Question #5-(2 points) A group of engineering students decided to see whether ca

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Question #5-(2 points) A group of engineering students decided to see whether cars that supposedly do not need high-octane gasoline get more miles per gallon using regular or high-octane gas. They test several cars (under similar road surface, weather, and other driving conditions), using both types of gas in each car at different times. The mileage for each gas type for each car is reported in the table below. What is the maximal probability of committing type I error during the hypothesis test process? #1 15 18 Car #3 20 25 #4 35 34 #5 27 30 Regular 23 21 High-octane (a) 0.10 (b) 0.15 (c) 0.20 (d) 0.30 (e) 0.40 Question #6-(2 points) An experiment was conducted to assess the effect of using magnets at the filler point in the manufacture of coffee filter packs. Weight of filter packs in grams was used to evaluate the effect. 31 packs produced with magnets (X1 19.72 g and Siz = 1.58 g2) are to be compared against 41 packs produced without magnets ( 2-20.40 g and 52-035 g2). What is the 95% confidence interval estimate of the difference between the average weights of filter packs made with and without magnets and based on the results, what is the effect of adding magnets on the weight of filter packs (9.12)? (a) [-1.12, -0.23]; increasing the weight (b) [-1.12,-0.23]; reducing the weight (c) [-1.17,-0.19]; reducing the weight (d) [-1.17, -0.19]; increasing the weight (e) [-1.16, -0.20]; reducing the weight

Explanation / Answer

Question 5

Here we will use the paired t test. The difference table is

Average difference d =1.6

standard deviation of difference sd = 2.966

Here stadnarrd error of difference se0 = s/ sqrt(n) = 2.97/ sqrt(5) =  1.3264

Here test statistic

t = d/ (se0) = 1.6/ 1.3264 = 1.2063

so to have a type I error we shall reject the null hypothesis even if it is true and for maximum type I error we shall reject the null hypothesis at the given t.

Pr(t > 1.2063; dF = 5 -1 =4 ; one tailed) = 0.15

so 0.15 is the maximum type I Error.

question 6

95% confidence interval = (x1 - x2) +- tdF,0.05 sqrt [s12/n1 + s22 /n2]

HEre we will use degree of freedomfor unequal variances.

here dF= 41 by degree of freedom calculation for unequal variances

tdf,0.05  = 2.0195

95% confidence interval = (x1 - x2) +- tdF,0.05 sqrt [s12/n1 + s22 /n2]

= (19.72 - 20.40) +- 2.0195 * sqrt [1.58/31 + 0.35/41]

= -0.68 +- 0.4926

= (-1.17, -0.19) that reduce the weights

Regular High Oct Diff. 15 18 3 23 21 -2 20 25 5 35 34 -1 27 30 3 average 1.6 Std. dev. 2.97

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