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daser333 [38]
3 years ago
9

Data for the estimated number of HIV cases worldwide between 1988 and 1996 suggests exponential growth in the number of cases ov

er this time period. An appropriate transformation of the data (base-ten logarithms were used) produced a strongly linear scatterplot, and a regression analysis was performed on the transformed data.
Here is the computer output:
Predictor Constant Coef SE Coef T F
-151.281 2.038 -74.22 0.000
Year 0.076577 0.001023 74.84 0.000
S= 0.00792598 R-Sq= 99.9% R-Sg (adj ) =99.9%
Note that HIV cases are in millions.
Which of the following value is closest to the value that this model predicts for the number of HIV cases in 1998?
(a) 1.72 million cases
(b) 2.18 million cases
(c) 17.2 million cases
(d) 52.5 million cases
(e) The predicted value is negative because we are extrapolating beyond the range o f the explanatory variable
Mathematics
1 answer:
Eduardwww [97]3 years ago
3 0

Answer:

d. 52.5 million cases

Step-by-step explanation:

from the data output in the question, we have the regression equation to be

log y = -151.281 + 0.076577x

y is the number of hiv cases which is in millions

x is the year of the hiv case

x = 1998

we put the value of x in the regression equation

log y = -151.281 + 0.076577(1998)

log y = -151.281 + 153.000846

log y  = 1.719846

since the result is in log form,

y=10^{1.791846}

y = 52.5 million cases

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3 years ago
A given population proportion is .25. What is the probability of getting each of the following sample proportions
anyanavicka [17]

This question is incomplete, the complete question is;

A given population proportion is .25. What is the probability of getting each of the following sample proportions

a) n = 110 and = p^ ≤ 0.21, prob = ?

b) n = 33 and p^ > 0.24, prob = ?

Round all z values to 2 decimal places. Round all intermediate calculation and answers to 4 decimal places.)

Answer:

a) the probability of getting the sample proportion is 0.1660

b) the probability of getting the sample proportion is 0.5517

Step-by-step explanation:

Given the data in the questions

a)

population proportion = 0.25

q = 1 - p = 1 - 0.25 = 0.75

sample size n = 110

mean = μ = 0.25

S.D = √( p( 1 - p) / n ) = √(0.25( 1 - 0.25) / 110 ) √( 0.1875 / 110 ) = 0.0413

Now, P( p^ ≤ 0.21 )

= P[ (( p^ - μ ) /S.D) < (( 0.21 - μ ) / S.D)

= P[ Z < ( 0.21 - 0.25 ) / 0.0413)

= P[ Z < -0.04 / 0.0413]

= P[ Z < -0.97 ]

from z-score table

P( X ≤ 0.21 ) = 0.1660

Therefore, the probability of getting the sample proportion is 0.1660

b)

population proportion = 0.25

q = 1 - p = 1 - 0.25 = 0.75

sample size n = 33

mean = μ = 0.25

S.D = √( p( 1 - p) / n ) = √(0.25( 1 - 0.25) / 33 ) = √( 0.1875 / 33 ) = 0.0754

Now, P( p^ > 0.24 )  

= P[ (( p^ - μ ) /S.D) > (( 0.24 - μ ) / S.D)

= P[ Z > ( 0.24 - 0.25 ) / 0.0754 )

= P[ Z > -0.01 / 0.0754  ]

= P[ Z > -0.13 ]

= 1 - P[ Z < -0.13 ]

from z-score table

{P[ Z < -0.13 ] = 0.4483}

1 - 0.4483

P( p^ > 0.24 )  = 0.5517

Therefore, the probability of getting the sample proportion is 0.5517

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