Answer to Question #184434 in Statistics and Probability for Sunny Paul Paulose

Question

An ambulance service claims that it takes on the average 8.9 minutes to reach its destination in emergency calls. To check on this claim, the agency which licenses ambulance services has then timed on 50 emergency calls, getting a mean of 9.3 minutes with a standard deviation of 1.8 minutes. Does this constitute evidence that the figure claimed is two low at 1 per cent significance level?

Answer 1

Answer:

$\bar{X}=9.3, μ=8.9, s=1.8, n=50$

(i) Null Hypothesis (H_o): μ₁=μ₂

Alternate Hypothesis (H_a): μ₁≠μ₂

(ii) Test statistic

$Z=\cfrac{\bar{X}- \mu}{s/ \sqrt{n}}= \cfrac{9.3-8.9}{1.8 / \sqrt{50} }=1.571$

(iii) Level of significance at 1%, that is, α=1% or α=0.01

(iv) Critical Value⟹ The value of z_a at 1% level of significance from the table is 0.1161.

(v) Decision: Since the computed value of |z|=1.571 is less than the critical value z_a=0.1161, the null hypothesis is accepted.

z>z_a

 Therefore, the claim is not acceptable at 0.01% LOS (level of significance).

From the analysis, this does not constitute evidence that the figure claimed is too low at 1 per cent significance level.