Mathematics, 15.04.2020 15:30 fatherbamboo

A random sample of 1800 adult women indicated that 841 of them asked for medical assistance last time they felt sick. The same sized sample of men indicated that 793 of them asked for medical assistance. Use the z-values rounded to three decimal places to obtain the answers. a) Construct a 98% confidence interval for the difference in the two tests. Round your answers to five decimal places (e. g. 98.76543).

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Answer from: carog24

$(0.467-0.441) - 2.33 \sqrt{\frac{0.467(1-0.467)}{1800} +\frac{0.441(1-0.441)}{1800}}=-0.0127$

$(0.467-0.441) + 2.33 \sqrt{\frac{0.467(1-0.467)}{1800} +\frac{0.441(1-0.441)}{1800}}=0.0647$

And the 98% confidence interval would be given (-0.0127;0.0647).

We are confident at 98% that the difference between the two proportions is between $-0.0127 \leq p_A -p_B \leq 0.0647$

Step-by-step explanation:

Previous concepts

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".

Solution to the problem

p_A represent the real population proportion for women

$\hat p_A =\frac{841}{1800}=0.467$ represent the estimated proportion for women

n_A=1800 is the sample size required for women

p_B represent the real population proportion for men

$\hat p_B =\frac{793}{1800}=0.441$ represent the estimated proportion for men

n_B=1800 is the sample size required for Brand B

represent the critical value for the margin of error

The population proportion have the following distribution

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

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 98% confidence interval the value of $\alpha=1-0.98=0.02$ and $\alpha/2=0.01$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.33$

And replacing into the confidence interval formula we got:

$(0.467-0.441) - 2.33 \sqrt{\frac{0.467(1-0.467)}{1800} +\frac{0.441(1-0.441)}{1800}}=-0.0127$

$(0.467-0.441) + 2.33 \sqrt{\frac{0.467(1-0.467)}{1800} +\frac{0.441(1-0.441)}{1800}}=0.0647$

And the 98% confidence interval would be given (-0.0127;0.0647).

We are confident at 98% that the difference between the two proportions is between $-0.0127 \leq p_A -p_B \leq 0.0647$

Answer from: Quest

4+n=5

step-by-step explanation:

this is because the four is four hours and the n represents the 1 extra hour over the weekend.

b= i'm srry i don't know that one.

4+n=5 n=1 because if vernon plays 5 hours a week then 4+1=5 hours.

hope this : )

Answer from: Quest

You need the internet to you

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A random sample of 1800 adult women indicated that 841 of them asked for medical assistance last tim...