Suppose that 73.2% of all adults with type 2 diabetes also suffer from hypertension. After developing a new drug to treat type 2 diabetes, a team of researchers at a pharmaceutical company wanted to know if their drug had any impact on the incidence of hypertension for diabetics who took their drug. The researchers selected a random sample of 1000 participants who had been taking their drug as part of a recent large-scale clinical trial and found that 718 suffered from h The researchers want to use a one-sample z-test for a population have hypertension while taking their new drug, p, is different from the proportion of all type 2 diabetics who have hypertension. They decide to use a significance level of a 0.01. to see if the proportion of type 2 diabetics who Determine if the requirements for a one-sample z-test for a proportion been met. If the requirements have not been met leave the rest of the questions blank. O The requirements have not been met because there are two samples: the type 2 diabetics with hypertension who are not taking the drug and the ones who are taking the drug O The requirements have been met because the sample was appropriately selected, the variable of interest is categorical with two possible outcomes, and the sampling distribution is approximately normal. O The requirements have not been met because the population standard deviation is unknown. O The requirements have been met because the sample was appropriately selected, the variable of interest is categorical with two possible outcomes, and the sample size is at least 30.

The obtained p-value is greater than the significance level at which the test was performed at, hence, we fail to reject the null hypothesis & conclude that there is no significant evidence that the proportion of type 2 diabetics that have hypertension while taking the new drug is different from the proportion of all type 2 diabetics who have hypertension.

No significant difference between the population proportion of type 2 diabetics with hypertension while using the new drug and the population proportion of all type 2 diabetics with hypertension.

Step-by-step explanation:

The full complete question is attached to this solution

The only major requirements for using the one sample z-test is that the population is approximately normal at least. And the population standard deviation is known. For this question, the conditions of approximate normality for binomial distribution is satisfied;

np = 718 ≥ 10

And np(1-p) = 1000×0.718×0.282 = 201 ≥ 10

But, no information on the population standard deviation is known. But we can carry on with the test because the sample size is large enough for the p-value obtained from t-test statistic will be approximately equal to the p-value obtained from the z-test statistic.

b) For hypothesis testing, we first clearly state our null and alternative hypothesis.

The null hypothesis is that there is no significant evidence that the proportion of type 2 diabetics that have hypertension while taking the new drug is different from the proportion of all type 2 diabetics who have hypertension.

And the alternative hypothesis is that there is significant evidence that the proportion of type 2 diabetics that have hypertension while taking the new drug is different from the proportion of all type 2 diabetics who have hypertension.

Mathematically, the null hypothesis is

H₀: p = 0.732

The alternative hypothesis is

Hₐ: p₀ ≠ 0.732

To do this test, we will use the z-distribution because, the degree of freedom is so large, it is large enough for the p-value obtained from t-test statistic will be approximately equal to the p-value obtained from the z-test statistic.

So, we compute the z-test statistic

z = (x - μ)/σₓ

x = sample proportion of type 2 diabetics with hypertension while using the drug = p =(718/1000) = 0.718

μ = p₀ = proportion of all type 2 diabetics with hypertension = 0.732

σₓ = standard error of the sample proportion = √[p(1-p)/n]

where n = Sample size = 1000

p = 0.718

σₓ = √[0.718×0.282/1000] = 0.0142294062 = 0.01423

z = (0.718 - 0.732) ÷ 0.01423

z = -0.984 = -0.98

checking the tables for the p-value of this z-statistic

Note that this test is a two-tailed test because we're checking in both directions, hence the not equal to sign, (≠) in the alternative hypothesis.

p-value (for z = -0.98, at 0.01 significance level, with a two tailed condition) = 0.327086

The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 1% = 0.01

p-value = 0.327086

0.327086 > 0.01

Hence,

p-value > significance level

This means that we fail to reject the null hypothesis & conclude that there is no significant evidence that the proportion of type 2 diabetics that have hypertension while taking the new drug is different from the proportion of all type 2 diabetics who have hypertension.

Hope this Helps!!!

joaobezerra joaobezerra · Answer 2 · 2021-11-28T18:01:50+01:00

Using the normal approximation to the binomial, it is found that the correct option is:

The requirements have been met because the sample was appropriately selected, the variable of interest is categorical with two possible outcomes, and the sampling distribution is approximately normal.

The binomial distribution, which is the probability of exactly x successes on n repeated trials, with p probability of success on each trial(each trial either is a success or it is a failure), can be approximated to the normal if in a sample, there are at least 10 successes and at least 10 failures.

In this problem:

Random sample of 1000, hence, appropriately selected.

For each person, there are two possible outcomes, either they suffered from hypertension, or they did not, hence, the variable is categorical.
718 have hypertension, 282 have not, at least 10 of each, hence, the sampling distribution is approximately normal.

Thus, the correct option is:

The requirements have been met because the sample was appropriately selected, the variable of interest is categorical with two possible outcomes, and the sampling distribution is approximately normal.

A similar problem is given at https://brainly.com/question/24261244