In an experiment to determine whether there is a systematic difference between the weights obtained with two different mass balances, six specimens were weighed, in grams, on each balance. The following data were obtained: Specimen A B
1 13.76 13.74
2 12.47 12.45
3 10.09 10.08
4 8.91 8.92
5 13.57 13.54
6 12.74 12.75
Can you conclude that the mean weight differs between the two balances?
i). State the null and alternative hypotheses.
ii). Compute the test statistic.
iii). State a conclusion using the a =0.05 level of significance.

H0: μd=0 Ha: μd≠0

t= 0.07607

On the basis of this we conclude that the mean weight differs between the two balances.

Step-by-step explanation:

The null and alternative hypotheses as

H0: μd=0 Ha: μd≠0

Significance level is set at ∝= 0.05

The critical region is t ( base alpha by 2 with df=5) ≥ ± 2.571

The test statistic under H0 is

t = d/ sd/ √n

Which has t distribution with n-1 degrees of freedom

Specimen        A               B           d = a - b         d²

1                     13.76        13.74         0.02           0.004

2                    12.47        12.45          0.02         0.004

3                    10.09        10.08           0.01        0.001

4                       8.91       8.92          -0.01          0.001

5                     13.57      13.54           0.03        0.009

6                     12.74      12.75          -0.01        0.001

∑                                                      0.06         0.0173

d`= ∑d/n= 0.006/6= 0.001

sd²= 1/6( 0.0173- 0.006²/6) = 1/6 ( 0.017294) = 0.002882

sd= 0.05368

t= 0.001/ 0.05368/ √6

t= 0.18629/2.449

t= 0.07607

Since the calculated value of t= 0.07607 does not falls in the rejection region we therefore accept the null hypothesis at 5 % significance level . On the basis of this we conclude that the mean weight differs between the two balances.

step-by-step explanation:

we are given a graph for system of inequalities.

in order to write the system of inequalities for the given graph, we need to find the inequality for each of the line separately.

let us find the slope of dotted line first.

taking two coordinates on the graph.

let us say (0,0) and (3,1).

let us find the slope of the line now.

slope of the dotted line would be =.

y-intercept of dotted line is y=0 only.

plugging slope m= and b=0 in slope-intercept form y=mx+b, we get

.

the line is shaded up of that dotted line.

therefore, for shading up we would take greater than inequality symbol > .

and because it's a dotted line we would take only > symbol in place of = sign in above equation of line.

therefore, inequality for dotted line would be

let us take two points (3,1) and (0,4) on the solid line.

slope of the solid line would be =.

for the solid line we have y-intercept b= 4.

plugging slope m= -1 and b=4 in slope-intercept form y=mx+b, we get

y = -1x + 4.

the line is shaded down of that solid line.

therefore, for shading down we would take less than inequality symbol < .

and because it's a solid line we would take ≤ symbol in place of = sign in above equation of line.

therefore, inequality for solid line would be

a

step-by-step explanation: