2.2 Shilunga’s weight loss centre provided a loss for 72% of its participants. Shilunga decides to test a new weight loss programme on a random sample of 140 participants and found that 91 participants lost weight. Should the centre continue using its new programme? Test an appropriate hypothesis using 1% level of significance. [8]

To determine whether Shilunga's weight loss center should continue using the new program, we can perform a hypothesis test for a proportion. Here's how to approach the problem:

### 1. **State the Hypotheses**

The center currently has a success rate of 72%, which is the proportion of participants who lose weight under the old program. We need to test if the new program's success rate is significantly different from 72%.

- **Null Hypothesis (H₀):** The success rate of the new program is 72%. \( p = 0.72 \)
- **Alternative Hypothesis (H₁):** The success rate of the new program is different from 72%. \( p \neq 0.72 \)

This is a two-tailed test since we are testing for any difference (either increase or decrease) in the success rate.

### 2. **Select the Significance Level**

The significance level \( \alpha \) is given as 1%, or 0.01.

### 3. **Calculate the Sample Proportion (\(\hat{p}\))**

The sample proportion \( \hat{p} \) is the proportion of participants in the sample who lost weight:

\[
\hat{p} = \frac{\text{Number of successes}}{\text{Sample size}} = \frac{91}{140} \approx 0.65
\]

### 4. **Determine the Test Statistic**

For a proportion test, the test statistic is calculated using the formula:

\[
z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}
\]

Where:
- \( p_0 = 0.72 \) (the proportion under the null hypothesis)
- \( \hat{p} \approx 0.65 \)
- \( n = 140 \)

Substituting the values:

\[
z = \frac{0.65 - 0.72}{\sqrt{\frac{0.72 \times 0.28}{140}}} = \frac{-0.07}{\sqrt{\frac{0.2016}{140}}} = \frac{-0.07}{\sqrt{0.00144}} = \frac{-0.07}{0.03795} \approx -1.844
\]

### 5. **Determine the Critical Value**

For a two-tailed test at the 1% significance level, the critical z-values are approximately:

\[
\text{Critical value} = \pm 2.576
\]

This means that if the calculated z-value is less than -2.576 or greater than 2.576, we would reject the null hypothesis.

### 6. **Make the Decision**

- If the calculated z-value is within the range \([-2.576, 2.576]\), we fail to reject the null hypothesis.
- If the calculated z-value falls outside this range, we reject the null hypothesis.

Here, the calculated \( z \approx -1.844 \), which is within the range \([-2.576, 2.576]\). Therefore, we **fail to reject the null hypothesis**.

### 7. **Conclusion**

At the 1% significance level, there is not enough evidence to suggest that the success rate of the new program is different from the 72% success rate of the old program. Based on this test, the center may not have sufficient reason to switch to or continue using the new program.

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Would you like a more detailed breakdown or have any questions?

### Related Questions:
1. What would be the effect of using a 5% significance level instead of 1%?
2. How does sample size affect the reliability of this test?
3. What if the new program’s success rate was exactly 72%, how would that affect the outcome?
4. Could a different test be used to compare the two programs, and if so, which one?
5. How can the center improve the power of this test?

### Tip:
When interpreting the results of hypothesis tests, consider both the statistical significance and the practical significance of the findings, especially in real-world decision-making contexts.

Explore the hypothesis testing for proportions using Shilunga's weight loss center case study. Analyze whether the new weight loss program should be continued based on statistical significance at 1% level. Details include calculation of z-test statistic, sample proportion, and decision criteria.

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