Math  /  Data & Statistics

Question1
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%5 \% significance level.
1 What is the value of the TEST STATISTIC?
Note: Enter XXX.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 134.784 is entered as 134.78,35.295134.78,35.295 is entered as 35.30,.274935.30, .2749 is entered as 0.27,.56500.27,-.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 was the improvement just random chance? Watch out! Don't mix up the *sample* mean and standard deviation with the *population* mean and standard deviation!
Also, make sure to use the correct z-score for a one-tailed test.

STEP 2

1. Set up the Hypothesis Test
2. Calculate the Standard Error
3. Calculate the Z-score
4. Find the p-value

STEP 3

We're testing if the training program *improved* scores, so we're looking for a significant increase.
This means we'll use a **one-tailed test**.

STEP 4

Our **null hypothesis** (H0H_0) is that the training program didn't do anything, and the average score is still the historical mean: μ=518\mu = 518.

STEP 5

Our **alternative hypothesis** (H1H_1) is that the training program *did* improve scores: μ>518\mu > 518.

STEP 6

Our **significance level**, also known as alpha, is α=0.05\alpha = 0.05.
This means we're willing to accept a 5% chance of concluding the training program worked when it actually didn't.

STEP 7

The **standard error** measures how much the sample mean is likely to vary from the population mean.
It's like the standard deviation of the sample means!

STEP 8

The formula for the standard error is SE=σnSE = \frac{\sigma}{\sqrt{n}}, where σ\sigma is the **population standard deviation** (σ=114\sigma = 114) and nn is the **sample size** (n=50n = 50).

STEP 9

Plugging in the values, we get SE=1145016.10SE = \frac{114}{\sqrt{50}} \approx 16.10.
So, our standard error is approximately **16.10**.

STEP 10

The **z-score** tells us how many standard errors our sample mean is away from the historical mean.
A bigger z-score means a bigger difference, and a bigger difference makes it more likely the training program actually worked!

STEP 11

The formula for the z-score is z=xˉμSEz = \frac{\bar{x} - \mu}{SE}, where xˉ\bar{x} is the **sample mean** (xˉ=550\bar{x} = 550), μ\mu is the **population mean** (μ=518\mu = 518), and SESE is the **standard error** we just calculated (SE16.10SE \approx 16.10).

STEP 12

Plugging in the values, we get z=55051816.103216.101.99z = \frac{550 - 518}{16.10} \approx \frac{32}{16.10} \approx 1.99.
Our z-score is approximately **1.99**!

STEP 13

The **p-value** is the probability of getting a sample mean as extreme as ours (or even more extreme) *if the null hypothesis is true*.
A small p-value means it's unlikely the improvement was just random chance.

STEP 14

Since our z-score is approximately 1.99, we need to find the area to the *right* of 1.99 in the standard normal distribution table (because it's a one-tailed test).
This gives us a p-value of approximately **0.02**.

STEP 15

Our p-value (0.02) is *less* than our significance level (0.05).
This means we *reject* the null hypothesis!

STEP 16

The test statistic is **1.99**.

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