Math / Data & Statistics

QuestionCollege tuition: The mean annual tuition and fees in the 2013-2014 academic year for a sample of 13 private colleges in California was $\$ 33,000$ with a standard deviation of $\$ 7300$ . 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 less than $\$ 35,000$ ? Use the $\alpha=0.01$ level of significance and the $P$ -value method and Excel.
Part 1 of 5 (a) State the appropriate null and alternate hypotheses. $\begin{array}{l} H_{0}: \mu=35,000 \\ H_{1}: \mu<35,000 \end{array}$
This hypothesis test is a left-tailed $\quad \mathbf{}$ test. Part 2 of 5 Skip Part Check Save For Later Submit Assi Q 2024 McGraw Hill LLC. All Rights Reserved. Terms of Use I Privacy Center

Studdy Solution

STEP 1

What is this asking? We want to find out if the average tuition for private colleges in California is actually less than $\$35,000$ based on a small sample of 13 colleges. Watch out! We're dealing with a sample of colleges, not all of them, so we need to be careful about how confident we are in our conclusion.
Also, we're only checking if the average is less than $\$35,000$ , not different in any direction.

STEP 2

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

STEP 3

Our null hypothesis ( $H_0$ ) is that the true average tuition is actually $\$35,000$ .
We write this as $H_0: \mu = \$35,000$ .

STEP 4

Our alternative hypothesis ( $H_1$ ) is what we're trying to prove: that the true average tuition is less than $\$35,000$ .
We write this as $H_1: \mu < \$35,000$ .

STEP 5

The significance level ( $\alpha$ ) is $\alpha = 0.01$ .
This means we're willing to accept a 1% chance of being wrong if we reject the null hypothesis.

STEP 6

We know the sample mean ( $\bar{x}$ ) is $\$33,000$ , the sample standard deviation ( $s$ ) is $\$7,300$ , and the sample size ( $n$ ) is 13.

STEP 7

Since we're dealing with a small sample and we don't know the true population standard deviation, we'll use the t-statistic.
The formula is: $t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}$ where $\mu$ is the value we're testing against (from our null hypothesis).

STEP 8

Plugging in our values, we get: $t = \frac{\$33,000 - \$35,000}{\frac{\$7,300}{\sqrt{13}}}$ $t = \frac{-\$2,000}{\frac{\$7,300}{3.60555}}$ $t = \frac{-\$2,000}{\$2,024.85}$ $t \approx -0.9877$ So our test statistic is approximately $-0.9877$ .

STEP 9

The degrees of freedom ( $df$ ) is $n - 1 = 13 - 1 = 12$ .

STEP 10

We'll use Excel's `T.DIST` function to find the p-value.
Since it's a left-tailed test, we want the area to the left of our test statistic.
In Excel, we'd type `=T.DIST(-0.9877, 12, TRUE)`.
This gives us a p-value of approximately $0.1716$ .

STEP 11

Our p-value ( $0.1716$ ) is greater than our significance level ( $0.01$ ).

STEP 12

Since the p-value is greater than the significance level, we fail to reject the null hypothesis.
This means we don't have enough evidence to say that the average tuition is less than $\$35,000$ .

STEP 13

We cannot conclude that the mean tuition and fees for private institutions in California is less than $\$35,000$ at the $\alpha = 0.01$ level of significance.

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Studdy Solution

STEP 1

STEP 2

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

STEP 3

Our null hypothesis ( $H_0$ ) is that the true average tuition is actually $\$35,000$ .
We write this as $H_0: \mu = \$35,000$ .

STEP 4

Our alternative hypothesis ( $H_1$ ) is what we're trying to prove: that the true average tuition is less than $\$35,000$ .
We write this as $H_1: \mu < \$35,000$ .

STEP 5

The significance level ( $\alpha$ ) is $\alpha = 0.01$ .
This means we're willing to accept a 1% chance of being wrong if we reject the null hypothesis.

STEP 6

We know the sample mean ( $\bar{x}$ ) is $\$33,000$ , the sample standard deviation ( $s$ ) is $\$7,300$ , and the sample size ( $n$ ) is 13.

STEP 7

Since we're dealing with a small sample and we don't know the true population standard deviation, we'll use the t-statistic.
The formula is: $t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}$ where $\mu$ is the value we're testing against (from our null hypothesis).

