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Maurinko [17]
2 years ago
6

The Pew Research Center reported in 2018 that 68% of U.S. adults rated "reducing health care costs" as a national priority. This

year a survey is conducted with a national sample of 1,505 U.S. adults selected by a combination of landline and cell phone random digit dials. This year's survey finds that 70% of the sample says that "reducing health care costs" is a national priority.
We test the following hypotheses:

H0: The proportion of U.S. adults this year who rate "reducing health care costs" as a national priority is still 0.68.
Ha: The proportion of U.S. adults this year who rate "reducing health care costs" as a national priority is greater than 0.68.

The P-value is 0.045. At a 5% significance level, we would reject the null hypothesis and concluded that the proportion of U.S. adults this year who rate "reducing health care costs" as a national priority is greater than 0.68.

Explain the P-value in the context of this issue.
Mathematics
1 answer:
Zanzabum2 years ago
8 0

Answer:

interpretation of p-value

With the information from the exercise we can deduce that the number of adults in North America who are going to reduce the costs of medical care, considering this as a national priority, are within a probability of 0.045, however the real data tells us that 0.68 of adults are in favor of this reduction

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a) p represent the real population proportion of people who showed partial or complete resistance to the antibiotic

b) \hat p=\frac{973}{1714}=0.568 represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

c) We are confident (95%) that the true proportion of individuals in Florida diagnosed with a strep infection was tested for resistance to the antibiotic penicillin present partial or complete resistance to the antibiotic is between 54.5% to 59.1%.  

Step-by-step explanation:

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".

Part a

Description in words of the parameter p

p represent the real population proportion of people who showed partial or complete resistance to the antibiotic

\hat p represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

n=1714 is the sample size required

z_{\alpha/2} represent the critical value for the margin of error

The population proportion have the following distribution  

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

Part b

Numerical estimate for p

In order to estimate a proportion we use this formula:

\hat p =\frac{X}{n} where X represent the number of people with a characteristic and n the total sample size selected.

\hat p=\frac{973}{1714}=0.568 represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

Part c

The confidence interval for a proportion is given by this formula

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

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.96

And replacing into the confidence interval formula we got:

0.568 - 1.96 \sqrt{\frac{0.568(1-0.568)}{1714}}=0.545

0.568 + 1.96 \sqrt{\frac{0.56(1-0.56)}{1714}}=0.591

And the 95% confidence interval would be given (0.545;0.591).

We are confident that about 54.5% to 59.1% of individuals in Florida diagnosed with a strep infection was tested for resistance to the antibiotic penicillin present partial or complete resistance to the antibiotic.  

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