The fire department tested two types of fire extinguishers on gasoline fires. Th

Question

The fire department tested two types of fire extinguishers on gasoline fires. They set 10 gallons of gasoline ablaze and then timed how long it took to extinguish the blaze using one extinguisher. The process was repeated 6 times for two types of fire extinguishers (Type 1 and Type 2). Assuming the population variances are equal:

a) Use a 5% significance interval to test the claim that it takes longer to put out the fire using a Type 2 fire extinguisher (I finished this problem correctly I believe. I attached a photo of my work).

b) Form a 90% confidence interbal for for the difference of the true means in the time needed to put out a fire using two fire extinguishers

The p-value needs to be in this format: ?<p<? obtained from a t-table (I attached this for you) by using degrees of freedom (5) and alpha. Thank you for your help.

TYPE N MEAN STD. DEVIATION STD. ERROR MEAN Type 1 6 12.8333 1.47196 0.60093 Type 2 6 15.3333 4.32049 1.7638 ses, indebet norm 15-3335 2 mI n7 6 ( Fail to reject 6) There is not S5fv to surport that the meantime of Eype is lese than the Mean time of tyrelD

Explanation / Answer

a)

Below are the null and alternate hypothesis

H0: mu1 - m2 = 0

Ha: mu1 - mu2 < 0

n1 = 6

n2 = 6

x1bar = 12.8333

x2bar = 15.3333

sigma1 (s1) = 1.4720

sigma2 (s2) = 4.3205

As population variances are equal, we use pooled t-test

Pooled estimate of variance

Sp = sqrt(((n1-1)*s1^2+(n2-1)*s2^2)/(n1+n2-2))

Sp = sqrt((5*1.47196^2 + 5*4.32049^2)/10)

Sp = 3.2275

SE = 3.2275*sqrt(1/6+1/6)

SE = 1.8634

Test statistics, t = (12.8333 - 15.3333)/1.8634

t = -1.3416

df = 10

p-value = 0.1187

alpha(0.1) < p-value < alpha(0.05)

p-value > alpha(0.05)

As p-value is greater than the significance level of 0.05, we fail to reject the null hypothesis

There is no significant evidence to support that mean time of type 1 is less than the mean time of type 2

b) CI

x1bar - x2bar = -2.50

SE = 1.8634

CI = 95%

DF = n1+n2-2 = 10

t-value = 1.8125

ME = t*SE = 1.8125*1.8634 = 3.3773

Confidence Interval (x1bar - x2bar) +/- ME

Lower bound = -5.87732

Upper bound = 0.87732

Confidence Interval (-5.8773 , 0.8773 )

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