Question

a survey of 2280 adults in a certain large country aged 18 and older conducted by reputable polling organization founded that 424 have donated blood in the past two years. Complete Parts (a) through (C).
a) obtain a point estimate for the population proportion of adults in the country age 18 and older who have donated blood in the past two years.

b) verify that the requirements for constructing a confidence interval.
the sample [ a) can be assumed to be b)is stated to not be. c) cannot be assumed to be. d) is stated to be. ] a simple random sample, the value of [ a)np b) p(1-p) c) np(1-p) d) p.] is [ ]. which is [ a) greater than. b) less then] 10. & the [ a) population size b) sample population c) sample size d) population proportion] [ a)can be assumed to be. b) is stated to not be c) cannot be assumed to be d) is stated to] less than or equal to 5% of the [a) sample proportion b) population proportion c) population size d) sample size.]

c. construct and interpret a 90% confidence interval for the population proportion of adults in the country who have donated blood in the past two years. Select the correct Choice below and fill in any answer box within your choice.( type integer or decimal rounded to three decimal places as needed. Use ascending order)
A) we are [ ]% confident the proportion of adults in the country aged 18 and older who have donated blood in the past two years is between [ ] and [ ].
B) there is a [ ]% chance the proportion of adults in the country aged 18 and older who have donated blood in the past 2 years is between [ ] and [ ].

Answer 1

Answer:

a) The best estimator for the population proportion is the sample proportion since $E(\hat p) = p$.

$\hat p=\frac{424}{2280}=0.186$ estimated proportion of people who donated blood in the past two years

b) We need to check the conditions in order to use the normal approximation.

Random sample assumed

$np=2280*0.186=424.08 \geq 10$

$n(1-p)=2280*(1-0.186)=1855.92 \geq 10$

We have all the conditions to create the confidence interval

c) $0.186 - 1.64\sqrt{\frac{0.186(1-0.186)}{2280}}=0.173$

$0.186 + 1.64\sqrt{\frac{0.186(1-0.186)}{2280}}=0.199$

The 90% confidence interval would be given by (0.173;0.199)

And the correct conclusion is:

A) we are [90]% confident the proportion of adults in the country aged 18 and older who have donated blood in the past two years is between [0.173] and [0.199].

Step-by-step explanation:

Notation and definitions

$X=424$ number of people who donated blood in the past two years

$n=2280$ random sample taken

$\hat p=\frac{424}{2280}=0.186$ estimated proportion of people who donated blood in the past two years

$p$ true population proportion of people who donated blood in the past two years

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

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

Part a

The best estimator for the population proportion is the sample proportion since $E(\hat p) = p$.

$\hat p=\frac{424}{2280}=0.186$ estimated proportion of people who donated blood in the past two years

Part b

We need to check the conditions in order to use the normal approximation.

Random sample assumed

$np=2280*0.186=424.08 \geq 10$

$n(1-p)=2280*(1-0.186)=1855.92 \geq 10$

We have all the conditions to create the confidence interval

Part c

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 90% of confidence, our significance level would be given by $\alpha=1-0.90=0.1$ and $\alpha/2 =0.05$. And the critical value would be given by:

$z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64$

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:

$0.186 - 1.64\sqrt{\frac{0.186(1-0.186)}{2280}}=0.173$

$0.186 + 1.64\sqrt{\frac{0.186(1-0.186)}{2280}}=0.199$

The 90% confidence interval would be given by (0.173;0.199)

And the correct conclusion is:

A) we are [90]% confident the proportion of adults in the country aged 18 and older who have donated blood in the past two years is between [0.173] and [0.199].