The Ciguetera toxin is a kind of poisonous element that is found in the Barracuda fish.
This toxin is present in the Barracuda because it consumes other fishes that feeds on algae that produces such toxins
The poison has no effect on the fishes but it is very harmful to humans.
When ingested it can cause a person to have nausea, pain and other cardia related problems.
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A foodborne illness caused by a variety of reef fish such as barracuda, grouper, moray eel, amberjack, sea bass, and sturgeon is known as ciguatera fish poisoning, also known simply as ciguatera.
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:
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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%.
Answer:
Some of these concerns are HIV, transferred infections, health problems, and mentality issues :3
Explanation:
:3
Answer: Monitor your blood pressure over time
If you need to monitor the progress of your blood pressure, domiciliary self-measurement will give you a detailed profile over time. This is particularly important when it comes to assessing the efficacy of antihypertensive drugs prescribed by your doctor.
Explanation:
Hope I help!!❤️
a. True
b. False
The right option is; b. False
Strength training is a form of exercise that builds the strength, and size of skeletal muscles by using resistance to stimulate muscular contraction. Strength training perform many important beneficial functions in the body such as improved muscle, bone, tendon, and ligament, increased joint function, improved cardiac function and increased metabolism.
relax and twitch
tighten and contract
contract and relax
The answer is tighten and contract.
In order for the students to pull the rope their muscles will need to tighten and contract so they are able to continuously pull with all their strength and power. Relaxed or twitched muscles will result in release of the rope.
To pull the rope, the students' muscles must tighten and contract in order for them to continuouslypull with all of their strength and power. The rope will be released if your muscles are relaxed or twitched. The correct option is b.
A band or bundle of fibrous tissue in a human or animal body that can contract, causing movement or maintaining the position of body parts.
To pull the rope, the students' muscles must tighten and contract in order for them to continuouslypull with all of their strength and power. The rope will be released if your muscles are relaxed or twitched.
Thus, the correct option is b.
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