Question

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The Scottsdale fire department aims to respond to fire calls in 7 minutes or less, on average. Response times are normally distributed with a standard deviation of 1 minute 42 seconds. Would a sample of 18 fire calls with a mean response time of 7 minutes 25 seconds provide sufficient evidence to show that the goal is not being met at a = .01? (a) Choose the appropriate hypotheses. a. Haus 7 min vs Hf u> 7 min. Reject Ho if calc> 2.326 b. Hauz 7 min vs Hu<7 min, Reject Hoif calc>2.326 Ob (b) Calculate the value of the test statistic. (Round your answer to 3 decimal places.) calc D (c) Use Excel to find the p-value. (Round your answer to 4 decimal places.) p-value (d) If a = 01, we would Fail to reject the null hypothesis. Reject the null hypothesis.

Answer 1

Solution:

Given:

Standard Deviation =

0=1 min 42 sec

1 min 42 sec = 1 + (42/60) = 1 + 0.7 = 1.7 minutes

thus

0 = 1.7 min

Sample Size = n = 18

Sample mean =

T= 7 min 25 sec

7 min 25 sec = 7 + (25/60) = 7 + 0.4167 = 7.4167 min.

Level of significance = 0.01

Since statement of the hypothesis is: The Scottsdale fire department aims to responds to fire calls in 7 minutes or less, then alternative hypothesis is Mean > 7 minutes

Thus hypothesis of the study are:

Ho :μ<7 Vs H, :μ>7

Find z critical value at 0.01 level of significance for right tailed test:

Since this is right tailed test , we use probability = 1 - 0.01 = 0.99

Use excel command:

=NORM.S.INV(probability)

=NORM.S.INV(0.99)

=2.326

Thus correct option is:

a) , Reject H0, if z_calc > 2.326

Ho :μ<7 Vs H, :μ>7

Part b) calculate z test statistic:

urlo

7.4167 – 7 1.7/18

10.4167 17/4.242641

0.4167 10.4006938

2= 1.040

Part c) p-value:

p-value = P( Z> z calculated)

p-value = P( Z> 1.040 )

p-value =1 - P( Z< 1.040 )

Thus

=1-NORM.S.DIST(1.040,TRUE)

=0.1492

Thus

p-value = 0.1492

Part d)

Since p-value = 0.1492 > 0.01 level of significance , we fail to reject null hypothesis H0.