Question

As part of a biological research project, researchers need to quantify the density of a certain type of malignant cell in blood. In order to assure the accuracy of measurement, two experienced researchers each make a sequence of separate counts of the number of such cells in the same blood sample. The 7 counts of the first researcher have a mean of 140.2 and a standard deviation of 17, while the 13 counts of the second researcher have a mean of 134.2 and a standard deviation of 15.1.

(a) Use a level 0.99 pooled variance confidence interval to compare the mean counts of the two researchers:

?≤μ1−μ2≤ ?

(b) Does the interval suggest that there is a difference in the mean counts of the two researchers?

Answer 1

a) The 99% confidence interval would be given by
`-15.277 \leq \mu_1 -\mu_2 \leq 27.277`

b) No, since the confidence interval contains the 0 we don't have enough evidence to conclude that we have significant differences between the two means.

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 =140.2` represent the sample mean 1

`\bar X_2 =134.2` represent the sample mean 2

n1=7 represent the sample 1 size

n2=13 represent the sample 2 size

`s_1 =17` sample standard deviation for sample 1

`s_2 =15.1` 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) S_p \sqrt{(1)/(n_1)+(1)/(n_2)}` (1)

And the pooled variance can be founded with the following formula:

`s^2_p=((n_x -1)s_x^2 +(n_y-1)s_y^2)/(n_x +n_y -2)`

`s^2_p=((7 -1)17^2 +(13-1)15.1^2)/(7 +13 -2)=248.34`

`S_p =15.759` the pooled deviation

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

`\bar X_1 -\bar X_2 =140.2-134.2=6`

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=7+13-2=18`

Since the Confidence is 0.99 or 99%, the value of
`\alpha=0.01` and
`\alpha/2 =0.005`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,18)".And we see that
`t_(\alpha/2)=2.88`

The standard error is given by the following formula:

`SE=S_p \sqrt{(1)/(n_1)+(1)/(n_2)}`

And replacing we have:

`SE=15.759\sqrt{(1)/(7)+(1)/(13)}=7.388`

Part a Confidence interval

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

`6-2.88(15.759)\sqrt{(1)/(7)+(1)/(13)}=-15.277`

`6+2.88(15.759)\sqrt{(1)/(7)+(1)/(13)}=27.277`

So on this case the 99% confidence interval would be given by
`-15.277 \leq \mu_1 -\mu_2 \leq 27.277`

Part b Does the interval suggest that there is a difference in the mean counts of the two researchers?

No, since the confidence interval contains the 0 we don't have enough evidence to conclude that we have significant differences between the two means.