131k views
2 votes
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?

1 Answer

2 votes

Answer: the value for the associated test statistic is 1.2653

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

User Saltycrane
by
7.6k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories