Question

A marketing researcher conducts a test to determine whether the number of chocolate chips per cookie differs significantly across two brands of chocolate chip cookies. A random sample of 10 Chewy Delight cookies reveals a mean of 6.4 chocolate chips per cookie and a standard deviation of 1.1 chocolate chips per cookie. A random sample of 11 Chips Destroy cookies reveals a mean of 5.6 chocolate chips per cookie and a standard deviation of 1.7 chocolate chips per cookie. Assuming unequal population variances, what is the calculated value for the associated test statistic?

Answer 1

Answer: the value for the associated test statistic is 1.2653

Step-by-step explanation:

Given that;

sample size one n₁ = 10

mean one x"₁ = 6.4

standard deviation one S₁ = 1.1

sample size two n₁ = 11

mean two x"₂ = 5.6

standard deviation one S₁ = 1.7

H₀ : μ₁ = μ₂

H₁ : μ₁ ≠ μ₂

Pooled Variance

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

we substitute

= √( { [(10 - 1) × (1.1)² + (11 - 1) × (1.7)²] / (10 + 11 - 2)} × (1/10 + 1/11))

= √( { [(9) × 1.21 + (10) × 2.89]  / (19) } × (0.1909))

= √({[ 39.79 ]  / 19} × (0.1909))  

= √( 2.0942 × 0.1909)

= √( 0.39978 )

= 0.63228

Now Test Statistics will be;

t = ( x"₁ - x"₂) / sp

we substitute

t = ( 6.4 - 5.6) / 0.63228

t = 0.8 / 0.63228

t = 1.2653

Therefore the value for the associated test statistic is 1.2653