Question

A sample of 50 impossible whoppersandwiches have a mean fat content of 37.6g and standard deviation of 4.9g . A sample of 58 regular whopper sandwiches have a mean fat content of 39.3g and standard deviation of 2.7g. Use a 0.05 significance level to test the claim that the mean fat content for the impossible whopper is less than the mean fat content for the regular whopper.
(Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal.)

USE A 0.05 SIGNIFICANCE LEVEL TO TEST THE CLAIM:

H0: μ1____ μ2For Q17(circle one: equal to, less than, greater than, not equal to )

H1 : μ1____ μ2For Q18(circle one: equal to, less than, greater than, not equal to )

Test Statistic t=__________ Q19(ROUND TO 2 DECIMAL PLACES)

P-value: p =_____________ Q20(ROUND TO 4 DECIMAL PLACES)

Decision/Conclusion: _________________________________Q21(CIRCLE ONE BELOW) Reject H0, There is sufficient evidence to warrant the rejection of the claim.

OR

Fail to Reject H0, There is not sufficient evidence to warrant the rejection of the claim.

Construct an appropriate confidence interval ( ______________ Q22, _______________ Q23) (ROUND TO 3 DECIMAL PLACES)

Answer 1

Answer:

Step-by-step explanation:

H0: μ1____ μ2For Q17(circle one: equal to, less than or equal to, greater than or equal to, not equal to )

H1 : μ1____ μ2For Q18(circle one: equal to, less than, greater than, not equal to )

Formula for calculating the the test:

t = [ (x₁ - x₂) - d ] / sqrt[(s₁²/n₁) + (s₂²/n₂)]

where x₁ = 37.6, x₂ = 39.3, d = 0 assuming equality = 37.6 - 39.3 = -1.7

s₁² = 4.92 = 24.01, s₂² = 2.72 = 7.29, n₁ = 50, n₂ = 58.

thus, t = [ (-1.7) - (0)] / √(24.01/50) + (7.29/58)

    t = (-1.7) / (0.4802 + 0.1257)

   t = -1.7 / 0.6059

   t = -2.8057. = -2.81

P value:

Lets find the degree of freedom:

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

DF = (24.01/50 + 7.29/58)² / {[(24.01 / 50)² / 50-1)] + (7.29/58)² / 58-1)]}

DF = (0.4802 + 0.1257)² /{[ (0.4802)²/49] + (0.1257)²/57)]}

DF = (0.6059)²/ (0.00471 + 0.00028)

DF = 0.3671 / 0.00499 = 75.37

Thus, P value = 0.9968 using a t distribution calculator.

Decision/Conclusion: Since the P value 0.9968 is greater than α (.05), Fail to Reject H0, There is not sufficient evidence to warrant the rejection of the claim.