \approx \textbf{0.500}$ .

STEP 9

Now, plug everything into the z-score formula: $z = \frac{0.441 - 0.500}{\sqrt{0.456(1 - 0.456)(\frac{1}{658} + \frac{1}{236})}} \approx \frac{-0.059}{\sqrt{0.456 \cdot 0.544 \cdot 0.0057}} \approx \frac{-0.059}{0.038} \approx \textbf{-1.55}$ Our test statistic is approximately $\textbf{-1.55}$ .

STEP 10

The critical value is like a cutoff point.
If our test statistic is beyond this value, it's so unlikely that we reject the null hypothesis.
Since this is a two-tailed test with $\alpha = 0.025$ , we'll look up the critical value for $\frac{\alpha}{2} = 0.0125$ in a z-table (or use a calculator).

STEP 11

The critical value for a two-tailed test at $\alpha = 0.025$ is approximately $\textbf{±2.24}$ .

STEP 12

Our test statistic ( $-1.55$ ) is between the critical values ( $-2.24$ and $2.24$ ).
This means it's not extreme enough to reject the null hypothesis.

STEP 13

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Studdy Solution

STEP 1

What is this asking? Are the percentages of online college applications the same in 2004 and 2008? Watch out! Don't mix up the years or the number of online vs. total applications!

STEP 2

1. Set up the hypothesis test2. Calculate the pooled proportion3. Calculate the test statistic4. Find the critical value5. Make a decision

STEP 3

STEP 4

The **significance level** is α=0.025\alpha = 0.025α=0.025.This means there's a 2.5% chance we'll reject the null hypothesis even if it's true.It's like setting the sensitivity on our difference-detector!

STEP 5

STEP 6

STEP 7

STEP 8

First, let's calculate the individual proportions: p^2004=290658≈0.441\hat{p}_{2004} = \frac{290}{658} \approx \textbf{0.441}p^​2004​=658290​≈0.441 and p^2008=118236≈0.500\hat{p}_{2008} = \frac{118}{236} \approx \textbf{0.500}p^​2008​=236118​≈0.500.

STEP 9

STEP 10

STEP 11

The **critical value** for a two-tailed test at α=0.025\alpha = 0.025α=0.025 is approximately ±2.24\textbf{±2.24}±2.24.

STEP 12

Our **test statistic** (−1.55-1.55−1.55) is between the **critical values** (−2.24-2.24−2.24 and 2.242.242.24).This means it's not extreme enough to reject the null hypothesis.

STEP 13

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Start learning now

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

1. Set up the hypothesis test
2. Calculate the pooled proportion
3. Calculate the test statistic
4. Find the critical value
5. Make a decision

The significance level is $\alpha = 0.025$ .
This means there's a 2.5% chance we'll reject the null hypothesis even if it's true.
It's like setting the sensitivity on our difference-detector!

First, let's calculate the individual proportions: $\hat{p}_{2004} = \frac{290}{658} \approx \textbf{0.441}$ and $\hat{p}_{2008} = \frac{118}{236} \approx \textbf{0.500}$ .

The critical value for a two-tailed test at $\alpha = 0.025$ is approximately $\textbf{±2.24}$ .

Our test statistic ( $-1.55$ ) is between the critical values ( $-2.24$ and $2.24$ ).
This means it's not extreme enough to reject the null hypothesis.