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?

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

thus, the first answer choice, y = sec x, is the correct one