Math / Data & Statistics

QuestionCollege tuition: The mean annual tuition and fees for a sample of 15 private colleges in California was $\$ 37,500$ with a standard deviation of $\$ 7850$ . A dotplot shows that it is reasonable to assume that the population is approximately normal. Can you conclude that the mean tuition and fees for private institutions in California is greater than $\$ 35,000$ ? Use the $\alpha=0.05$ level of significance and the $P$ -value method with the TI-84 Plus calculator.
Part: $0 / 5$
Part 1 of 5 (a) State the appropriate null and alternate hypotheses. $\begin{array}{l} H_{0}: \square \\ H_{1}: \mu>35,000 \end{array}$
This hypothesis test is a right-tailed $\square$ test.

Studdy Solution

STEP 1

What is this asking? We want to find out if the average tuition for private colleges in California is more than $\$35,000$ based on a sample of 15 colleges. Watch out! Don't mix up the sample mean and the population mean!
We're trying to figure something out about all private colleges in California, not just the ones in our sample.

STEP 2

1. Set up the hypotheses
2. Calculate the test statistic
3. Find the p-value
4. Make a decision

STEP 3

Our null hypothesis ( $H_0$ ) is what we're trying to disprove.
It's like the default assumption.
In this case, it's that the average tuition is not greater than $\$35,000$ , meaning it's less than or equal to $\$35,000$ .
So, we write: $H_0: \mu \le 35,000$

STEP 4

The alternative hypothesis ( $H_1$ ) is what we think might be true instead of the null hypothesis.
Here, we're testing if the average tuition is greater than $\$35,000$ .
So we write: $H_1: \mu > 35,000$

STEP 5

Since our alternative hypothesis is $\mu > 35,000$ , we're looking for values in the right tail of our distribution.
This means we're doing a right-tailed test!

STEP 6

We'll use the t-test statistic because we don't know the population standard deviation.
The formula is: $t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}$ Where: * $\bar{x}$ is the sample mean (the average tuition of our 15 colleges). * $\mu_0$ is the hypothesized population mean (the $\$35,000$ we're testing against). * $s$ is the sample standard deviation. * $n$ is the sample size (number of colleges in our sample).

STEP 7

We know: * $\bar{x} = 37,500$ * $\mu_0 = 35,000$ * $s = 7,850$ * $n = 15$ Let's plug these values into our formula: $t = \frac{37,500 - 35,000}{\frac{7,850}{\sqrt{15}}}$

STEP 8

$t = \frac{2,500}{\frac{7,850}{3.873}}$ $t = \frac{2,500}{2,026.8}$ $t \approx 1.233$ So, our test statistic is approximately $1.233$ .

STEP 9

The p-value tells us the probability of getting our sample results (or even more extreme results) if the null hypothesis were actually true.
A small p-value makes us doubt the null hypothesis.

STEP 10

We'll use a TI-84 Plus calculator to find the p-value.
We'll use the `T-Test` function.
Input the values: * $\mu_0 = 35,000$ * $\bar{x} = 37,500$ * $s = 7,850$ * $n = 15$ * $\mu > \mu_0$ (because it's a right-tailed test)

STEP 11

The calculator gives us a p-value of approximately $0.118$ .

STEP 12

Our significance level ( $\alpha$ ) is $0.05$ .
We compare this to our p-value.

STEP 13

Since our p-value ( $0.118$ ) is greater than our significance level ( $0.05$ ), we fail to reject the null hypothesis.

STEP 14

We do not have enough evidence to conclude that the mean tuition and fees for private institutions in California is greater than $\$35,000$ .

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Studdy Solution

STEP 1

STEP 2

1. Set up the hypotheses
2. Calculate the test statistic
3. Find the p-value
4. Make a decision

STEP 3

Our null hypothesis ( $H_0$ ) is what we're trying to disprove.
It's like the default assumption.
In this case, it's that the average tuition is not greater than $\$35,000$ , meaning it's less than or equal to $\$35,000$ .
So, we write: $H_0: \mu \le 35,000$

STEP 4

The alternative hypothesis ( $H_1$ ) is what we think might be true instead of the null hypothesis.
Here, we're testing if the average tuition is greater than $\$35,000$ .
So we write: $H_1: \mu > 35,000$

STEP 5

Since our alternative hypothesis is $\mu > 35,000$ , we're looking for values in the right tail of our distribution.
This means we're doing a right-tailed test!

STEP 6

STEP 7

We know: * $\bar{x} = 37,500$ * $\mu_0 = 35,000$ * $s = 7,850$ * $n = 15$ Let's plug these values into our formula: $t = \frac{37,500 - 35,000}{\frac{7,850}{\sqrt{15}}}$

STEP 8

$t = \frac{2,500}{\frac{7,850}{3.873}}$ $t = \frac{2,500}{2,026.8}$ $t \approx 1.233$ So, our test statistic is approximately $1.233$ .

STEP 9

The p-value tells us the probability of getting our sample results (or even more extreme results) if the null hypothesis were actually true.
A small p-value makes us doubt the null hypothesis.

