A math teacher claims that she has developed a review course that increases the scores of students on the math portion of a college entrance exam. Based on data from the administrator of the exam, scores are normally distributed with u=524. The teacher obtains a random sample of 2000 students, puts them through the review class, and finds that the mean math score of the 2000 students is 532 with a standard deviation of 117. Complete parts (a) through (d) below. (a) state the full and alterative hypoteses. Ho H = 524, H₁: H > 524 (b) Test the hypothesis at the a=0.10 level of significance. Is a mean math score of 532 statistically significantly higher than 524? Conduct a hypothesis test using the P-value approach. Find the test statistic. to = 0 (Round to two decimal places as needed.)

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**Hypothesis Testing: Evaluating the Effectiveness of a Review Course**

A math teacher claims that she has developed a review course that increases the scores of students on the math portion of a college entrance exam. Based on data from the administrator of the exam, scores are normally distributed with a population mean (μ) of 524. The teacher obtains a random sample of 2000 students, puts them through the review class, and finds that the mean math score of the 2000 students is 532 with a standard deviation of 117. Complete parts (a) through (d) below.

### (a) State the null and alternative hypotheses:

- Null Hypothesis (H₀): μ = 524
- Alternative Hypothesis (H₁): μ > 524

### (b) Test the hypothesis at the α = 0.10 level of significance. Is a mean math score of 532 statistically significantly higher than 524? Conduct a hypothesis test using the P-value approach.

- Find the test statistic (t₀):
  \[ t₀ = \]
  (Round to two decimal places as needed.)

### Explanation of Process:

1. **Null and Alternative Hypotheses:**
   - **H₀:** μ = 524 (The mean score after the review course is equal to the general mean score).
   - **H₁:** μ > 524 (The mean score after the review course is greater than the general mean score, indicating the course is effective).

2. **Level of Significance:**
   - α = 0.10, which represents a 10% risk of concluding that a difference exists when there is no actual difference.

3. **Test Statistic:**
   - Calculated using the formula for the test statistic in hypothesis testing for the mean with a known standard deviation:
     \[
     t₀ = \frac{\bar{x} - \mu}{\left(\frac{\sigma}{\sqrt{n}}\right)}
     \]
     where:
     - \( \bar{x} \) is the sample mean (532),
     - \( \mu \) is the population mean (524),
     - \( \sigma \) is the population standard deviation (117),
     - \( n \) is the sample size (2000).

Given the sample mean (532), population mean (524), population standard deviation (117), and sample size (2000), calculate
Transcribed Image Text:**Hypothesis Testing: Evaluating the Effectiveness of a Review Course** A math teacher claims that she has developed a review course that increases the scores of students on the math portion of a college entrance exam. Based on data from the administrator of the exam, scores are normally distributed with a population mean (μ) of 524. The teacher obtains a random sample of 2000 students, puts them through the review class, and finds that the mean math score of the 2000 students is 532 with a standard deviation of 117. Complete parts (a) through (d) below. ### (a) State the null and alternative hypotheses: - Null Hypothesis (H₀): μ = 524 - Alternative Hypothesis (H₁): μ > 524 ### (b) Test the hypothesis at the α = 0.10 level of significance. Is a mean math score of 532 statistically significantly higher than 524? Conduct a hypothesis test using the P-value approach. - Find the test statistic (t₀): \[ t₀ = \] (Round to two decimal places as needed.) ### Explanation of Process: 1. **Null and Alternative Hypotheses:** - **H₀:** μ = 524 (The mean score after the review course is equal to the general mean score). - **H₁:** μ > 524 (The mean score after the review course is greater than the general mean score, indicating the course is effective). 2. **Level of Significance:** - α = 0.10, which represents a 10% risk of concluding that a difference exists when there is no actual difference. 3. **Test Statistic:** - Calculated using the formula for the test statistic in hypothesis testing for the mean with a known standard deviation: \[ t₀ = \frac{\bar{x} - \mu}{\left(\frac{\sigma}{\sqrt{n}}\right)} \] where: - \( \bar{x} \) is the sample mean (532), - \( \mu \) is the population mean (524), - \( \sigma \) is the population standard deviation (117), - \( n \) is the sample size (2000). Given the sample mean (532), population mean (524), population standard deviation (117), and sample size (2000), calculate
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