Math / Data & Statistics

Question1. SAT scores for High School $A$ are approximately normally distrio deviation 105. SAT scores for High School B are approximately normally distributed with mean 1200 and standard deviation 125. A random sample of 15 SAT scores from High School A is selected and the sample mean $\bar{x}_{A}$ is calculated. A separate random sample of 12 SAT scores from High School B is selected and the sample mean $\bar{x}_{B}$ is calculated. a. Describe the shape of the sampling distribution of $\bar{x}_{A}-\bar{x}_{B}$ . Justify. b. What is the mean of the sampling distribution of $\bar{x}_{A}-\bar{x}_{B}$ ? Interpret c. What is the standard deviation of the sampling distribution of $\bar{x}_{A}-\bar{x}_{B}$ ? Be sure to check the $10 \%$ condition. d. What is the probability that the sample mean of High School $A$ is greater than the sample mean from High School B? Wordle is a popular online game in the New York Times. To win, you must correctly guess a 5-letter word in 6 attempts or less. The hardest word to date was JAZZY. For all people that correctly solved JAZZY, the mean was 5.82 attempts with a standard deviation of 0.56 attempts. a. Can we assume the population shape for number of attempts to correctly solve JAZZY is approximately normal? Why or why not? b. Separate random samples of 32 AP Statistics students and 45 AP Calculus students who correctly solved JAZZY will be selected, and the average number of attempts to solve will be calculated for each sample ( $\bar{x}_{S}$ and $\bar{x}_{C}$ ). Assuming both groups have a mean of 5.82 attempts and a standard deviation of 0.56 attempts, describe the sampling distribution of $\bar{x}_{S}-\bar{x}_{C}$ . c. Find $P\left(\bar{x}_{S}-\bar{x}_{C} \leq-0.3\right)$ . d. Suppose the sample mean for the AP Statistics students is 5.1 attempts and the sample mean for the AP Calculus students is 5.4 attempts. Does this result provide convincing evidence that AP Statistics students can correctly guess the word JAZZY in fewer attempts than AP Calculus students, on average?

Studdy Solution

STEP 1

What is this asking? We're comparing SAT scores and Wordle attempts between two groups, using fancy statistical tools to see how likely one group is to outperform the other! Watch out! Don't mix up the formulas for standard deviation of a sample mean versus the difference of two sample means.
Also, remember to check conditions before making conclusions!

STEP 2

1. SAT Score Comparison
2. Wordle Attempts Comparison

STEP 3

Describe the shape: The sampling distribution of $\bar{x}_{A} - \bar{x}_{B}$ will be approximately normal.
This is because both individual distributions are approximately normal, and the difference of two independent normal distributions is also normal!

STEP 4

Calculate the mean: The mean of the sampling distribution is simply the difference of the individual means. $\mu_{\bar{x}_{A} - \bar{x}_{B}} = \mu_{A} - \mu_{B} = 1150 - 1200 = \mathbf{-50}$ .
This means we expect the sample mean of School A to be 50 points lower than School B, on average.

STEP 5

Check the 10% condition and calculate the standard deviation: We need to check if our samples are less than 10% of their respective populations.
Assuming each school has more than $15 \cdot 10 = 150$ and $12 \cdot 10 = 120$ students respectively, which is very likely, the 10% condition is met.
Now, we can calculate the standard deviation: $\sigma_{\bar{x}_{A} - \bar{x}_{B}} = \sqrt{\frac{\sigma_{A}^2}{n_{A}} + \frac{\sigma_{B}^2}{n_{B}}} = \sqrt{\frac{105^2}{15} + \frac{125^2}{12}} = \sqrt{735 + 1302.083} \approx \sqrt{2037.083} \approx \mathbf{45.134}$

STEP 6

Calculate the probability: We want to find $P(\bar{x}_{A} > \bar{x}_{B})$ , which is the same as $P(\bar{x}_{A} - \bar{x}_{B} > 0)$ .
We standardize our difference: $z = \frac{0 - (-50)}{45.134} \approx \frac{50}{45.134} \approx 1.108$ Using a z-table or calculator, $P(Z > 1.108) \approx \mathbf{0.134}$ .
So, there's about a 13.4% chance that the sample mean from School A is higher than School B.

STEP 7

Assess normality: We can't definitively say the population distribution is normal just because we have a sample mean and standard deviation.
We'd need more information about the overall distribution shape.

