Answer Choices: 1/12 1/2 1/6 2/3
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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.
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