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3 years ago

What ethical issues have been raised by the scientific community's commitment to ensuring that experimental treatments are not

Mathematics

1 answer:

e-lub [12.9K]3 years ago

8 0

Answer:

They include;

1. Fewer chances of determining how effective the treatment plan would be.

2. Inability of every patient to access the experimental treatment.

3. Difficulties in making knowledgeable decisions on the treatment plan.

4. Determining that the experimental treatments are offered with the right motive.

Step-by-step explanation:

In medical treatment administration, it is standard practice that drugs undergo clinical trials on test animals before they are administered to patients. However, some sicknesses are without known drugs for treatment or may have drugs that are still undergoing experiments and trials. In such cases, patients may want to be treated with experimental drugs.

Ethical issues such as the above-listed can arise from this. The foremost of them all is the fact that the treatment might prove ineffective thus causing more problems to the patient.

You might be interested in

Calculate the sample standard deviation and sample variance for the following frequency distribution of hourly wages for a sampl

lidiya [134]

Answer:

(a) The sample variance is 16.51

(a) The sample standard deviation is 4.06

Step-by-step explanation:

Given

$\begin{array}{cc}{Class} & {Frequency} & 8.26 - 10.00 & 20 &10.01-11.75 & 38 &11.76 - 13.50& 36 & 13.51-15.25 &25&15.26-17.00 &27 &\ \end{array}$

Solving (a); The sample variance.

First, calculate the class midpoints.

This is the mean of the intervals.

i.e.

$x_1 = \frac{8.26+10.00}{2} = \frac{18.26}{2} = 9.13$

$x_2 = \frac{10.01+11.75}{2} = \frac{21.76}{2} = 10.88$

$x_3 = \frac{11.76+13.50}{2} = \frac{25.26}{2} = 12.63$

$x_4 = \frac{13.51+15.25}{2} = \frac{28.76}{2} = 14.38$

$x_5 = \frac{15.26+17.00}{2} = \frac{32.26}{2} = 16.13$

So, the table becomes:

$\begin{array}{ccc}{Class} & {Frequency} & {x} & 8.26 - 10.00 & 20&9.13 &10.01-11.75 & 38 &10.88&11.76 - 13.50& 36 &12.63& 13.51-15.25 &25&14.38&15.26-17.00 &27 &16.13\ \end{array}$

Next, calculate the mean

$\bar x = \frac{\sum fx}{\sum f}$

$\bar x = \frac{20*9.13 + 38 * 10.88+36*12.63+25*14.38+27*16.13}{20+38+36+25+27}$

$\bar x = \frac{1845.73}{146}$

$\bar x = 12.64$

Next, the sample variance is:

$\sigma^2 = \frac{\sum f(x - \bar x)^2}{\sum f - 1}$

So, we have:

$\sigma^2 = \frac{20*(9.13-12.63)^2 + 38 * (10.88-12.63)^2 +...........+27 * (16.13 -12.63)^2}{20+38+36+25+27-1}$

$\sigma^2 = \frac{2393.6875}{145}$

$\sigma^2 = 16.51$

The sample standard deviation is:

$\sigma = \sqrt{\sigma^2}$

$\sigma = \sqrt{16.51}$

$\sigma = 4.06$

6 0

2 years ago

Can some one explain in an easy way pls

Helga [31]

What needs to be done first is to add up females and males that have passed.

42 + 14 = 56

so out of 56 students who passed 42 females passed 42/56 = 3/4 = 0.75

out of 56 students who passed, 14 males passed which turns into 14/56 = 1/4 = 0.25

check work; 0.75 + 0.25 = 1.00

NOW WE ARE DOING FAILS.

15 + 5 = 20

so out of 20 students who failed, 15 females failed so it turns into 15/20 = 3/4 = 0.75

out of 20 students who failed, 5 males failed. 5/20 = 1/4 = 0.25

check work; 0.75 + 0.25 = 1.00

i hope this helped! :)

7 0

2 years ago

Suppose that a market research firm is hired to estimate the percent of adults living in a large city who have cell phones. 500

Alborosie

Answer:

The 95% confidence interval estimate for the true proportion of adults residents of this city who have cell phones is (0.81, 0.874).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.