b. False
nformation for how many patients took each drug or combination of drugs is summarized below in the two tables. Use these to answer questions a) -d)
Table 1. Summary of performance of drug A: UTI rates among those taking and not taking drug ADid not take Drug ADid take Drug A
total UTI
759
887
164 No UTI
441 312 753Total 1200 1200
2400Table 2. Summary of performance of drug B and C: recovery status after 1 week of taking medications.
Did not take Drug A
Did take Drug A
Drug B
Drug C
Drug B
Drug C
Recovered
191
209
221
244
Not Recovered
189
170
223
199
Total
380
379
444
443
a. Use the above Table 1 to determine if Drug A was useful in preventing UTIs. In other words, is the proportion of those having taking Drug A but still getting a UTI equal to average rate of UTI for this population (living in an assisted living home) of 74%. Use hypothesis testing to test our hypothesis and use the confidence interval approach with a significance level of α=0.01.
b. Using Table 2, let’s examine the rate of UTI recovery among Drug C (conventional antibiotics). The manufacturer of Drug C claims it has a success rate (recovery within a week) of 55%. Use our data to see if this success rate is true: test if our recovery rate of those taking Drug C, regardless of whether the person took Drug A or not, is the same or different than 55%. Use hypothesis testing and the p-value approach with an α=0.05.
c. Similarly, let’s examine Drug B’s performance. Repeat our hypothesis among Drug B: test if our recovery rate of those taking drug B is different than 55% (regardless of whether the patient took Drug A or not). Use hypothesis testing and p-value approach with an α=0.1.
Answer:
(View Below)
Explanation:
Let's tackle each part of the question step by step:
a. **Testing the Effectiveness of Drug A:**
We want to test if the proportion of patients who took Drug A and still got a UTI is equal to the average rate of UTIs for this population (74%). We can use a hypothesis test for proportions. Here are the hypotheses:
- **Null Hypothesis (H0):** The proportion of patients who took Drug A and got a UTI is equal to 74%.
- **Alternative Hypothesis (Ha):** The proportion of patients who took Drug A and got a UTI is not equal to 74%.
We'll perform a two-tailed test at a significance level of α = 0.01.
Using the provided data:
- Proportion of UTIs among those who took Drug A = 887 / 1200 ≈ 0.7392
- Proportion of UTIs among those who did not take Drug A = 759 / 1200 ≈ 0.6325
We can calculate the standard error for the difference in proportions and perform the hypothesis test. I'll calculate the z-score and p-value for you:
Z = (0.7392 - 0.6325) / √[0.6325 * (1 - 0.6325) / 1200] ≈ 2.8413
Now, looking up the z-score in a standard normal distribution table, we find the critical values for a two-tailed test at α = 0.01 to be approximately ±2.576.
Since our calculated z-score (2.8413) is greater than the critical value (2.576), we can reject the null hypothesis.
Therefore, there is evidence to suggest that Drug A is useful in preventing UTIs because the proportion of patients who took Drug A and still got a UTI is significantly different from the average rate of UTIs for this population.
b. **Testing the Recovery Rate of Drug C:**
We want to test if the recovery rate for Drug C is different from the claimed success rate of 55%. We can use a hypothesis test for proportions. Here are the hypotheses:
- **Null Hypothesis (H0):** The recovery rate of those taking Drug C is equal to 55%.
- **Alternative Hypothesis (Ha):** The recovery rate of those taking Drug C is different from 55%.
We'll perform a two-tailed test at a significance level of α = 0.05.
Using the provided data:
- Proportion of recovery among those taking Drug C = (221 + 244) / 443 ≈ 0.9955
We can calculate the standard error for the proportion and perform the hypothesis test. I'll calculate the z-score and p-value for you:
Z = (0.9955 - 0.55) / √[0.55 * (1 - 0.55) / 443] ≈ 18.3841
The critical values for a two-tailed test at α = 0.05 are approximately ±1.96.
Since our calculated z-score (18.3841) is much greater than the critical value (1.96), we can reject the null hypothesis.
Therefore, there is strong evidence to suggest that the recovery rate for Drug C is different from the claimed success rate of 55%.
c. **Testing the Recovery Rate of Drug B:**
We want to test if the recovery rate for Drug B is different from the claimed success rate of 55%. We'll perform a two-tailed test at a significance level of α = 0.1.
Using the provided data:
- Proportion of recovery among those taking Drug B = (221 + 244) / 444 ≈ 0.9919
We can calculate the standard error for the proportion and perform the hypothesis test. I'll calculate the z-score and p-value for you:
Z = (0.9919 - 0.55) / √[0.55 * (1 - 0.55) / 444] ≈ 17.7503
The critical values for a two-tailed test at α = 0.1 are approximately ±1.645.
Since our calculated z-score (17.7503) is much greater than the critical value (1.645), we can reject the null hypothesis.
Therefore, there is strong evidence to suggest that the recovery rate for Drug B is different from the claimed success rate of 55%.
B. how well rested the athlete is before competition.
C. the methods by which the athlete prepared him/herself for competition.
D. the type of equipment the athlete uses during competition.