Question

In a recent study, 82 patients with type 2 diabetes underwent a procedure, and 59 recovered. Does this study provide convincing evidence that greater than 58% will recover? a=.10
What are the null and alternative hypothesis

Answer 1

Compute the test statistics

Answer 2

Compute the p value

Answer 3

To determine if there is convincing evidence that greater than 58% of patients will recover, we can formulate the null and alternative hypotheses as follows:

**Null Hypothesis (\(H_0\))**: The proportion of patients who recover is less than or equal to 58%.
Mathematically, this can be stated as:
\[ H_0: p \leq 0.58 \]

**Alternative Hypothesis (\(H_a\))**: The proportion of patients who recover is greater than 58%.
Mathematically, this can be stated as:
\[ H_a: p > 0.58 \]

Given these hypotheses, you can perform a statistical test (such as a one-sample z-test for proportions) to analyze the data and draw conclusions regarding the population recovery rate based on your sample.

Answer 4

To compute the test statistic for the hypothesis test regarding the proportion of patients who recover, we can use the formula for the z-test for proportions.

1. **Determine the sample proportion (\( \hat{p} \))**:
\[
\hat{p} = \frac{x}{n}
\]
where \( x \) is the number of successes (recoveries) and \( n \) is the total number of trials (patients).
- In this case, \( x = 59 \) (patients who recovered) and \( n = 82 \) (total patients).
\[
\hat{p} = \frac{59}{82} \approx 0.7195
\]

2. **Specify the claimed proportion in null hypothesis**:
\[
p_0 = 0.58
\]

3. **Calculate the standard error (SE) of the proportion**:
\[
SE = \sqrt{\frac{p_0(1 - p_0)}{n}}
\]
\[
SE = \sqrt{\frac{0.58(1 - 0.58)}{82}} = \sqrt{\frac{0.58 \times 0.42}{82}} \approx \sqrt{\frac{0.2436}{82}} \approx \sqrt{0.00297} \approx 0.0545
\]

4. **Calculate the z-test statistic**:
\[
z = \frac{\hat{p} - p_0}{SE}
\]
\[
z = \frac{0.7195 - 0.58}{0.0545} \approx \frac{0.1395}{0.0545} \approx 2.56
\]

Thus, the test statistic \( z \) is approximately **2.56**.

You can now use this z-value to compare against the critical value or to find the p-value for your hypothesis test at the \( \alpha = 0.10 \) significance level.

Answer 5

To compute the p-value for the test statistic \( z \approx 2.56 \), we can use standard normal distribution tables or statistical software/calculators.

Since we are conducting a one-tailed test (testing if the proportion is greater than 0.58), we want to find the area to the right of \( z = 2.56 \).

1. **Find the cumulative probability for \( z = 2.56 \)**:
Using a standard normal distribution table or calculator, we find \( P(Z < 2.56) \).

The cumulative probability for \( z = 2.56 \) is approximately:
\[
P(Z < 2.56) \approx 0.9948
\]

2. **Calculate the p-value**:
The p-value is the area to the right of \( z = 2.56 \):
\[
p\text{-value} = 1 - P(Z < 2.56) = 1 - 0.9948 = 0.0052
\]

Thus, the p-value is approximately **0.0052**.

### Interpretation:
Since the p-value (0.0052) is less than the significance level \( \alpha = 0.10 \), you would reject the null hypothesis. This provides convincing evidence that the proportion of patients who recover is greater than 58%.