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please answer all questions and use the formula sheet below for question 23.
20) statistics students must be t ssume that we want to estimate the mean 10 score for the pogaoe that the sample mean is wit

STATISTICS FORMULAS P-P) P(l- P) ,2, 囮zーーコ(x.-X1)sz, i.», po](F.-x.)±z, 应-.jr-r- n, n 10e n+n +n-2 P1-P N, N2 (尼-尼) 11b 12a
20) statistics students must be t ssume that we want to estimate the mean 10 score for the pogaoe that the sample mean is within statistics shudents. How many 95% confidence that the sample mean ts within andomly selected for 10 tests if we want 2 10 points of the population mean? Note that 10 tests are typicalls standard devintion is 15 error Ey? (a) Decrease ow to improve confidence interval (i.e. make it tighter by decreasing the margine o increase the level of confidence, (b) Decrease n and decrease the level of confidence. (e) Increase n and/or t ce the level of confidence. (d) Increase n and increase the level of confidence 22) How (a) Increase the level of confidence 1-alpha and decrease the margin of error E. (b) Increase the level ce 1-alpha and increase the margin of error E. (c) Decrease the levet of confidence 1-alpha and decrease to decrease the si of error E12ccrease the sample size needed to get certain accuracy (i.e sample size needed to get particular margin error E)? of confidence rgin of error E. (d) Decrease the level of confidence 1-alpha ot increase the margin of error E n a test of effectiveness of Echinacea against cold viruses, 47 out of 72 subjects treated with Echinacea developed virus. In the placebo group 85 out of 100 subjects developed virus. Use 0.01 significance level to test the claim that the placebo and drug groups differ. (Note here it is easier to use calculator rather than conduct all the steps) 23) What are the formulas, parameters, resulting statistic, and p-value describing this problem (a) 9c, 9d Ps)-0.65, Ps 0.85, Spoed0.0025, -3.02, p-value-0.77 (b) 10c, 10d, 10e Ps:-О.65, PS2-0.85, P-o0025, z--3.02, p-value-0.77 (c) 9c, 9d, 1ofr x, -0.65, X, 0.85, Spoled0.77,-3.02, p-value- 0.0025 f X,-0.65, X2-0.85, spooled-o77, t=-3.02, p-value-0.0025 (e) 10c, 10d, 10e Psi 0.65, P 0.85, P-0.77,-3.02, p-value 0.0025
STATISTICS FORMULAS P-P) P(l- P) ,2, 囮zーーコ(x.-X1)sz,' 'i.», po](F.-x.)±z, 应-.jr-r- n, n 10e n+n +n-2 P1-P N, N2 (尼-尼) 11b 12a! L,소 where E [120) ee.eo-Ecp-μ.»
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Answer #1

20. Population standard deviation, σ =15

Margin of error, E = 2

Significance level, α = 1-0.95= 0.05

Critical value, z = NORM.S.INV(0.05/2) = 1.96

Sample size, n = (z² * σ²) / E²

=216.09 = 216

closest value to this answer is option a) so Answer a) 284

21. Answer c) Increase n and/or reduce the level of confidence.

22. Answer d) Decrease the level of confidence 1-alpha or increase the margin of error E

23. Answer e)

Sample 1:

n1 =72

x1 = 47

p̂1 =0.6528

Sample 2:

n2 =100

x2 =85

p̂2 =0.85

α = 0.01

Null and Alternative hypothesis:

Ho : p1 = p2

H1 : p1 ≠ p2

Pooled proportion:

p̄ = (x1+x2)/(n1+n2)

=0.77

Test statistic:

z =(p̂1 - p̂2)/(√ [p̄*(1-p̄)*(1/n1+1/n2)])

= -3.02

p-value = 2*(1-NORM.S.DIST(ABS(-3.0204, 1) =0.0025

Decision:

p-value < α, Reject the null hypothesis

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