Question
Vertical banded gastroplasty is a surgical procedure that reduces the volume of the stomach in order to produce weight loss. In a recent study, 82 patients with Type 2 diabetes underwent this procedure, and 59 of them experienced a recovery from diabetes. Does this study provide convincing evidence that greater than 63% of those with diabetes who undergo this surgery will recover from diabetes? Use the a=0.10 level of significance and the P-value method.
Hv0 : [?]
Hv1 : [?]
Left/right or two tailed test?
Hv0 : [?]
Hv1 : [?]
Left/right or two tailed test?
Answers
Answer
(b) Compute the value of the test statistic. Round the answer to two decimal places.
Answer
No z=1.68
use this to compute the P-Value. round to at least 4 decimal places
use this to compute the P-Value. round to at least 4 decimal places
Answer
Do we reject Hv0?
Answer
e) State a conclusion.
There [is or isnt?] enough evidence to conclude that the percentage of people with diabetes who undergo the surgery and recover from diabetes is greater than 63%.
There [is or isnt?] enough evidence to conclude that the percentage of people with diabetes who undergo the surgery and recover from diabetes is greater than 63%.
Answered by
GPT-4o mini
To analyze whether greater than 63% of patients with Type 2 diabetes recover from diabetes after undergoing vertical banded gastroplasty, we need to set up the hypotheses.
### Hypotheses:
- **Null Hypothesis (H0)**: The proportion of patients who recover from diabetes, \( p \), is equal to 0.63 (i.e., \( p = 0.63 \)).
- **Alternative Hypothesis (H1)**: The proportion of patients who recover from diabetes, \( p \), is greater than 0.63 (i.e., \( p > 0.63 \)).
### Statement:
- \( H_0: p = 0.63 \)
- \( H_1: p > 0.63 \)
### Type of Test:
Since we are interested in whether the recovery rate is **greater than** 63%, this is a **one-tailed test** (specifically, a right-tailed test).
Once we have the hypotheses and the type of test defined, we can calculate the test statistic and determine the p-value to see if we can reject the null hypothesis at the alpha level of 0.10.
### Hypotheses:
- **Null Hypothesis (H0)**: The proportion of patients who recover from diabetes, \( p \), is equal to 0.63 (i.e., \( p = 0.63 \)).
- **Alternative Hypothesis (H1)**: The proportion of patients who recover from diabetes, \( p \), is greater than 0.63 (i.e., \( p > 0.63 \)).
### Statement:
- \( H_0: p = 0.63 \)
- \( H_1: p > 0.63 \)
### Type of Test:
Since we are interested in whether the recovery rate is **greater than** 63%, this is a **one-tailed test** (specifically, a right-tailed test).
Once we have the hypotheses and the type of test defined, we can calculate the test statistic and determine the p-value to see if we can reject the null hypothesis at the alpha level of 0.10.
Answered by
GPT-4o mini
To compute the test statistic for this proportion test, we'll use the formula for the z-test statistic for a single proportion:
\[
z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}
\]
Where:
- \(\hat{p}\) = sample proportion of successes (those who recovered)
- \(p_0\) = hypothesized population proportion under the null hypothesis
- \(n\) = sample size
From the information provided:
- The number of patients who recovered from diabetes, \(x = 59\)
- The total number of patients studied, \(n = 82\)
- The hypothesized proportion, \(p_0 = 0.63\)
First, we need to calculate the sample proportion \(\hat{p}\):
\[
\hat{p} = \frac{x}{n} = \frac{59}{82} \approx 0.7183
\]
Now we can compute the z-test statistic:
1. Calculate the standard deviation for the null hypothesis proportion:
\[
\sqrt{\frac{p_0(1 - p_0)}{n}} = \sqrt{\frac{0.63 \times (1 - 0.63)}{82}} = \sqrt{\frac{0.63 \times 0.37}{82}} = \sqrt{\frac{0.2331}{82}} \approx \sqrt{0.002840} \approx 0.0533
\]
2. Compute the z value:
\[
z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} = \frac{0.7183 - 0.63}{0.0533} \approx \frac{0.0883}{0.0533} \approx 1.60
\]
### Conclusion:
The value of the test statistic \(z\) is approximately **1.60** (rounded to two decimal places).
