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 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

Answer 1

a)
`\S^2_p =((6-1)(1.47196)^2 +(6 -1)(4.32049)^2)/(6 +6 -2)=10.417`

And the deviation would be just the square root of the variance:

`S_p=3.227`

And now we can calculate the statistic:

`t=\frac{(15.3333 -12.8333)-(0)}{3.227\sqrt{(1)/(6)+(1)/(6)}}=1.342`

Now we can calculate the degrees of freedom given by:

`df=6+6-2=10`

And now we can calculate the p value using the altenative hypothesis:

`p_v =P(t_(10)>1.342) =0.105`

If we compare the p value obtained and using the significance level assumed
`\alpha=0.05` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

b)
`(\bar X_2 -\bar X_1) \pm t_(\alpha/2)* S_p \sqrt{(1)/(n_!) +(1)/(n_2)}`

And replacing we got:

`(15.8333-12.8333) -1.812 \sqrt{(1)/(6) +(1)/(6)}= 1.954`

`(15.8333-12.8333) +1.812 \sqrt{(1)/(6) +(1)/(6)}= 4.046`

Explanation:

Part a

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

When we have two independent samples from two normal distributions with equal variances we are assuming that

`\sigma^2_1 =\sigma^2_2 =\sigma^2`

And the statistic is given by this formula:

`t=\frac{(\bar X_1 -\bar X_2)-(\mu_(1)-\mu_2)}{S_p\sqrt{(1)/(n_1)+(1)/(n_2)}}`

Where t follows a t distribution with
`n_1+n_2 -2` degrees of freedom and the pooled variance
`S^2_p` is given by this formula:

`\S^2_p =((n_1-1)S^2_1 +(n_2 -1)S^2_2)/(n_1 +n_2 -2)`

This last one is an unbiased estimator of the common variance
`\sigma^2`

The system of hypothesis on this case are:

Null hypothesis:
`\mu_2 \leq \mu_1`

Alternative hypothesis:
`\mu_2 > \mu_1`

Or equivalently:

Null hypothesis:
`\mu_2 - \mu_1 \leq 0`

Alternative hypothesis:
`\mu_2 -\mu_1 > 0`

Our notation on this case :

`n_1 =6` represent the sample size for group 1

`n_2 =5` represent the sample size for group 2

`\bar X_1 =12.8333` represent the sample mean for the group 1

`\bar X_2 =15.3333` represent the sample mean for the group 2

`s_1= 1.47196` represent the sample standard deviation for group 1

`s_2=4.32049` represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

`\S^2_p =((6-1)(1.47196)^2 +(6 -1)(4.32049)^2)/(6 +6 -2)=10.417`

And the deviation would be just the square root of the variance:

`S_p=3.227`

And now we can calculate the statistic:

`t=\frac{(15.3333 -12.8333)-(0)}{3.227\sqrt{(1)/(6)+(1)/(6)}}=1.342`

Now we can calculate the degrees of freedom given by:

`df=6+6-2=10`

And now we can calculate the p value using the altenative hypothesis:

`p_v =P(t_(10)>1.342) =0.105`

If we compare the p value obtained and using the significance level assumed
`\alpha=0.05` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

Part b

For the confidence interval we know that the confidence is 90% so then the value os
`\alpha = 0.1` and
`\alpha/2 =0.05`, the degrees of freedom are given by:

`df= n_1 + n_2 -2 = 6+6-2= 10`

And the crtitical value for this case would be
`t_(critc)= 1.812`

The confidence interval for the difference of means is given by:

`(\bar X_2 -\bar X_1) \pm t_(\alpha/2)*S_p \sqrt{(1)/(n_!) +(1)/(n_2)}`

And replacing we got:

`(15.8333-12.8333) -1.812 \sqrt{(1)/(6) +(1)/(6)}= 1.954`

`(15.8333-12.8333) +1.812 \sqrt{(1)/(6) +(1)/(6)}= 4.046`