HW 3 1)q8.36 (b) 2)q9.8 3)q9.19 (b) 4)q9.22 (b) 5)q9.46. ======= please help me

Question

          

HW 3    

1)q8.36 (b)

2)q9.8

3)q9.19 (b)

4)q9.22 (b)

5)q9.46.

=======
please help me with this

(( in 20l subject ))
the book I study from is ( statistics:a first course by B.M.perles & J.E.freund)

using Chebyshev's theorem, about the probability that the error will be less than S18? The mean of a random sample of size n an infinite population having the standard deviation ?-2. What can be asserted about the probability the error will be less than 0.25 using Chebyshev's theorem 8.35 100 is used to estimate the mean of 8.36 The supervisor of a waste disposal truck that collects waste in a residential aste pickups neighborhood wants to estimate the mean number of cubic fee of waste collect at these residences. Suppose that the supervisor uses a sample of 30 w and knows from experience that the standard deviation is 3 cubic feet for such data. Based on the central limit theorem; with what probability can the supervisor assert that the error will be

Explanation / Answer

9.19

a)

CI for = 95%

n = 49

mean = 25.1

z-value of 95% CI = 1.9600

std. dev. = 8.5

SE = std.dev./sqrt(n)

= 8.5/sqrt(49)

= 1.21429

ME = z*SE

= 1.96 * 1.21429

= 2.37996

Lower Limit = Mean - ME

= 25.1 - 2.37996

= 22.72004

Upper Limit = Mean + ME

= 25.1 + 2.37996

= 27.47996

95% CI (22.72 , 27.48 )

b)

CI for = 99%

n = 49

mean = 25.1

z-value of 99% CI = 2.5758

std. dev. = 8.5

SE = std.dev./sqrt(n)

= 8.5/sqrt(49)

= 1.21429

ME = z*SE

= 1.96 * 1.21429

= 3.12779

Lower Limit = Mean - ME

= 25.1 - 2.37996

= 21.97221

Upper Limit = Mean + ME

= 25.1 + 2.37996

= 28.22779

99% CI (21.9722 , 28.2278 )

9.22)

a)

CI for = 90%

n = 60

mean = 72.3

z-value of 90% CI = 1.6449

std. dev. = 8.5

SE = std.dev./sqrt(n)

= 8.5/sqrt(60)

= 1.09735

ME = z*SE

= 1.6449 * 1.09735

= 1.80497

Lower Limit = Mean - ME

= 72.3 - 1.80497

= 70.49503

Upper Limit = Mean + ME

= 72.3 + 1.80497

= 74.10497

90% CI (70.495 , 74.105 )

b)

CI for = 90%

n = 60

mean = 82.3

z-value of 90% CI = 1.6449

std. dev. = 8.5

SE = std.dev./sqrt(n)

= 8.5/sqrt(60)

= 1.09735

ME = z*SE

= 1.6449 * 1.09735

= 1.80497

Lower Limit = Mean - ME

= 82.3 - 1.80497

= 80.49503

Upper Limit = Mean + ME

= 82.3 + 1.80497

= 84.10497

90% CI (80.495 , 84.105 )

9.46)

n = 500

p = 0.15

z-value of 95% CI = 1.9600

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

= sqrt (0.15 * (1 -0.15) /500)

= 0.01597

ME = z*SE

= 1.96 * 0.01597

= 0.03130

Lower Limit = p - ME

= 0.15 - 0.03130

= 0.11870

Upper Limit = p + ME

= 0.15 + 0.03130

= 0.18130

95% CI (0.1187 , 0.1813 )

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