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Chapter 4 .

An ambulance service claims that it takes on an average 8.9 minutes to reach the destination in emergency calls. To check this the licensing agency has then timed on 50 emergency calls, getting a mean of 9.3 minutes with a standard deviation of 1.6 minutes. Is this acceptable at 5% LOS?

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Solution:

$\bar{X}=9.3, \ \mu=8.9, \ s=1.6, \ n=50$

(i) Null Hypothesis $(H_{o}): \ \mu_{1}=\mu_{2}$

Alternate Hypothesis $(H_{a}): \ \mu_{1} \ne \mu_{2}$

(ii) Test statistic

$Z=\cfrac{\bar{X}- \mu}{s/ \sqrt{n}}= \cfrac{9.3-8.9}{1.6 / \sqrt{50} }=1.7678$

(iii) Level of significance at 5%, i.e., $\alpha=5\% $ or $\alpha=0.05$

(iv) Critical Value $\Longrightarrow$ The value of $z_{a}$ at 5% level of significance from the table is 1.96.

(v) Decision: Since the computed value of $|z|=1.7678$ is less than the critical value $z_{a}=1.96,$ the null hypothesis is accepted.

$z \lt z_{a}$

$\therefore$ The claim is acceptable at 5% LOS.

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