Can someone please help

A professor wants to know whether or not there is a difference in comprehension of a lab assignment among students depending on

san4es73 [151]

Answer:

It is concluded that no difference exists in the comprehension of the lab based on the test scores.

Step-by-step explanation:

Let the populations be normally distributed with a populations standard deviation of 5.32 points for both the text and visual illustrations.

The above statement tells us that the independent t or two sample t test will be performed as both have equal variances and are normally distributed.

This can be easily done through excel.

The following table is obtained

t-Test: Two-Sample Assuming Equal Variances

Text Visual Illustrations

Mean 70.28 75.08

Variance 304.48 228.58

Observations 15 15

Pooled Variance Sp²= 266.53

Pooled Standard Deviation = Sp = 16.33

Hypothesized Mean Difference = x1`-x2`= 0

df = n1+n2-2= 15+15-2= 30-2= 28

t Stat -0.805188239

P(T<=t) two-tail 0.427495979

t Critical two-tail 1.701130908

Let the null and alternate hypotheses be

H0 : u1-u2= 0 against the claim Ha: u1-u2≠0

There is no difference between the means

against the claim

that there is a difference in means of a lab assignment among students depending on if the instructions are given all in text, or if they are given primarily with visual illustrations.

The significance level is ∝= 0.1

The d.f is n1+n2-2= 15+15-2= 30-2=28

This is a two tailed test and the critical region is t (0.025) (28) ≥ 1.7011 and t (0.025) (28) ≤ - 1.7011.

The test statistic is

t= x1-x2/ Sp √1/n1+ 1/n2

t= 70.28 -75.08/ 16.33√1/15 +1/15

t= -4.8/5.963

t= -0.8049811 ( minute difference from excel result due to rounding)

Since the calculated value of t= -0.8049 does not fall in the critical region we conclude that there is no difference in means of a lab assignment among students depending on if the instructions are given all in text, or if they are given primarily with visual illustrations.

We accept the null hypothesis.

The p- value is 0.427495979.

8 0

3 years ago

The length of triangle base is 36. The line which is parallel to the base divides the triangle into two equal area parts. Find a

Soloha48 [4]

Answer:

$18\sqrt{2}$

Step-by-step explanation:

Consider triangle ABC with the base AC=36. Let the line DE be parallel to the line AC. If DE||AC, then triangles ABC and DBE are similar. If k is the scale factor of these triangles, then

$\dfrac{DE}{AC}=\dfrac{BG}{BF}=k.$

Thus,

$DE=k\cdot AC=36k,\\ \\BG=k\cdot BF.$

The area of the triangle ABC is

$A_{ABC}=\dfrac{1}{2}AC\cdot BF=\dfrac{1}{2}\cdot 36\cdot BF=18BF.$

The area of the triangle DBE is

$A_{DBE}=\dfrac{1}{2}DE\cdot BG=\dfrac{1}{2}\cdot 36k\cdot kBF=18k^2BF.$

Since line DE divides triangle ABC in two equal area parts, we have that

$A_{DBE}=\dfrac{1}{2}A_{ABC},\\ \\18k^2BF=\dfrac{1}{2}\cdot 18BF,\\ \\k^2=\dfrac{1}{2},\\ \\k=\dfrac{1}{\sqrt{2}}$

and

$DE=\dfrac{1}{\sqrt{2}}\cdot 36=18\sqrt{2}.$

4 0

3 years ago

PLZ HELP ASAP !!! 25 POINTS !!!

jolli1 [7]

the first picture number 1 is a function

the second one the third one

but I'm not sure

7 0

3 years ago

Ham costs $4.48 per pound. Cheese costs $6.36 per pound. You buy 1.5 pounds of ham and 0.75 pound of cheese. How much more do yo

irinina [24]

$6.72= 1.5 lbs of ham
$4.77= 0.75 lbs of cheese
$6.72 - $ 4.77= $1.95
The answer is $1.95
I think

5 0

4 years ago

Read 2 more answers

Complete this sentence. The melting point of a sample will ____________ when the size of the sample increases.

s344n2d4d5 [400]

Answer: stay the same

The melting point is a temperature when the solid object become liquid. The melting point is not influenced by how many samples need to be melt, but by the kind of the object itself. Water still freeze/melt at 0 Celcius no matter how many liters the sample size.
Increase in sample size only increase the force needed.

3 0

3 years ago