Question

9 of 12 (2 complete) L J Score: 0 of 2 pts Question Help ? for mean cholesterol content of a hamburger from two fast food chains and confidence interval constructhion formula below Assume ?:"97mg, s:"2.22 mg":-8 8.2.24 Constrct a 90% confidence merval tr u,-us2 with the sample statstics the populations are approximately normal with unequal variances content of a 122 mg ,391 mg,n 1 Stats Confidence irten? when variances are not equal d1 isthe trnaler of n,-1or":-1 Enter the endpoints of the intervai ?<»,-u2 <?Rond to the nearest neger as needed )

Answer 1

Step 1: Find ?/2
Level of Confidence = 90%
? = 100% - (Level of Confidence) = 10%
?/2 = 5% = 0.05


Step 2: Find degrees of freedom and t_?/2
Degrees of freedom = smaller of (n₁ - 1 , n₂ - 1 ) = smaller of (16 , 7) = 7

Calculate t_?/2 by using t-distribution with degrees of freedom (DF) = 7 and ?/2 = 0.05 as right-tailed area and left-tailed area.

Step 3: Calculate Confidence Interval

t_?/2 = 1.89455

Standard Error = ? (s?)²/n? + (s?)²/n? = ?1.5153 = 1.2309

Lower Bound = (x?? - x??) - t_?/2•(? (s?)²/n? + (s?)²/n? ) = (122 - 97) - (1.89455)(1.2309) = 22.6678

Upper Bound = (x?? + x??) + t_?/2•(? (s?)²/n? + (s?)²/n? ) = (122 - 97) + (1.89455)(1.2309) = 27.3321

Confidence Interval = (22.6678174147, 27.3321825853)



Interpretation of a confidence interval:
Since we do not know if the confidence interval (22.6678, 27.3321) contains (?? - ??) or not, we are only 90% confident that (22.6678, 27.3321) contains (?? - ??).