STEP 10

We'll use a TI-84 Plus calculator to find the p-value.
We'll use the `T-Test` function.
Input the values: * $\mu_0 = 35,000$ * $\bar{x} = 37,500$ * $s = 7,850$ * $n = 15$ * $\mu > \mu_0$ (because it's a right-tailed test)

STEP 11

The calculator gives us a p-value of approximately $0.118$ .

STEP 12

Our significance level ( $\alpha$ ) is $0.05$ .
We compare this to our p-value.

STEP 13

Since our p-value ( $0.118$ ) is greater than our significance level ( $0.05$ ), we fail to reject the null hypothesis.

STEP 14

We do not have enough evidence to conclude that the mean tuition and fees for private institutions in California is greater than $\$35,000$ .

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Studdy Solution

STEP 1

STEP 2

1. Set up the hypotheses2. Calculate the test statistic3. Find the p-value4. Make a decision

STEP 3

STEP 4

The **alternative hypothesis** (H1H_1H1​) is what we think might be true instead of the null hypothesis.Here, we're testing if the average tuition is *greater* than $35,000\$35,000$35,000.So we write: H1:μ>35,000H_1: \mu > 35,000H1​:μ>35,000

STEP 5

Since our alternative hypothesis is μ>35,000\mu > 35,000μ>35,000, we're looking for values in the *right tail* of our distribution.This means we're doing a **right-tailed test**!

STEP 6

STEP 7

We know: * xˉ=37,500\bar{x} = 37,500xˉ=37,500 * μ0=35,000\mu_0 = 35,000μ0​=35,000 * s=7,850s = 7,850s=7,850 * n=15n = 15n=15 Let's plug these values into our formula: t=37,500−35,0007,85015t = \frac{37,500 - 35,000}{\frac{7,850}{\sqrt{15}}}t=15​7,850​37,500−35,000​

STEP 8

t=2,5007,8503.873t = \frac{2,500}{\frac{7,850}{3.873}}t=3.8737,850​2,500​ t=2,5002,026.8t = \frac{2,500}{2,026.8}t=2,026.82,500​ t≈1.233t \approx 1.233t≈1.233So, our **test statistic** is approximately 1.2331.2331.233.

STEP 9

The **p-value** tells us the probability of getting our sample results (or even more extreme results) if the null hypothesis were actually true.A small p-value makes us doubt the null hypothesis.

STEP 10

We'll use a TI-84 Plus calculator to find the p-value.We'll use the `T-Test` function.Input the values: * μ0=35,000\mu_0 = 35,000μ0​=35,000 * xˉ=37,500\bar{x} = 37,500xˉ=37,500 * s=7,850s = 7,850s=7,850 * n=15n = 15n=15 * μ>μ0\mu > \mu_0μ>μ0​ (because it's a right-tailed test)

STEP 11

The calculator gives us a p-value of approximately 0.1180.1180.118.

STEP 12

Our **significance level** (α\alphaα) is 0.050.050.05.We compare this to our p-value.

STEP 13

Since our p-value (0.1180.1180.118) is *greater* than our significance level (0.050.050.05), we **fail to reject the null hypothesis**.

STEP 14

We do not have enough evidence to conclude that the mean tuition and fees for private institutions in California is greater than $35,000\$35,000$35,000.

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1. Set up the hypotheses
2. Calculate the test statistic
3. Find the p-value
4. Make a decision

The alternative hypothesis ( $H_1$ ) is what we think might be true instead of the null hypothesis.
Here, we're testing if the average tuition is greater than $\$35,000$ .
So we write: $H_1: \mu > 35,000$

Since our alternative hypothesis is $\mu > 35,000$ , we're looking for values in the right tail of our distribution.
This means we're doing a right-tailed test!

We know: * $\bar{x} = 37,500$ * $\mu_0 = 35,000$ * $s = 7,850$ * $n = 15$ Let's plug these values into our formula: $t = \frac{37,500 - 35,000}{\frac{7,850}{\sqrt{15}}}$

$t = \frac{2,500}{\frac{7,850}{3.873}}$ $t = \frac{2,500}{2,026.8}$ $t \approx 1.233$ So, our test statistic is approximately $1.233$ .

The p-value tells us the probability of getting our sample results (or even more extreme results) if the null hypothesis were actually true.
A small p-value makes us doubt the null hypothesis.

We'll use a TI-84 Plus calculator to find the p-value.
We'll use the `T-Test` function.
Input the values: * $\mu_0 = 35,000$ * $\bar{x} = 37,500$ * $s = 7,850$ * $n = 15$ * $\mu > \mu_0$ (because it's a right-tailed test)

The calculator gives us a p-value of approximately $0.118$ .

Our significance level ( $\alpha$ ) is $0.05$ .
We compare this to our p-value.

Since our p-value ( $0.118$ ) is greater than our significance level ( $0.05$ ), we fail to reject the null hypothesis.

We do not have enough evidence to conclude that the mean tuition and fees for private institutions in California is greater than $\$35,000$ .