Mathematics, 18.12.2019 19:31 manco6

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?

Answers: 2

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Answers

Answer from: Usman458

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.

Step-by-step 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:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2} S_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$ (1)

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

$s^2_p=\frac{(n_x -1)s_x^2 +(n_y-1)s_y^2}{n_x +n_y -2}$

$s^2_p=\frac{(7 -1)17^2 +(13-1)15.1^2}{7 +13 -2}=248.34$

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:

$SE=S_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$

And replacing we have:

$SE=15.759\sqrt{\frac{1}{7}+\frac{1}{13}}=7.388$

Part a Confidence interval

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

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

$6+2.88(15.759)\sqrt{\frac{1}{7}+\frac{1}{13}}=27.277$

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.

Answer from: Quest

3; 9 is 8 to -4

step-by-step explanation:

Answer from: Quest

answer: 1.06 km per sec

step-by-step explanation:

39.6 m=63.72 km

60 min in an hour so 63.72/60=1.06

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