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    Loreto is a manager at the Da Vinci health centre. If the mean rate of patients arriving at the health centre exceeds 1.5 per minute then Loreto will employ extra staff. It is assumed that the number of patients arriving in any given time period follows a Poisson distribution.

    Question
    HLPaper 2

    Loreto is a manager at the Da Vinci health centre. If the mean rate of patients arriving at the health centre exceeds 1.5 per minute then Loreto will employ extra staff. It is assumed that the number of patients arriving in any given time period follows a Poisson distribution.

    1.

    Write down null and alternative hypotheses for Loreto's test.

    [2]
    Verified
    Solution

    let XXX be the random variable "number of patients arriving in a minute", such that X∼Po⁡(m)X \sim \operatorname{Po}(m)X∼Po(m). H0:m=1.5\mathrm{H}_{0}: m=1.5H0​:m=1.5 A1 H1:m>1.5\mathrm{H}_{1}: m>1.5H1​:m>1.5 A1

    Note: Allow a value of 270 for mmm. Award at most A0A1 if it is not clear that it is the population mean being referred to e.g H0\mathrm{H}_{0}H0​: The number of patients is equal to 1.5 every minute H1\mathrm{H}_{1}H1​: The number of patients exceeds 1.5 every minute. Referring to the "expected" number of patients or the use of μ\muμ or λ\lambdaλ is sufficient for A1A1.

    2.

    Using the data from Loreto's sample, perform the hypothesis test at a 5% significance level to determine if Loreto should employ extra staff.

    [5]
    Verified
    Solution

    under H0\mathrm{H}_{0}H0​ let YYY be the number of patients in 3 hours Y∼Po⁡(270)Y \sim \operatorname{Po}(270)Y∼Po(270) (A1) P(Y≥320)(=1−P(Y≤319))=0.00166(0.00165874)\mathrm{P}(Y \geq 320)(=1-\mathrm{P}(Y \leq 319))=0.00166(0.00165874)P(Y≥320)(=1−P(Y≤319))=0.00166(0.00165874) (M1)A1 since 0.00166<0.050.00166<0.050.00166<0.05 R1 (reject H0\mathrm{H}_{0}H0​) Loreto should employ more staff A1

    3.

    Write down null and alternative hypotheses for this test.

    [2]
    Verified
    Solution

    H0\mathrm{H}_{0}H0​: The probability of a patient waiting less than 20 minutes is 0.950.950.95 A1 H1\mathrm{H}_{1}H1​: The probability of a patient waiting less than 20 minutes is less than 0.95 A1

    4.

    Perform the test, clearly stating the conclusion in context.

    [5]
    Verified
    Solution

    under H0\mathrm{H}_{0}H0​ let WWW be the number of patients waiting more than 20 minutes W∼B(150,0.05)W \sim \mathrm{B}(150,0.05)W∼B(150,0.05) (A1) P(W≥11)=0.132(0.132215…)\mathrm{P}(W \geq 11)=0.132(0.132215 \ldots)P(W≥11)=0.132(0.132215…) (M1)A1 since 0.132>0.10.132>0.10.132>0.1 R1 (fail to reject H0\mathrm{H}_{0}H0​) insufficient evidence to suggest they are not meeting their target A1

    Note: Do not accept "they are meeting target" for the A1. Accept use of B(150,0.95)\mathrm{B}(150,0.95)B(150,0.95) and P(W≤139)\mathrm{P}(W \leq 139)P(W≤139) and any consistent use of a random variable, appropriate ppp-value and significance level.

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