Question

Suppose a television news broadcast reports that the proportion of people in the United States who are living with a particular disease is 0.09. A team of biomedical students examined a random sample of 527 medical records and found that 34 of them had this disease. They constructed the following 95% z z‑confidence interval for the proportion, p p, of people in the United States who have this disease. 0.0435 < p < 0.0855 0.0435

Answer 1

Answer:

On this case the 0.09 value is not included on the interval so we can say that the statement on the television news broadcast reports, is a value away from the real proportion at least at 95% of confidence.

Step-by-step explanation:

1) Notation and definitions

$X=34$ number of people living at USA with a particular disease.

$n=527$ random sample taken

$\hat p=\frac{34}{527}=0.0645$ estimated proportion of people living at USA with a particular disease

$p$ true population proportion of people living at USA with a particular disease.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$. And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The confidence interval for the mean is given by the following formula:  

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.0645 - 1.96\sqrt{\frac{0.0645(1-0.0645)}{527}}=0.0435$

$0.0645 + 1.96\sqrt{\frac{0.0645(1-0.0645)}{527}}=0.0855$

The 95% confidence interval would be given by (0.0435;0.0855)

On this case the 0.09 value is not included on the interval so we can say that the statement on the television news broadcast reports, is a value away from the real mean at least at 95% of confidence.