Mathematics, 04.12.2019 02:31 costel8532

Asurvey of 1010 college seniors working towards an undergraduate degree was conducted. each student was asked, "are you planning or not planning to pursue a graduate degree? " of the 1010 surveyed, 658 stated that they were planning to pursue a graduate degree. construct and interpret a 98% confidence interval for the proportion of college seniors who are planning to pursue a graduate degree. round to the nearest thousandth.

Answers: 3

Show answers

Answers

Answer from: glowbaby123

The 98% confidence interval would be given (0.616;0.686).

We are confident at 98% that the true proportion of people that they were planning to pursue a graduate degree is between (0.616;0.686).

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".

Description in words of the parameter p

represent the real population proportion of people that they were planning to pursue a graduate degree

$\hat p$ represent the estimated proportion of people that they were planning to pursue a graduate degree

n=1010 is the sample size required

$z_{\alpha/2}$ represent the critical value for the margin of error

The population proportion have the following distribution

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

Numerical estimate for p

In order to estimate a proportion we use this formula:

$\hat p =\frac{X}{n}$ where X represent the number of people with a characteristic and n the total sample size selected.

$\hat p=\frac{658}{1010}=0.651$ represent the estimated proportion of people that they were planning to pursue a graduate degree

Confidence interval

The confidence interval for a proportion is given by this formula

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

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:

$0.651 - 2.33 \sqrt{\frac{0.651(1-0.651)}{1010}}=0.616$

$0.651 + 2.33 \sqrt{\frac{0.651(1-0.651)}{1010}}=0.686$

And the 98% confidence interval would be given (0.616;0.686).

We are confident at 98% that the true proportion of people that they were planning to pursue a graduate degree is between (0.616;0.686).

Answer from: Quest

He na a

Answer from: Quest

Claude debussy's music was associated with answer choice 3, impressionist

Another question on Mathematics

Mathematics, 21.06.2019 16:00

Question part points submissions used suppose that 100 lottery tickets are given out in sequence to the first 100 guests to arrive at a party. of these 100 tickets, only 12 are winning tickets. the generalized pigeonhole principle guarantees that there must be a streak of at least l losing tickets in a row. find l.

Answers: 2

Answer

Mathematics, 21.06.2019 17:50

Which of these triangle pairs can be mapped to each other using two reflections?

Answers: 1

Answer

Mathematics, 21.06.2019 19:00

Simplify. −4x^2 (5x^4−3x^2+x−2) −20x^6−12x^4+8x^3−8x^2 −20x^6+12x^4−4x^3+8x^2 −20x^8+12x^4−4x^2+8x −20x^6+12x^4+4x^3−8x^2

Answers: 1

Answer

Mathematics, 21.06.2019 20:30

Which expression is equivalent to (4 +6i)^2? ? -20 + 48i 8 + 12i 16 - 36i 20 + 48i