Wildlife biologists inspect 153 deer taken by hunters and found 42 of them carrying Lyme disease ticks. Calculate and interpret a 90% confid

Question

Wildlife biologists inspect 153 deer taken by hunters and found 42 of them carrying Lyme disease ticks. Calculate and interpret a 90% confidence interval for the proportion of deer that carry Lyme disease ticks.

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Nick 5 months 2021-08-14T15:13:44+00:00 1 Answers 16 views 0

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    2021-08-14T15:15:21+00:00

    Answer:

    90 Confidence Interval for the proportion = [ 0.216,0.334]

    Step-by-step explanation:

    The confidence interval for proportion formula =

    p ± z × √p(1 – p)/n

    Where p = x/n

    x = 42

    n = number of samples = 153

    p = 42/153

    p = 0.2745098039 ≈ 0.275

    z = z score of 90% confidence interval = 1.645

    Confidence interval =

    0.275 ± 1.645 × √0.275 – ( 1 – 0.275)/153

    = 0.275 ± 1.645 × √0.0013031046

    = 0.275 ± 0.0593820985

    Confidence Interval

    = 0.275 – 0.0593820985

    = 0.2156179015

    ≈ 0.216

    = 0.275 + 0.0593820985

    = 0.3343820985

    ≈ 0.334

    90 Confidence Interval for the proportion = [ 0.216,0.334]

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