Mathematics, 21.12.2019 00:31 nuneybaby

Gardeners on the west coast of the united states are investigating the difference in survival rates of two flowering plants in drought climates. plant a has a survival rate of 0.75 and plant b has a survival rate of 0.44. the standard error of the difference in proportions is 0.082. what is the margin of error for a 99% confidence interval?

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Answer from: kam110800

$ME=z_{\alpha/2} SE= 2.58*0.082=0.21156$

The 99% confidence interval would be given (0.09844;0.5216).

We are confident at 99% that the difference between the two proportions is between $0.09844 \leq p_B -p_A \leq 0.5216$

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".

p_A represent the real population proportion for plant A

$\hat p_A =0.75$ represent the estimated proportion for plant A

n_A is the sample size required for plant A

p_B represent the real population proportion for plant b

$\hat p_B =0.44$ represent the estimated proportion for plant B

n_B is the sample size required for plant 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 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

The standard error is given by:

$SE=z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}=0.082$

The margin of error is:

$ME=z_{\alpha/2} SE= 2.58*0.082=0.21156$

And replacing into the confidence interval formula we got:

(0.75-0.44) - 2.58(0.082)=0.09844

(0.75 -0.44) + 2.58(0.082)=0.5216

And the 99% confidence interval would be given (0.09844;0.5216).

We are confident at 99% that the difference between the two proportions is between $0.09844 \leq p_B -p_A \leq 0.5216$

Answer from: Quest

i'm not sure i quite understand, could you provide more information?

Answer from: Quest

[2, 2.5)

(2.5, 3]

step-by-step explanation:

i think its these two not sure [2, 2.5)

(2.5, 3]

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