https://file.aiquickdraw.com/mathbot/file/17298139398982aazkhyx.jpg

Let's go through the questions from the image step by step.

### a. What type of test will be used in this problem?
- **Answer**: This is a **one-sample t-test**, since we are comparing the sample mean (37.6 hours) to the claimed population mean (37 hours) with an unknown population standard deviation.

### b. What evidence justifies the use of this test?
Here are the conditions that apply:
- **The sample size is larger than 30** (since n = 323).
- **The population standard deviation is not known** (we only have the sample standard deviation).
  
### c. Enter the null hypothesis for this test.
- \( H_0: \mu = 37 \)
  
This is the claim made by the internet service provider, i.e., the population mean is 37 hours per week.

### d. Enter the alternative hypothesis for this test.
- \( H_1: \mu \neq 37 \)
  
The competitor is testing whether the actual average is different from 37, making this a two-tailed test.

### e. Is the original claim located in the null or alternative hypothesis?
- **Answer**: The original claim (that users averaged 37 hours per week) is located in the **null hypothesis**.

### f. What is the test statistic for the given statistics?
- To compute the test statistic, we use the formula for the t-statistic:
  
  \[
  t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}
  \]

  Where:
  - \( \bar{x} = 37.6 \) (sample mean)
  - \( \mu = 37 \) (claimed mean)
  - \( s = 5.6 \) (sample standard deviation)
  - \( n = 323 \) (sample size)
  
  Plugging in these values:

  \[
  t = \frac{37.6 - 37}{\frac{5.6}{\sqrt{323}}} = \frac{0.6}{\frac{5.6}{17.97}} = \frac{0.6}{0.3116} \approx 1.925
  \]

### g. What is the p-value for this test?
- The degrees of freedom (df) is \( n - 1 = 323 - 1 = 322 \).
  
  Using a t-distribution table or calculator for \( t = 1.925 \) and df = 322, the p-value for a two-tailed test would be approximately **0.055**.

### h. What is the decision based on the given statistics?
- **Answer**: Since \( \alpha = 0.1 \) and the p-value is 0.055, we **fail to reject the null hypothesis**. The p-value is less than the significance level \( \alpha \), but it's not enough evidence to reject the null hypothesis.

### i. What is the correct interpretation of this decision?
- **Answer**: Using a **10% level of significance**, there **is not sufficient evidence** to reject the claim that users averaged 37 hours per week.

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Do you have any questions or need further clarification? Here are some related questions that can extend the discussion:

1. What would happen if the sample size was smaller than 30?
2. Why is the t-test used instead of a z-test in this case?
3. What would change if the population standard deviation was known?
4. How would the test change if this was a one-tailed test?
5. What are the implications of choosing a significance level of 0.1 instead of 0.05?

**Tip**: Always check the assumptions for the test you're conducting (e.g., normality, sample size) before calculating the test statistic!

An internet service provider claimed that its users averaged 37 hours per week. To test this claim, a competitor conducted a survey of 323 customers and found that the average time spent online was 37.6 hours per week with a standard deviation of 5.6 hours per week. Test the claim using α = 0.1.

Understand how to conduct a one-sample t-test to evaluate if the average time spent online by users differs from 37 hours per week. Using a sample size of 323 customers, this problem covers key statistical concepts including hypothesis testing, t-statistics, and p-values. Suitable for college-level statistics students.

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