Math / Data & Statistics

QuestionI
A preparation SAT test score can be assumed normally distributed with a historical mean of 518 and known standard deviation 114. In a training program, 50 students had an average score of 550 on the preparatory test. Test whether there has been improvement over the historical average. Use a $5 \%$ significance level.
1 What is the P-VALUE from Table A2 for testing HO? See two entry formats below.
Note: Two formats possible: 1) Enter LT01 if P-value is less than 0.01 2) 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

Studdy Solution

STEP 1

What is this asking? Did the training program actually help students improve their SAT scores, or could the higher average just be random chance? Watch out! Don't mix up one-tailed and two-tailed tests!
We're only checking for improvement (a higher score), not just any difference.

STEP 2

1. Set up the hypothesis test
2. Calculate the test statistic
3. Find the p-value

STEP 3

Our null hypothesis ( $H_0$ ) is that the training program didn't do anything special.
So, the average score is still the historical mean of $\mu = 518$ .
Mathematically, we write $H_0: \mu = 518$ .

STEP 4

The alternative hypothesis ( $H_1$ ) is what we're trying to prove: that the training program did improve scores.
So, the average score is now greater than the historical mean.
Mathematically, we write $H_1: \mu > 518$ .
This is a one-tailed test because we're only interested in scores being higher.

STEP 5

The significance level, often denoted by $\alpha$ , is how much evidence we need to reject the null hypothesis.
Here, $\alpha = 0.05$ or 5%.
This means we're willing to accept a 5% chance of being wrong when we reject the null hypothesis.

STEP 6

Since we know the population standard deviation ( $\sigma = 114$ ), we'll use a z-test.
The z-test is perfect for this situation!

STEP 7

The z-score tells us how many standard deviations the sample mean is away from the historical mean.
The formula is: $z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ Where $\bar{x}$ is the sample mean ( $\bar{x} = 550$ ), $\mu$ is the population mean under the null hypothesis ( $\mu = 518$ ), $\sigma$ is the population standard deviation ( $\sigma = 114$ ), and $n$ is the sample size ( $n = 50$ ).

STEP 8

$z = \frac{550 - 518}{\frac{114}{\sqrt{50}}}$

STEP 9

$z = \frac{32}{\frac{114}{7.071}}$ $z = \frac{32}{16.124}$

STEP 10

$z \approx 1.985$ So, our sample mean is almost two standard deviations above the historical mean!

STEP 11

The p-value is the probability of observing a sample mean as extreme as ours (or even more extreme) if the null hypothesis were true.
A small p-value means it's unlikely that the improvement happened by chance.

STEP 12

Since our z-score is $1.985$ , we look up the corresponding p-value in a z-table.
We are performing a one-tailed test, so we look at the one-tailed probabilities.
The value for $z = 1.98$ is $0.9761$ .
The value for $z = 1.99$ is $0.9767$ .
Since our z-score is between these two, we can estimate our p-value by linear interpolation: $0.9761 + (1.985 - 1.98) \cdot (0.9767 - 0.9761) = 0.9761 + 0.005 \cdot 0.0006 = 0.9761 + 0.000003 = 0.9764$ Since we are looking for the probability of a value greater than our z-score, we subtract this value from 1: $1 - 0.9764 = 0.0236$

STEP 13

Following the requested format, we round up to two decimal places: $0.0236$ becomes 0.03.

STEP 14

The p-value is 0.03.

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Studdy Solution

STEP 1

What is this asking? Did the training program actually help students improve their SAT scores, or could the higher average just be random chance? Watch out! Don't mix up one-tailed and two-tailed tests!
We're only checking for improvement (a higher score), not just any difference.

STEP 2

1. Set up the hypothesis test
2. Calculate the test statistic
3. Find the p-value

STEP 3

Our null hypothesis ( $H_0$ ) is that the training program didn't do anything special.
So, the average score is still the historical mean of $\mu = 518$ .
Mathematically, we write $H_0: \mu = 518$ .

STEP 4

The alternative hypothesis ( $H_1$ ) is what we're trying to prove: that the training program did improve scores.
So, the average score is now greater than the historical mean.
Mathematically, we write $H_1: \mu > 518$ .
This is a one-tailed test because we're only interested in scores being higher.

