Math / Data & Statistics

QuestionA 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 Table A2 CRITICAL VALUE for the problem? Note Answer Input Format for Critical Values.
Note: Enter X.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 1.645 is entered as $1.65,-2.326$ is entered as $-2.33,2.054$ is entered as $2.05,-1.282$ is entered as -1.28

Studdy Solution

STEP 1

What is this asking? We're checking if the average SAT score of 50 students in a training program is higher than the historical average of 518, using a significance level of $5\%$ . Watch out! Don't forget to use the correct critical value for a one-tailed test and ensure you round the critical value correctly!

STEP 2

1. Define the null and alternative hypotheses
2. Calculate the test statistic
3. Determine the critical value
4. Compare the test statistic to the critical value

STEP 3

Alright, let's start by setting up our hypotheses.
The null hypothesis $H_0$ is that there is no improvement, meaning the average score is still $518$ .
So, we write:
$H_0: \mu = 518$

STEP 4

Next, the alternative hypothesis $H_a$ suggests there is an improvement, meaning the average score is greater than $518$ .
So, we write:
$H_a: \mu > 518$

STEP 5

Now, let's calculate the test statistic!
We use the formula for the z-test statistic:
$z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ where: - $\bar{x} = 550$ is the sample mean, - $\mu = 518$ is the population mean, - $\sigma = 114$ is the standard deviation, - $n = 50$ is the sample size.

STEP 6

Plug in the values:
$z = \frac{550 - 518}{\frac{114}{\sqrt{50}}}$

STEP 7

Calculate the denominator first:
$\frac{114}{\sqrt{50}} \approx \frac{114}{7.071} \approx 16.12$

STEP 8

Now, calculate the entire expression:
$z = \frac{32}{16.12} \approx 1.98$

STEP 9

Since we're using a 5\% significance level for a one-tailed test, we need to find the critical value from the standard normal distribution table.
For a $5\%$ significance level, the critical value is approximately $1.645$ .

STEP 10

Now, let's compare our calculated z-value of $1.98$ to the critical value of $1.645$ .

STEP 11

Since $1.98 > 1.645$ , we reject the null hypothesis.
This suggests that there is a statistically significant improvement in the average score!

STEP 12

The critical value for the problem is $1.65$ .
Since the test statistic $1.98$ is greater than the critical value $1.65$ , there is evidence to suggest improvement over the historical average.

Was this helpful?

Studdy Solution

STEP 1

STEP 2

1. Define the null and alternative hypotheses
2. Calculate the test statistic
3. Determine the critical value
4. Compare the test statistic to the critical value

STEP 3

Alright, let's start by setting up our hypotheses.
The null hypothesis $H_0$ is that there is no improvement, meaning the average score is still $518$ .
So, we write:
$H_0: \mu = 518$

STEP 4

Next, the alternative hypothesis $H_a$ suggests there is an improvement, meaning the average score is greater than $518$ .
So, we write:
$H_a: \mu > 518$

STEP 5

Now, let's calculate the test statistic!
We use the formula for the z-test statistic:
$z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ where: - $\bar{x} = 550$ is the sample mean, - $\mu = 518$ is the population mean, - $\sigma = 114$ is the standard deviation, - $n = 50$ is the sample size.

STEP 6

Plug in the values:
$z = \frac{550 - 518}{\frac{114}{\sqrt{50}}}$

STEP 7

Calculate the denominator first:
$\frac{114}{\sqrt{50}} \approx \frac{114}{7.071} \approx 16.12$

STEP 8

Now, calculate the entire expression:
$z = \frac{32}{16.12} \approx 1.98$

STEP 9

Since we're using a 5\% significance level for a one-tailed test, we need to find the critical value from the standard normal distribution table.
For a $5\%$ significance level, the critical value is approximately $1.645$ .

STEP 10

Now, let's compare our calculated z-value of $1.98$ to the critical value of $1.645$ .

STEP 11

Since $1.98 > 1.645$ , we reject the null hypothesis.
This suggests that there is a statistically significant improvement in the average score!

STEP 12

The critical value for the problem is $1.65$ .
Since the test statistic $1.98$ is greater than the critical value $1.65$ , there is evidence to suggest improvement over the historical average.

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.

Studdy Solution

STEP 1

STEP 2

1. Define the null and alternative hypotheses2. Calculate the test statistic3. Determine the critical value4. Compare the test statistic to the critical value

STEP 3

Alright, let's start by setting up our hypotheses.The **null hypothesis** H0H_0H0​ is that there is no improvement, meaning the average score is still 518518518.So, we write:H0:μ=518 H_0: \mu = 518 H0​:μ=518

STEP 4

Next, the **alternative hypothesis** HaH_aHa​ suggests there is an improvement, meaning the average score is greater than 518518518.So, we write:Ha:μ>518 H_a: \mu > 518 Ha​:μ>518

STEP 5

STEP 6

Plug in the values:z=550−51811450z = \frac{550 - 518}{\frac{114}{\sqrt{50}}}z=50​114​550−518​

STEP 7

Calculate the denominator first:11450≈1147.071≈16.12\frac{114}{\sqrt{50}} \approx \frac{114}{7.071} \approx 16.1250​114​≈7.071114​≈16.12

STEP 8

Now, calculate the entire expression:z=3216.12≈1.98z = \frac{32}{16.12} \approx 1.98z=16.1232​≈1.98

STEP 9

Since we're using a **5\% significance level** for a one-tailed test, we need to find the critical value from the standard normal distribution table.For a 5%5\%5% significance level, the critical value is approximately 1.6451.6451.645.

STEP 10

Now, let's compare our calculated **z-value** of 1.981.981.98 to the critical value of 1.6451.6451.645.

STEP 11

Since 1.98>1.6451.98 > 1.6451.98>1.645, we reject the null hypothesis.This suggests that there is a statistically significant improvement in the average score!

STEP 12

The critical value for the problem is 1.651.651.65.Since the test statistic 1.981.981.98 is greater than the critical value 1.651.651.65, there is evidence to suggest improvement over the historical average.

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.

1. Define the null and alternative hypotheses
2. Calculate the test statistic
3. Determine the critical value
4. Compare the test statistic to the critical value

Alright, let's start by setting up our hypotheses.
The null hypothesis $H_0$ is that there is no improvement, meaning the average score is still $518$ .
So, we write:
$H_0: \mu = 518$

Next, the alternative hypothesis $H_a$ suggests there is an improvement, meaning the average score is greater than $518$ .
So, we write:
$H_a: \mu > 518$

Plug in the values:
$z = \frac{550 - 518}{\frac{114}{\sqrt{50}}}$

Calculate the denominator first:
$\frac{114}{\sqrt{50}} \approx \frac{114}{7.071} \approx 16.12$

Now, calculate the entire expression:
$z = \frac{32}{16.12} \approx 1.98$

Since we're using a 5\% significance level for a one-tailed test, we need to find the critical value from the standard normal distribution table.
For a $5\%$ significance level, the critical value is approximately $1.645$ .

Now, let's compare our calculated z-value of $1.98$ to the critical value of $1.645$ .

Since $1.98 > 1.645$ , we reject the null hypothesis.
This suggests that there is a statistically significant improvement in the average score!

The critical value for the problem is $1.65$ .
Since the test statistic $1.98$ is greater than the critical value $1.65$ , there is evidence to suggest improvement over the historical average.