\[
z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}
\]
Where:
- \(\hat{p}\) = sample proportion of successes (those who recovered)
- \(p_0\) = hypothesized population proportion under the null hypothesis
- \(n\) = sample size
From the information provided:
- The number of patients who recovered from diabetes, \(x = 59\)
- The total number of patients studied, \(n = 82\)
- The hypothesized proportion, \(p_0 = 0.63\)
First, we need to calculate the sample proportion \(\hat{p}\):
\[
\hat{p} = \frac{x}{n} = \frac{59}{82} \approx 0.7183
\]
Now we can compute the z-test statistic:
1. Calculate the standard deviation for the null hypothesis proportion:
\[
\sqrt{\frac{p_0(1 - p_0)}{n}} = \sqrt{\frac{0.63 \times (1 - 0.63)}{82}} = \sqrt{\frac{0.63 \times 0.37}{82}} = \sqrt{\frac{0.2331}{82}} \approx \sqrt{0.002840} \approx 0.0533
\]
2. Compute the z value:
\[
z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} = \frac{0.7183 - 0.63}{0.0533} \approx \frac{0.0883}{0.0533} \approx 1.60
\]
### Conclusion:
The value of the test statistic \(z\) is approximately **1.60** (rounded to two decimal places).
Answered by
GPT-4o mini
To compute the p-value for the z-test statistic of \( z = 1.68 \), we can use a standard normal distribution table or a calculator to find the area to the right of the z-value because we are conducting a one-tailed test.
1. **Finding the p-value**: The p-value represents the probability of observing a test statistic as extreme as, or more extreme than, \( z = 1.68 \) under the null hypothesis \( H_0 \).
2. Using a standard normal distribution table or calculator:
- We can find the area to the left of \( z = 1.68 \) and subtract it from 1 to find the right-tail area.
- Looking up \( z = 1.68 \) in the Z-table gives us the value of approximately \( 0.9535 \) (the area to the left).
- To find the area to the right (the p-value):
\[
\text{p-value} = 1 - P(Z < 1.68) = 1 - 0.9535 \approx 0.0465
\]
### Conclusion:
The p-value is approximately **0.0465** (rounded to four decimal places).
1. **Finding the p-value**: The p-value represents the probability of observing a test statistic as extreme as, or more extreme than, \( z = 1.68 \) under the null hypothesis \( H_0 \).
2. Using a standard normal distribution table or calculator:
- We can find the area to the left of \( z = 1.68 \) and subtract it from 1 to find the right-tail area.
- Looking up \( z = 1.68 \) in the Z-table gives us the value of approximately \( 0.9535 \) (the area to the left).
- To find the area to the right (the p-value):
\[
\text{p-value} = 1 - P(Z < 1.68) = 1 - 0.9535 \approx 0.0465
\]
### Conclusion:
The p-value is approximately **0.0465** (rounded to four decimal places).
Answered by
GPT-4o mini
To determine whether to reject the null hypothesis (\( H_0 \)), you need to compare the p-value to the significance level (\( \alpha \)).
In this case:
- The p-value we calculated is approximately **0.0465**.
- The significance level \( \alpha \) is given as **0.10**.
### Decision Rule:
- If the p-value \( < \alpha \), we reject \( H_0 \).
- If the p-value \( \geq \alpha \), we fail to reject \( H_0 \).
### Conclusion:
Since \( 0.0465 < 0.10 \), we reject the null hypothesis \( H_0 \).
This provides statistically significant evidence to support the alternative hypothesis \( H_1 \): that greater than 63% of patients with Type 2 diabetes who undergo vertical banded gastroplasty will recover from diabetes.
In this case:
- The p-value we calculated is approximately **0.0465**.
- The significance level \( \alpha \) is given as **0.10**.
### Decision Rule:
- If the p-value \( < \alpha \), we reject \( H_0 \).
- If the p-value \( \geq \alpha \), we fail to reject \( H_0 \).
### Conclusion:
Since \( 0.0465 < 0.10 \), we reject the null hypothesis \( H_0 \).
This provides statistically significant evidence to support the alternative hypothesis \( H_1 \): that greater than 63% of patients with Type 2 diabetes who undergo vertical banded gastroplasty will recover from diabetes.
Answered by
GPT-4o mini
There **is** enough evidence to conclude that the percentage of people with diabetes who undergo the surgery and recover from diabetes is greater than 63%.
This conclusion is based on the results of the hypothesis test, where we rejected the null hypothesis in favor of the alternative hypothesis at the \( \alpha = 0.10 \) significance level.
This conclusion is based on the results of the hypothesis test, where we rejected the null hypothesis in favor of the alternative hypothesis at the \( \alpha = 0.10 \) significance level.