A random sample of 109 teams playing in major international team sporting competitions has been selected to study corporate sponsorship and subsequent sporting success. Some of the variables recorded for each team are listed below.

Variable	Description
subid	Subject ID
sex	sex of players
numSponsors	number of distinct corporate sponsors
percentWin	the percentage of international games won
Address the following questions, assuming as necessary that all test assumptions are verified.
Part 1
A test is run to answer the research question, ’Is the average number of corporate sponsors equal to 38?’
Which test is correct to use, if we know the population standard deviation?	Answer 1 Question 1
 
The test statistic is -1.92. This means that the difference between the sample mean and the possible population mean, as stated in the null hypothesis (i.e. 38), is:	Answer 2 Question 1
 
The p-value of the test is 0.058. What is the decision of the test?

### Part 1:
The research question is whether the average number of corporate sponsors is equal to 38. 

**Answer 1 (Question 1):**
- The correct test to use, given that the population standard deviation is known, is a **z-test for a single mean**. A z-test is appropriate when:
  1. We know the population standard deviation.
  2. The sample size is reasonably large (which is the case with 109 teams).

**Answer 2 (Question 1):**
- The test statistic of \(-1.92\) indicates how many standard deviations the sample mean is away from the population mean (38). Since the test statistic is negative, it means the sample mean is **1.92 standard deviations less than** the hypothesized population mean of 38.

**p-value and Decision:**
- The p-value is **0.058**. This value represents the probability of observing a test statistic as extreme as \(-1.92\) (or more extreme) under the null hypothesis.

  **Decision:**
  - Typically, if the significance level (\(\alpha\)) is set at **0.05**, the p-value of **0.058** is **greater** than this threshold.
  - Therefore, the decision would be to **fail to reject the null hypothesis**. This means there is not enough evidence to conclude that the average number of corporate sponsors is different from 38.

Would you like more details on the reasoning process or any specific concept?

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Here are 5 questions to expand on this:
1. What are the key differences between a z-test and a t-test for a single mean?
2. How does sample size impact the choice of statistical test?
3. Can you explain the concept of the null and alternative hypotheses in hypothesis testing?
4. What would happen to the decision if the significance level were set to 0.10 instead of 0.05?
5. How does the p-value provide information about the strength of evidence against the null hypothesis?

**Tip:** When interpreting a p-value, it’s important to remember that it doesn’t prove or disprove a hypothesis. Instead, it provides a measure of the evidence against the null hypothesis.

A random sample of 109 teams playing in major international team sporting competitions has been selected to study corporate sponsorship and subsequent sporting success. Some of the variables recorded for each team are listed below.

Variable	Description
subid	Subject ID
sex	sex of players
numSponsors	number of distinct corporate sponsors
percentWin	the percentage of international games won

Address the following questions, assuming as necessary that all test assumptions are verified.

Part 1
A test is run to answer the research question, 'Is the average number of corporate sponsors equal to 38?' Which test is correct to use, if we know the population standard deviation?

The test statistic is -1.92. This means that the difference between the sample mean and the possible population mean, as stated in the null hypothesis (i.e., 38), is:

The p-value of the test is 0.058. What is the decision of the test?

Explore a hypothesis test to determine if the average number of corporate sponsors is equal to 38 for international sports teams. With a sample size of 109, a Z-test for a single mean is performed. Understand the implications of a test statistic of -1.92 and a p-value of 0.058 in relation to the null hypothesis.

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