1 answer

=========================================== f=25.09 is wrong. Olestra is a fat substitute approved by the FDA for use in...

Question:

Olestra is a fat substitute approved by the FDA for use in snack foods. Because there have been anecdotal reports of gastroin===========================================
1 An article describes an experiment in which several types of boxes were compared with respect to compression strength (lb).f=25.09 is wrong.

Olestra is a fat substitute approved by the FDA for use in snack foods. Because there have been anecdotal reports of gastrointestinal problems associated with olestra consumption, a randomized, double-blind, placebo-controlled experiment was carried out to compare olestra potato chips to regular potato chips with respect to GI symptoms. Among 500 individuals in the TG control group, 16.6% experienced an adverse GI event, whereas among the 500 individuals in the olestra treatment group, 14.8% experienced such an event. USE SALT (a) Carry out a test of hypotheses at the 5% significance level to decide whether the incidence rate of GI problems for those who consume olestra chips according to the experimental regimen differs from the incidence rate for the TG control treatment. (Use P, for the TG control group and P2 for the Olestra treatment group.) State the relevant hypotheses. Ho: P1 P2 = 0 H: 41 - My > 0 O Ho: P1-P2 > 0 H: P1 - P2 = 0 Ho: P1 P2 = 0 H: P2 - P2 = 0 Ho: P1 P2 < 0 H: P1 - P2 = 0 O Ho: P1 - P2 = 0 H.: P1 - P20 Calculate the test statistic and P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) z = 1.1064 P-value = State the conclusion in the problem context. Reject Ho. The data suggest a statistically significant difference between the incidence rates of GI problems between the two groups. Fail to reject Ho. The data suggest a statistically significant difference between the incidence rates of GI problems between the two groups. Fail to reject Ho. The data do not suggest a statistically significant difference between the incidence rates of GI problems between the two groups. Reject Ho. The data do not suggest a statistically significant difference between the incidence rates of GI problems between the two groups. (b) If the true percentages for the two treatments were 20% and 25%, respectively, what sample sizes (m = n) would be necessary to detect such a difference with probability 0.90? (Round your answer up to the nearest whole number.)
1 An article describes an experiment in which several types of boxes were compared with respect to compression strength (lb). The table below presents the results of a single-factor ANOVA experiment involving I = 4 types of boxes. Type of Box Compression Strength (lb) Sample Mean Sample SD 655.5 788.3 734.3 721.4 679.1 699.4 713.00 46.55 789.2 772.5 786.9 686.1 732.1 774.8 756.93 40.34 3 737.1 639.0 696.3 671.7 717.2 727.1 698.07 37.20 4 535.1 628.7 542.4 559.0 586.9 520.0 562.02 39.87 Suppose that the compression strength observations on the fourth type of box had been 645.1, 738.7, 652.4, 569.0, 595.9, and 530.0 (obtained by adding 110 to each previous Xa;). Assuming no change in the remaining observations, carry out an F test with a = 0.05. 2 State the appropriate hypotheses. Ho: 41 = 42 = 13 = 44 Ha: at least two u;'s are unequal O Ho: 41 = 42 = 13 = 44 He: all four ;'s are unequal Ho: M, * 42 43 44 Hy: at least two y's are equal 0 Hà: A. * 42 + 3 + 44 Hy all four 's are equal Calculate the test statistic. (Round your answer to two decimal places.) f=

Answers

1)

a)

1st group 2nd group
x=    83 74
n = 500 500
p̂=x/n= 0.1660 0.1480
estimated prop. diff =p̂1-p̂2    = 0.0180
pooled prop p̂ =(x1+x2)/(n1+n2)= 0.1570
std error Se=√(p̂1*(1-p̂1)*(1/n1+1/n2) = 0.0230
test stat z=(p̂1-p̂2)/Se = 0.78
P value   = 0.4354 (from excel:2*normsdist(-0.78)

b)

first sample proportion p1 = 0.2
second sample proportion p2= 0.25
0.05 level critical Z= 1.96
0.1 level critical Zβ= 1.28
n=((zα(√(p1+p2)*(q1+q2)/2)+zβ(√p2q2+p2q2))/(p1-p2))2= 1465


2)Applying ANOVA:

Source of Variation SS df MS F P-value
Between Groups 22781.2546 3 7593.7515 4.4882 0.0145
Within Groups 33838.9750 20 1691.9488
Total 56620.2296 23

f =4.49

.

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