STEP 8

Describe the sampling distribution: Assuming the means and standard deviations are correct for both groups, the sampling distribution of $\bar{x}_{S} - \bar{x}_{C}$ will be approximately normal (due to the Central Limit Theorem, since sample sizes are large enough) with a mean of $5.82 - 5.82 = \mathbf{0}$ and a standard deviation of: $\sqrt{\frac{0.56^2}{32} + \frac{0.56^2}{45}} \approx \sqrt{0.0098 + 0.007} \approx \sqrt{0.0168} \approx \mathbf{0.1296}$

STEP 9

Calculate the probability: We want $P(\bar{x}_{S} - \bar{x}_{C} \leq -0.3)$ .
Standardizing: $z = \frac{-0.3 - 0}{0.1296} \approx -2.315$ Using a z-table or calculator, $P(Z \leq -2.315) \approx \mathbf{0.0103}$ .
There's a tiny 1.03% chance of observing this difference if the true means are equal.

STEP 10

Analyze the evidence: The observed difference in sample means (5.1 - 5.4 = -0.3) is quite unlikely if there's no real difference between the groups.
This provides some evidence that Stats students might be slightly faster at Wordle, but we'd need more formal hypothesis testing to be sure!

STEP 11

For the SAT scores, there's a 13.4% chance the sample mean from School A is higher than School B.
The sampling distribution of the difference has a mean of -50 and a standard deviation of 45.134.
For Wordle, assuming equal means, there's a 1.03% chance of observing a difference of -0.3 or less between the sample means.
The observed difference provides some evidence that Stats students might be faster, but further testing is needed.

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Studdy Solution

STEP 1

STEP 2

1. SAT Score Comparison
2. Wordle Attempts Comparison

STEP 3

Describe the shape: The sampling distribution of $\bar{x}_{A} - \bar{x}_{B}$ will be approximately normal.
This is because both individual distributions are approximately normal, and the difference of two independent normal distributions is also normal!

STEP 4

Calculate the mean: The mean of the sampling distribution is simply the difference of the individual means. $\mu_{\bar{x}_{A} - \bar{x}_{B}} = \mu_{A} - \mu_{B} = 1150 - 1200 = \mathbf{-50}$ .
This means we expect the sample mean of School A to be 50 points lower than School B, on average.

STEP 5

STEP 6

STEP 7

Assess normality: We can't definitively say the population distribution is normal just because we have a sample mean and standard deviation.
We'd need more information about the overall distribution shape.

STEP 8

STEP 9

Calculate the probability: We want $P(\bar{x}_{S} - \bar{x}_{C} \leq -0.3)$ .
Standardizing: $z = \frac{-0.3 - 0}{0.1296} \approx -2.315$ Using a z-table or calculator, $P(Z \leq -2.315) \approx \mathbf{0.0103}$ .
There's a tiny 1.03% chance of observing this difference if the true means are equal.

STEP 10

Analyze the evidence: The observed difference in sample means (5.1 - 5.4 = -0.3) is quite unlikely if there's no real difference between the groups.
This provides some evidence that Stats students might be slightly faster at Wordle, but we'd need more formal hypothesis testing to be sure!

STEP 11

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Studdy Solution

STEP 1

STEP 2

1. SAT Score Comparison2. Wordle Attempts Comparison

STEP 3

**Describe the shape:** The sampling distribution of xˉA−xˉB\bar{x}_{A} - \bar{x}_{B}xˉA​−xˉB​ will be approximately normal.This is because both individual distributions are approximately normal, and the difference of two independent normal distributions is also normal!

STEP 4

STEP 5

STEP 6

STEP 7

**Assess normality:** We can't definitively say the population distribution is normal just because we have a sample mean and standard deviation.We'd need more information about the overall distribution shape.

STEP 8

STEP 9

STEP 10

**Analyze the evidence:** The observed difference in sample means (5.1 - 5.4 = -0.3) is quite unlikely if there's no real difference between the groups.This provides some evidence that Stats students might be slightly faster at Wordle, but we'd need more formal hypothesis testing to be sure!

STEP 11

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. SAT Score Comparison
2. Wordle Attempts Comparison

Describe the shape: The sampling distribution of $\bar{x}_{A} - \bar{x}_{B}$ will be approximately normal.
This is because both individual distributions are approximately normal, and the difference of two independent normal distributions is also normal!

Assess normality: We can't definitively say the population distribution is normal just because we have a sample mean and standard deviation.
We'd need more information about the overall distribution shape.

Analyze the evidence: The observed difference in sample means (5.1 - 5.4 = -0.3) is quite unlikely if there's no real difference between the groups.
This provides some evidence that Stats students might be slightly faster at Wordle, but we'd need more formal hypothesis testing to be sure!