Answer to Question #129431 in Statistics and Probability for Kimathi

Question

) An ambulance service claims that it takes, on average 8.9 minutes to reach its destination in
emergency calls. To check on this claim, the agency which licenses ambulance services had them
timed on 50 minutes emergency calls, getting a mean of 9.3 minutes with a standard deviation of
1.8 minutes. At the level of significance of 0.05, does this constitute evidence that the figure
claimed is too low? [5 Marks]

Answer 1

$\bar{X}=9.3, μ=8.9, s=1.6, n=50$

(i) Null Hypothesis (H_o): μ₁=μ₂

Alternate Hypothesis (H_a): μ₁≠μ₂

(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., α=5%  or α=0.05

(iv) Critical Value⟹ 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<z_a

∴ The claim is acceptable at 5% LOS.