Question

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.

Answer 1

Answer:

a) Option C is correct.

The requirements have not been met because the population standard deviation is unknown.

The null hypothesis is

H₀: p = 0.732

The alternative hypothesis is

Hₐ: p₀ ≠ 0.732

z-test statistic = -0.98

p-value = 0.327086

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!!!