STEP 8

Plugging in our values, we get: $t = \frac{\$33,000 - \$35,000}{\frac{\$7,300}{\sqrt{13}}}$ $t = \frac{-\$2,000}{\frac{\$7,300}{3.60555}}$ $t = \frac{-\$2,000}{\$2,024.85}$ $t \approx -0.9877$ So our test statistic is approximately $-0.9877$ .

STEP 9

The degrees of freedom ( $df$ ) is $n - 1 = 13 - 1 = 12$ .

STEP 10

We'll use Excel's `T.DIST` function to find the p-value.
Since it's a left-tailed test, we want the area to the left of our test statistic.
In Excel, we'd type `=T.DIST(-0.9877, 12, TRUE)`.
This gives us a p-value of approximately $0.1716$ .

STEP 11

Our p-value ( $0.1716$ ) is greater than our significance level ( $0.01$ ).

STEP 12

Since the p-value is greater than the significance level, we fail to reject the null hypothesis.
This means we don't have enough evidence to say that the average tuition is less than $\$35,000$ .

STEP 13

We cannot conclude that the mean tuition and fees for private institutions in California is less than $\$35,000$ at the $\alpha = 0.01$ level of significance.

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Studdy Solution

STEP 1

STEP 2

1. Set up the hypothesis test2. Calculate the test statistic3. Find the p-value4. Make a decision

STEP 3

Our **null hypothesis** (H0H_0H0​) is that the *true* average tuition is actually $35,000\$35,000$35,000.We write this as H0:μ=$35,000H_0: \mu = \$35,000H0​:μ=$35,000.

STEP 4

Our **alternative hypothesis** (H1H_1H1​) is what we're trying to prove: that the *true* average tuition is *less* than $35,000\$35,000$35,000.We write this as H1:μ<$35,000H_1: \mu < \$35,000H1​:μ<$35,000.

STEP 5

The **significance level** (α\alphaα) is α=0.01\alpha = 0.01α=0.01.This means we're willing to accept a 1% chance of being wrong if we reject the null hypothesis.

STEP 6

We know the **sample mean** (xˉ\bar{x}xˉ) is $33,000\$33,000$33,000, the **sample standard deviation** (sss) is $7,300\$7,300$7,300, and the **sample size** (nnn) is **13**.

STEP 7

STEP 8

STEP 9

The **degrees of freedom** (dfdfdf) is n−1=13−1=12n - 1 = 13 - 1 = 12n−1=13−1=12.

STEP 10

We'll use Excel's `T.DIST` function to find the p-value.Since it's a left-tailed test, we want the area to the *left* of our test statistic.In Excel, we'd type `=T.DIST(-0.9877, 12, TRUE)`.This gives us a **p-value** of approximately 0.17160.17160.1716.

STEP 11

Our **p-value** (0.17160.17160.1716) is *greater* than our **significance level** (0.010.010.01).

STEP 12

Since the p-value is greater than the significance level, we *fail* to reject the null hypothesis.This means we don't have enough evidence to say that the average tuition is *less* than $35,000\$35,000$35,000.

STEP 13

We cannot conclude that the mean tuition and fees for private institutions in California is less than $35,000\$35,000$35,000 at the α=0.01\alpha = 0.01α=0.01 level of significance.

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

Our null hypothesis ( $H_0$ ) is that the true average tuition is actually $\$35,000$ .
We write this as $H_0: \mu = \$35,000$ .

Our alternative hypothesis ( $H_1$ ) is what we're trying to prove: that the true average tuition is less than $\$35,000$ .
We write this as $H_1: \mu < \$35,000$ .

The significance level ( $\alpha$ ) is $\alpha = 0.01$ .
This means we're willing to accept a 1% chance of being wrong if we reject the null hypothesis.

We know the sample mean ( $\bar{x}$ ) is $\$33,000$ , the sample standard deviation ( $s$ ) is $\$7,300$ , and the sample size ( $n$ ) is 13.

The degrees of freedom ( $df$ ) is $n - 1 = 13 - 1 = 12$ .

We'll use Excel's `T.DIST` function to find the p-value.
Since it's a left-tailed test, we want the area to the left of our test statistic.
In Excel, we'd type `=T.DIST(-0.9877, 12, TRUE)`.
This gives us a p-value of approximately $0.1716$ .

Our p-value ( $0.1716$ ) is greater than our significance level ( $0.01$ ).

Since the p-value is greater than the significance level, we fail to reject the null hypothesis.
This means we don't have enough evidence to say that the average tuition is less than $\$35,000$ .

We cannot conclude that the mean tuition and fees for private institutions in California is less than $\$35,000$ at the $\alpha = 0.01$ level of significance.