STEP 5

The significance level, often denoted by $\alpha$ , is how much evidence we need to reject the null hypothesis.
Here, $\alpha = 0.05$ or 5%.
This means we're willing to accept a 5% chance of being wrong when we reject the null hypothesis.

STEP 6

Since we know the population standard deviation ( $\sigma = 114$ ), we'll use a z-test.
The z-test is perfect for this situation!

STEP 7

STEP 8

$z = \frac{550 - 518}{\frac{114}{\sqrt{50}}}$

STEP 9

$z = \frac{32}{\frac{114}{7.071}}$ $z = \frac{32}{16.124}$

STEP 10

$z \approx 1.985$ So, our sample mean is almost two standard deviations above the historical mean!

STEP 11

The p-value is the probability of observing a sample mean as extreme as ours (or even more extreme) if the null hypothesis were true.
A small p-value means it's unlikely that the improvement happened by chance.

STEP 12

STEP 13

Following the requested format, we round up to two decimal places: $0.0236$ becomes 0.03.

STEP 14

The p-value is 0.03.

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Studdy Solution

STEP 1

What is this asking? Did the training program actually help students improve their SAT scores, or could the higher average just be random chance? Watch out! Don't mix up one-tailed and two-tailed tests!We're only checking for *improvement* (a higher score), not just *any* difference.

STEP 2

1. Set up the hypothesis test2. Calculate the test statistic3. Find the p-value

STEP 3

Our *null hypothesis* (H0H_0H0​) is that the training program didn't do anything special.So, the average score is still the *historical mean* of μ=518\mu = 518μ=518.Mathematically, we write H0:μ=518H_0: \mu = 518H0​:μ=518.

STEP 4

STEP 5

The *significance level*, often denoted by α\alphaα, is how much evidence we need to reject the null hypothesis.Here, α=0.05\alpha = 0.05α=0.05 or 5%.This means we're willing to accept a 5% chance of being wrong when we reject the null hypothesis.

STEP 6

Since we know the *population standard deviation* (σ=114\sigma = 114σ=114), we'll use a *z-test*.The z-test is perfect for this situation!

STEP 7

STEP 8

z=550−51811450 z = \frac{550 - 518}{\frac{114}{\sqrt{50}}} z=50​114​550−518​

STEP 9

z=321147.071 z = \frac{32}{\frac{114}{7.071}} z=7.071114​32​ z=3216.124 z = \frac{32}{16.124} z=16.12432​

STEP 10

z≈1.985 z \approx 1.985 z≈1.985 So, our sample mean is almost **two standard deviations** above the historical mean!

STEP 11

The *p-value* is the probability of observing a sample mean as extreme as ours (or even more extreme) if the null hypothesis were true.A small p-value means it's unlikely that the improvement happened by chance.

STEP 12

STEP 13

Following the requested format, we round up to two decimal places: 0.02360.02360.0236 becomes **0.03**.

STEP 14

The p-value is **0.03**.

Get Studdy

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? Did the training program actually help students improve their SAT scores, or could the higher average just be random chance? Watch out! Don't mix up one-tailed and two-tailed tests!
We're only checking for improvement (a higher score), not just any difference.

1. Set up the hypothesis test
2. Calculate the test statistic
3. Find the p-value

Our null hypothesis ( $H_0$ ) is that the training program didn't do anything special.
So, the average score is still the historical mean of $\mu = 518$ .
Mathematically, we write $H_0: \mu = 518$ .

The significance level, often denoted by $\alpha$ , is how much evidence we need to reject the null hypothesis.
Here, $\alpha = 0.05$ or 5%.
This means we're willing to accept a 5% chance of being wrong when we reject the null hypothesis.

Since we know the population standard deviation ( $\sigma = 114$ ), we'll use a z-test.
The z-test is perfect for this situation!

$z = \frac{550 - 518}{\frac{114}{\sqrt{50}}}$

$z = \frac{32}{\frac{114}{7.071}}$ $z = \frac{32}{16.124}$

$z \approx 1.985$ So, our sample mean is almost two standard deviations above the historical mean!

The p-value is the probability of observing a sample mean as extreme as ours (or even more extreme) if the null hypothesis were true.
A small p-value means it's unlikely that the improvement happened by chance.

Following the requested format, we round up to two decimal places: $0.0236$ becomes 0.03.

The p-value is 0.03.