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

b) i)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.  

ii) The sample needs to be large enough  

$np_o =2280*0.186=424.08>10$  

$n(1-p_o)=2280*(1-0.186)=319.2>10$  

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

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:

1) Data given and notation  

n=2280 represent the random sample taken    

X=424 represent the people that donated blood in the past two years

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

$\alpha=0.05$ represent the significance level (no given, but is assumed)    

z would represent the statistic (variable of interest)    

$p_v$ represent the p value (variable of interest)    

p= population proportion of people that donated blood in the past two years

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.

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

b) verify that the requirements for constructing a confidence interval.the sample

Check for the assumptions that he sample must satisfy in order to use the confidence interval

i)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.  

ii) The sample needs to be large enough  

$np_o =2280*0.186=424.08>10$  

$n(1-p_o)=2280*(1-0.186)=319.2>10$  

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.

The confidence interval would be given by this formula

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

For the 90% confidence interval the value of $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$, with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.64$

And replacing into the confidence interval formula we got:

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

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

And the 90% confidence interval would be given (0.173;0.199).

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 ]

Answer 2

Refer the below solution for better understanding.

Step-by-step explanation:

Given :

Random Sample, n = 2280

People that donated blood in the past two years, X =424

Calculaton :

Significance level (no given, but is assumed),

$\alpha =0.05$  

Let z represent the statistic (variable of interest) and $p_v$  represent the p value (variable of interest)

Let p be the population proportion of people that donated blood in the past two years.

a)      $\widehat{p} = \dfrac{424}{2280}= 0.186$

estimated proportion of people that donated blood in the past two years.

b)

i) Random sample needs to be representative: On this case the problem not mention about it but we can assume it.

ii) Sample needs to be large enough

$n\timesp_0=2280\times0.186=424.08>10$

$n(1-p_0)= 2280(1-0.186)=319.8>10$

c)  Confidence interval is given by,

$\widehat{p} \;\pm\;z_\frac{\alpha }{2}\sqrt{\dfrac{\widehat{p}(1-\widehat{p})}{n}}$

$\alpha =1-0.9=0.1$

$\dfrac{\alpha }{2}=0.05$

$z_\frac{\alpha }{2}=1.64$

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

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

90% confidence interval is given, (0.173,0.199).

A) We are 90% confident that 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

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