Question

A paint manufacturer made a modification to a paint to speed up its drying time. Independent simple random samples of 11 cans of type A (the original paint) and 9 cans of type B (the modified paint) were selected and applied to similar surfaces. The drying times, in hours, were recorded.

The summary statistics are as follows.
Type A Type B
x1 = 76.3 hrs x2 = 65.1 hrs
s1 = 4.5 hrs s2 = 5.1 hrs
n1 = 11 n2 = 9
The following 98% confidence interval was obtained for μ1 - μ2, the difference between the mean drying time for paint cans of type A and the mean drying time for paint cans of type B:
4.90 hrs < μ1 - μ2 < 17.50 hrs
What does the confidence interval suggest about the population means?

Answer 1

We can conclude that the drying time in hours for type A is significantly higher than the drying time for type B. And the margin above it's between 4.90 and 17.5 hours at 2% of significance.

Explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

`\bar X_1 =76.3` represent the sample mean 1

`\bar X_2 =65.1` represent the sample mean 2

n1=11 represent the sample 1 size

n2=9 represent the sample 2 size

`s_1 =4.5` sample standard deviation for sample 1

`s_2 =5.1` sample standard deviation for sample 2

`\mu_1 -\mu_2` parameter of interest.

Confidence interval

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

`(\bar X_1 -\bar X_2) \pm t_(\alpha/2)\sqrt{(s^2_1)/(n_1)+(s^2_2)/(n_2)}` (1)

The point of estimate for
`\mu_1 -\mu_2` is just given by:

`\bar X_1 -\bar X_2 =76.3-65.1=11.2`

In order to calculate the critical value
`t_(\alpha/2)` we need to find first the degrees of freedom, given by:

`df=n_1 +n_2 -1=11+9-2=18`

Since the Confidence is 0.98 or 98%, the value of
`\alpha=0.02` and
`\alpha/2 =0.01`, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.01,18)".And we see that
`t_(\alpha/2)=\pm 2.55`

The standard error is given by the following formula:

`SE=\sqrt{(s^2_1)/(n_1)+(s^2_2)/(n_2)}`

And replacing we have:

`SE=\sqrt{(4.5^2)/(11)+(5.1^2)/(9)}=2.175`

Confidence interval

Now we have everything in order to replace into formula (1):

`11.2-2.55\sqrt{(4.5^2)/(11)+(5.1^2)/(9)}=5.65`

`11.2+2.55\sqrt{(4.5^2)/(11)+(5.1^2)/(9)}=16.75`

So on this case the 98% confidence interval would be given by
`5.65 \leq \mu_1 -\mu_2 \leq 16.75`

But let's assume that the confidence interval given is true 4.90 hrs < μ1 - μ2 < 17.50 hrs

What does the confidence interval suggest about the population means?

We can conclude that the drying time in hours for type A is significantly higher than the drying time for type B. And the margin above it's between 4.90 and 17.5 hours at 2% of significance.