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Material Type: Exam; Class: Basic Engg 3:Problty&Stats; Subject: Basic Engineering; University: Wayne State University; Term: Fall 2010;
Typology: Exams
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were marketable after 240 days. The null hypothesis is that radiation was not beneficial. The alternative hypothesis is that radiation does improve the shelf-life (i.e. how long it can be sold) of a garlic bulb. The level of confidence to be used is 0.. What is the rejection region to be used, what is the value of the test statistic and do you reject or not? a) rejection |z| > 1.88, test stat. z = 4.20 reject b) rejection |z| > 2.33, test stat. z = 2.15 do not reject c) rejection |z| > 1.65, test stat. z = 2.67 reject d) rejection |z| > 1.88, test stat. z = 1.15 do not reject e) rejection |z| > 2.33, test stat. z = 4.20 reject
Mixture B .72 .69 .87 .78. Mixture C .62 1.08 1.07 .99.
Complete the above table. Explain whether or not the data indicate that the mixtures (treatments) are a statistically significant factor in the degree of soiling at the 0.05 level. Write your explanation below.
4 Measurements Hour 1 2 3 4
x ± Z 2
x ± t 2
p ˆ^ ± Z 2
n
pq = 133/539 ± 2.33* 539
p= (153+119)/(180+180) = 0.
test statistic Z =
α (^) = F10,10,0.025= 3. Test statistic F = S 12 / S 22 = 277.3^2 /205.9^2 = 1. Do not reject Null Hypothesis since F is not greater than Fα
Sp^2 = 6 8 2
Test Statistic t =
Rejection region t > t0.025 (2.179, From tables) Reject since test statistic t > t0.
4 Measurements