Two teaching methods and their effects on science test scores are being reviewed. A random sample of 18 students, taught in traditional lab sessions, had a mean test score of 78.3 with a standard deviation of 6.4 . A random sample of 11 students, taught using interactive simulation software, had a mean test score of 84.3 with a standard deviation of 5.3 . 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? Let μ 1 be the mean test score for the students taught in traditional lab sessions and μ 2 be the mean test score for students taught using interactive simulation software. Use a significance level of α=0.1 for the test. Assume that the population variances are equal and that the two populations are normally distributed. Step 2 of 4 :
Compute the value of the t test statistic. Round your answer to three decimal places.

the value of the t test statistic is - 3.419

Step-by-step explanation:

Given that;

n₁ = 18, u₁ = 78.3, s₁ = 6.4

n₂ = 11, u₂ = 84.3, s₂ = 5.3

α = 0.1

Now The hypothesis are;

H₀ : u₁ = u₂

H₁ : u₁ < u₂

To compute the value of the t test statistic;

t = [(x₁ - x₂) / s × √(1/n₁ + 1/n₂)]

where

s = √ [ ((n₁-1) × s₁² + (n₂ - 1 ) × s₂²) / ( n₁ + n₂ - 2)]

s = √ [ ((18-1) × 6.4² + (11 - 1 ) × 5.3²) / ( 18 + 11 - 2)]

s = √ [ (7 × 40.96 + 10 × 28.09 ) / 27 ]

s = √ [ (286.72 + 280.9) / 27 ]

s = √(567.62/27)

s = √21.0229

s = 4.585

Now  t test statistics t = [(x₁ - x₂) / s × √(1/n₁ + 1/n₂)]

t = [(78.3 - 84.3) / 4.585 × √(1/18 + 1/11)]

t = -6 / (4.585 × 0.3827)

t = - 6 / 1.7546795

t = - 3.419

Therefore the value of the t test statistic is - 3.419

as as level of significance α  = 0.1

df = 18+11-2 = 27

∴ T(csal) = t(0.1, 27) = -1.313

That is

t(statistics) < t(cal)

{ - 3.419  < -1.313 }

the answer is mp

step-by-step explanation:

sorry

step-by-step explanation:

sorry, but im pretty sure its a now that i look at it