Hello. I need some help with my statistics homework. I\'m completely lost as far

Question

Hello. I need some help with my statistics homework. I'm completely lost as far as what to do. could you show me every question step by step please? thank you.

Free Response: Pro 1. One difficulty in measuring the nesting success of birds is that the researchers vide your response on the specified regions after each question. Please show your wot t your solutions. Answers provided without necessary support will not be sco v support will not be scored. must count the number of eggs in the nest, which is disturbing to the parents. the researcher does not harm the birds, the flight of the bird might alert pred the presence of a nest. To see if researcher activity might degrade nesting su nest sturvival of 120 nests that had their eggs counted, was recorded. Sixty nests failed (meaning the parent abandoned the nest). Assuming that it is reasot regard the 120 nests in the sample as representative of the population of the eggs have been counted. n though ccess, t the poru that itive of t (a) (1 point) What is the proportion (population characteristic) of interest? (b) (1 point) Compute the sample proportion p

Explanation / Answer

Let X be the random variable that number of nests that have eggs counted that are then abandoned.

a) P be the population proportion of nests that have eggs counted that are then abandoned.

b) p^ be the sample proprotion of nests that have eggs counted that are then abandoned.

Given that, x = 65

n = 120

sample proportion (p^) = x/n = 65/120 = 0.54

c) Check normality assuptions for p^.

np^ > 10 and n(1-p^) > 10

np^ = 120*0.54 = 65 > 10 and n(1-p^) = 120*(1-0.54) = 55 > 10

d) Standard error of p^.

SE(p^) = sqrt((p^*(1-p^)) / n)

= sqrt[0.54*(1-0.54) / 120] = 0.045

e) Now we have to find 95% confidence interval for p.

95% CI for p is,

p^ -E < p < p^ + E

where p^ = 0.54

E is margin of error.

E = Zc*SE(p^)

Zc is the critical value for normal distribution.

C is confidence level = 95% = 0.95

For 95% confidence level = 1.96

E = 1.96*0.045 = 0.0882

lower bound = p^ - E = 0.54 - 0.0882 = 0.4518

upper bound = p^ + E = 0.54 + 0.0882 = 0.6282

95% confidence interval for p is (0.4518, 0.6282)

f) Now we have to find 98% confidence interval for p.

C = confidence level = 9% = 0.98

For 98% confidence level = Zc = 2.33

E = Zc* SE(p^) = 2.33*0.045 = 0.1049

98% CI for p is (0.54 - 0.1049, 0.54 + 0.1049).

98% CI for p is (0.4352, 0.6449)

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