A random sample of 17 students, taught in traditional lab sessions, had a mean test score of 71.5 with a standard deviation of 4.8. A random sample of 13 students, taught using interactive simulation software, had a mean test score of 78.5 with a standard deviation of 6. Do these results support the claim that the mean science test score is lower for students taught in traditional lab sessions than it is for students taught using interactive simulation software

The calculated value of t= -0.9468 does not lie in the critical region t= -1.701. Therefore we accept our null hypothesis that mean science test score is lower for students taught in traditional lab sessions than it is for students taught using interactive simulation software at the 0.05 significance level

Step-by-step explanation:

Categories         Sample Size           Mean           Standard D

Traditional Lab        17                        71.5                    4.8

Software                 13                        78. 5                    6

We are using Student's t- test considering that the population variance is equal.

We formulate null and alternate hypotheses are

H0 : u1 < u2 against Ha: u1 ≥ u 2

The Significance level alpha is chosen to be ∝ = 0.05

The critical region t ≤ -t (0.05, 28) = -1.701

Degrees of freedom is calculated df = υ= n1+n2- 2= 17+13-2= 28

Here the difference between the sample means is x`1- x`2= 71.5-78.5= -7

The pooled estimate for the common variance σ² is

Sp² = 1/n1+n2 -2 [ ∑ (x1i - x1`)² + ∑ (x2j - x`2)²]

= 1/17+13-2 [ 71.5²+78.5²]

Sp = 20.0664

The test statistic is

t = (x`1- x` ) /. Sp √1/n1 + 1/n2

t= -7/ 20.0664 √1/17+ 1/13

t= -7/7.393

t= -0.9468

The calculated value of t= -0.9468 does not lie in the critical region t= -1.701. Therefore we accept our null hypothesis that mean science test score is lower for students taught in traditional lab sessions than it is for students taught using interactive simulation software at the 0.05 significance level

the quotient is 4x^2 - 27x + 167 with a remainder of -996

step-by-step explanation:

in synthetic division, use -6 as the divisor and use the coefficients {4, -3, 5, 6}:

-6     )   4     -3     5     6

                -24   162 -1002

          4   -27   167   -996

thus, the quotient is 4x^2 - 27x + 167 with a remainder of -996.

note:   your "4x°-3x+5x+6" is ambiguous / incomplete.   i had to assume that you meant '4x^3 - 3x^2 + 5x + 6."

3-4

step-by-step explanation:

the answer is 3-4 because both 45 and 60 has a multiple of 15.

45/3=15

60/4=15

therefore the odds against selecting a red is 4-3