An article describes a study of 48 males with pierced tongues. Suppose that it is reasonable to regard this sample as a random sample from the population of all males with pierced tongues. The researchers found receding gums, which can lead to tooth loss, in 25 of the study participants. (a) Suppose you are interested in learning about the value of p, the proportion of all males with pierced tongues who have receding gums. This proportion can be estimated using the sample proportion, p. What is the value of p for this sample? (Round your answer to two decimal places.) (b) Based on what you know about the sampling distribution of p, is it reasonable to think that this estimate is within 0.05 of the actual value of the population proportion? Explain why or why not. (Hint: See Example 8.4. Round your answer to three decimal places.) G--Select--- Areasonable to think that this estimate is within 0.05 of the actual value of the population proportion. The sampling distribution of ô --Select--A approximately normal, so the margin of error for this estimate is | , which is ----Select-- than 0.05.

The estimator of the population proportion is the sample proportion, you calculate it as the division of the number of success of the trial "x" and the total of the sample "n":

p'= x/n= 25/48= 0.52

b)

The hypothesis is that "the estimate is within 0.05 of the actual value of the population proportion" symbolically p' ± 0.05p

Thanks to the central limit theorem you can approximate the distribution of the sample proportion to normal: p' ≈ N(p;[tex]\frac{p(1-p)}{n}[/tex]).

To study the possible range of values of a population parameter is best to construct a confidence interval.

Let's say, for example, that you can work with a confidence level of 1 - α: 0.95

The formula for the CI is:

p' ± [tex]Z_{1-\alpha /2}[/tex] * [tex]\sqrt{\frac{p'(1-p')}{n} }[/tex]

Where [tex]Z_{1-\alpha /2}[/tex] * [tex]\sqrt{\frac{p'(1-p')}{n} }[/tex] is the margin of error of the interval. By calculating the margin of error you can see if the sample proportion is within 0.05 of the actual proportion or not:

[tex]Z_{1-\alpha /2}= Z_{0.975}= 1.965[/tex]

1.965*[tex]\sqrt{\frac{0.52*0.48}{48} }[/tex]= 0.14

With a confidence level of 95%, you can expect the value of the sample proportion to be within 0.14 of the actual value of the population proportion.

The calculated margin of error is greater than the suspected one, so you cannot say that is reasonable to believe that the estimate is within 0.05 of the parameter.

I hope this helps!

akindeleot akindeleot · Answer 2 · 2020-04-11T05:20:06+02:00

Answer:

(a)

point estimate of p is sample proportion

sample proportion = p^ = x/n = successes/total

∴

p = x/n = 25/48 = 0.5208333 = 0.52

(b)

since n = 48. large sample,

sampling distribution follows normal distribution

margin of error = zcrit*sqrt(p*(1-p)/n)

=1.96*sqrt( 0.5208333*(1- 0.5208333)/48)

=0.141328

=0.141

Yes reasonable.

The sampling distribution of p^ is approximately normal,so the margin of error of this estimate is 0.141,which is greater than 0.05