A manager at a local discount gym believes that less than 20% of gym members use the gym, at least 5 days a week. She randomly selects 100 gym members and tracks (using the electronic login system at the door) how many days they used the gym over the 2-week period. The following are the results:
2	3	10	4	2	3	8	4	8	10
5	0	6	3	9	13	6	3	12	5
3	3	5	1	5	9	8	5	8	2
6	4	4	2	12	1	3	3	2	12
7	3	14	2	8	5	2	6	1	5
6	9	6	8	10	1	11	3	2	1
5	4	1	2	3	13	7	4	8	3
7	4	3	2	10	3	1	7	11	8
4	7	6	7	8	11	7	6	3	2
5	0	4	6	5	12	2	10	1	2
Test the manager's claim at the 2.5% level of significance.
Standard Normal Distribution Table
a. Calculate the test statistic.
z=
 
Round to two decimal places if necessary
Enter 0 if normal approximation to the binomial cannot be used
b. Determine the critical value(s) for the hypothesis test.
 +
Round to two decimal places if necessary
Enter 0 if normal approximation to the binomial cannot be used
c. Conclude whether to reject the null hypothesis or not based on the test statistic.
Reject
Fail to Reject
Cannot Use Normal Approximation to Binomial

The manager wants to test the claim that **less than 20%** of the gym members use the gym at least 5 days a week. This is a hypothesis test for a proportion, which we'll solve using a **one-tailed z-test** for the proportion \( p \).

### Hypotheses:
- **Null hypothesis (H₀):** \( p = 0.20 \) (20% of gym members use the gym at least 5 days a week)
- **Alternative hypothesis (H₁):** \( p < 0.20 \) (Less than 20% of gym members use the gym at least 5 days a week)

### Step 1: Data Preparation
Let's first count how many members used the gym at least 5 days a week. We count any number greater than or equal to 5.

#### Members who used the gym at least 5 days in the 2-week period:
\[
\{10, 8, 8, 10, 5, 6, 9, 13, 6, 12, 5, 5, 9, 8, 5, 8, 6, 12, 7, 14, 8, 5, 6, 8, 10, 11, 5, 7, 6, 7, 8, 11, 7, 6, 5, 12, 7, 8, 5, 10, 12\}
\]
There are **40 members** who used the gym at least 5 days in the 2-week period (out of 100 members).

### Step 2: Sample Proportion \( \hat{p} \)
The sample proportion \( \hat{p} \) is calculated as:

\[
\hat{p} = \frac{40}{100} = 0.40
\]

### Step 3: Check Normal Approximation Condition
For the normal approximation to the binomial to hold, both \( np \geq 5 \) and \( n(1-p) \geq 5 \) should be satisfied, where \( n = 100 \) and \( p = 0.20 \).

\[
np = 100 \times 0.20 = 20 \quad \text{and} \quad n(1-p) = 100 \times 0.80 = 80
\]
Since both values are greater than 5, the normal approximation can be used.

### Step 4: Test Statistic
The formula for the z-test statistic is:

\[
z = \frac{\hat{p} - p}{\sqrt{\frac{p(1 - p)}{n}}}
\]
Where:
- \( \hat{p} = 0.40 \) (sample proportion)
- \( p = 0.20 \) (hypothesized proportion)
- \( n = 100 \)

Substitute the values:

\[
z = \frac{0.40 - 0.20}{\sqrt{\frac{0.20(1 - 0.20)}{100}}} = \frac{0.20}{\sqrt{\frac{0.20 \times 0.80}{100}}} = \frac{0.20}{\sqrt{\frac{0.16}{100}}} = \frac{0.20}{0.04} = 5.00
\]

So, the **test statistic** is \( z = 5.00 \).

### Step 5: Critical Value
For a one-tailed test at the 2.5% significance level (\( \alpha = 0.025 \)), the critical z-value can be found from standard normal distribution tables. For a left-tailed test:

\[
z_{\text{critical}} = -1.96
\]

### Step 6: Decision
Since the test statistic \( z = 5.00 \) is **greater than** the critical value \( z_{\text{critical}} = -1.96 \), we **fail to reject** the null hypothesis. 

### Conclusion:
There is not enough evidence to support the manager's claim that less than 20% of the members use the gym at least 5 days a week. The data suggests that 40% of the gym members use the gym at least 5 days a week.

---

Would you like further details or have any questions?

Here are 5 related questions for further exploration:
1. How does changing the sample size affect the test statistic?
2. What would the conclusion be if we used a different significance level?
3. How does a two-tailed test differ from a one-tailed test in this case?
4. What are Type I and Type II errors in the context of this hypothesis test?
5. How can we use confidence intervals to support the results of this hypothesis test?

**Tip:** Always ensure the sample size is large enough to justify the use of normal approximation for binomial tests.

A manager at a local discount gym believes that less than 20% of gym members use the gym at least 5 days a week. She randomly selects 100 gym members and tracks (using the electronic login system at the door) how many days they used the gym over the 2-week period. The following are the results: [data provided]. Test the manager's claim at the 2.5% level of significance.

This page covers a hypothesis test to determine if less than 20% of gym members use the gym at least 5 days a week. Using a sample of 100 members, a Z-test for proportions is applied, with a significance level of 2.5%. Learn about hypothesis testing, Z-statistics, and normal approximation to binomial distribution.

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