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VARVARA [1.3K]
3 years ago
7

PTC is a substance that has a strong bitter taste for some people and is tasteless for others. The ability to taste PTC is inher

ited. About 76% of people from a certain country can taste PTC, for example. You want to estimate the proportion of people with at least one grandparent from this country who can taste PTC. (Round your answers up to the nearest person.)
(a) Starting with the 76% estimate, how large a sample must you collect in order to estimate the proportion of PTC tasters within ±0.01 with 90% confidence?
(b) Estimate the sample size required if you made no assumptions about the value of the proportion who could taste PTC?
Mathematics
1 answer:
lana66690 [7]3 years ago
3 0

Answer:

a) n=\frac{0.76(1-0.76)}{(\frac{0.01}{1.64})^2}=4905.83  

And rounded up we have that n=4906

b) For this case since we don't have prior info we need to use as estimatro for the proportion \hat p =0.5

n=\frac{0.5(1-0.5)}{(\frac{0.01}{1.64})^2}=6724  

And rounded up we have that n=6724

Step-by-step explanation:

We need to remember that the confidence interval for the true proportion is given by :  

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

Part a

The estimated proportion for this case is \hat p =0.76

Our interval is at 90% of confidence, and the significance level is given by \alpha=1-0.90=0.1 and \alpha/2 =0.05. The critical values for this case are:

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

The margin of error for the proportion interval is given by this formula:  

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

The margin of error desired is given ME =\pm 0.01 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

Replacing we got:

n=\frac{0.76(1-0.76)}{(\frac{0.01}{1.64})^2}=4905.83  

And rounded up we have that n=4906

Part b

For this case since we don't have prior info we need to use as estimatro for the proportion \hat p =0.5

n=\frac{0.5(1-0.5)}{(\frac{0.01}{1.64})^2}=6724  

And rounded up we have that n=6724

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3 years ago
This is a proportion. What is the value of X?
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x = 15 or x = - \frac{2}{5}

Step-by-step explanation:

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5x² - 73x - 30 = 0 ← in standard form

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the factors are - 75 and + 2

Use these factors to split the middle term

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5x(x - 15) + 2(x - 15) = 0 ← take out the factor (x - 15)

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x - 15 = 0 ⇒ x = 15

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99% of all confidence intervals with a 99% confidence level should contain the population parameter of interest. true or false
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The statement that 99% of all confidence intervals with a 99% confidence level should contain the population parameter of interest is false.

A confidence interval (CI) is essentially a range of estimates for an unknown parameter in frequentist statistics. The most frequent confidence level is 95%, but other levels, such 90% or 99%, are infrequently used for generating confidence intervals.

The confidence level is a measurement of the proportion of long-term associated CIs that include the parameter's true value. This is closely related to the moment-based estimate approach.

In a straightforward illustration, when the population mean is the quantity that needs to be estimated, the sample mean is a straightforward estimate. The population variance can also be calculated using the sample variance. Using the sample mean and the true mean's probability.

Hence we can generally infer that the given statement is false.

To learn more about confidence intervals visit:

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