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There is a study on two different approaches to improve the hand function of st...
Nov 29, 2024
Solution by Steps
step 1
To test if more than 65% of stroke patients have significant hand function improvement in hospital settings, we set up the null hypothesis H0:p1≤0.65 H_0: p_1 \leq 0.65 and the alternative hypothesis H_a: p_1 > 0.65 . The sample proportion is p1^=6590≈0.7222 \hat{p_1} = \frac{65}{90} \approx 0.7222
step 2
The standard error (SE) is calculated using the formula SE=p0(1−p0)n SE = \sqrt{\frac{p_0(1 - p_0)}{n}} , where p0=0.65 p_0 = 0.65 and n=90 n = 90 . Thus, SE=0.65(1−0.65)90≈0.0514 SE = \sqrt{\frac{0.65(1 - 0.65)}{90}} \approx 0.0514
step 3
The test statistic z z is calculated using z=p1^−p0SE z = \frac{\hat{p_1} - p_0}{SE} . Substituting the values, we get z=0.7222−0.650.0514≈1.40 z = \frac{0.7222 - 0.65}{0.0514} \approx 1.40
step 4
Using the standard normal distribution, we find the p-value corresponding to z=1.40 z = 1.40 . The p-value is approximately 0.0808 0.0808 . Since 0.0808 > 0.05 , we fail to reject the null hypothesis
step 5
Therefore, there is not enough evidence to support the claim that more than 65% of stroke patients have significant hand function improvement in hospital settings
1 Answer
B
Solution by Steps
step 1
Let p1 p_1 be the proportion of patients with significant improvement in hospital settings and p2 p_2 for telehealth. The null hypothesis is H0:p1=p2 H_0: p_1 = p_2 and the alternative hypothesis is Ha:p1≠p2 H_a: p_1 \neq p_2
step 2
The pooled proportion p^ \hat{p} is calculated as p^=65+4590+80=110170≈0.6471 \hat{p} = \frac{65 + 45}{90 + 80} = \frac{110}{170} \approx 0.6471 . The standard error for the difference in proportions is SE=p^(1−p^)(1n1+1n2) SE = \sqrt{\hat{p}(1 - \hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}
step 3
Substituting the values, we find SE≈0.6471(1−0.6471)(190+180)≈0.0865 SE \approx \sqrt{0.6471(1 - 0.6471)\left(\frac{1}{90} + \frac{1}{80}\right)} \approx 0.0865
step 4
The test statistic z z is calculated as z=p1^−p2^SE=0.7222−0.56250.0865≈1.85 z = \frac{\hat{p_1} - \hat{p_2}}{SE} = \frac{0.7222 - 0.5625}{0.0865} \approx 1.85 . The corresponding p-value is approximately 0.0643 0.0643
step 5
Since 0.0643 > 0.05 , we fail to reject the null hypothesis. Thus, there is not enough evidence to conclude that there is a significant difference between p1 p_1 and p2 p_2
2 Answer
B
Solution by Steps
step 1
Type I error occurs when we reject the null hypothesis when it is actually true. In this context, it would mean concluding that there is a significant difference in proportions when there is none
step 2
The probability of Type I error is denoted by α \alpha . In this case, if we decide to reject the null hypothesis, α=0.05 \alpha = 0.05
3 Answer
B
Solution by Steps
step 1
The 95% confidence interval for p1−p2 p_1 - p_2 is given as (0.017,0.302) (0.017, 0.302) . Since the interval does not include 0, it indicates that there is a significant difference between the two proportions
step 2
Therefore, we conclude that the population proportion p1 p_1 (hospital settings) is larger than p2 p_2 (telehealth)
4 Answer
A
Key Concept
Hypothesis Testing and Confidence Intervals
Explanation
Hypothesis testing helps determine if there is enough evidence to support a claim, while confidence intervals provide a range of values for estimating population parameters.
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