STA 106 Homework 3 Solutions

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STA 106 Homework 3 Solutions


Book Problems

1.  (a)  The non-centrality parameter is : φ = ′ i ni (a(µ)i ′µ ′ )1

where we use our sample values to estimate σ , µi , and µ 2 . So, = ′ i ′ (8(38 一 32)2 + 10(32 一 32)2 + 6(24 一 32)2 ) = 3.3599043

(b)  From the table, we use d.f{numγ = 2, d.f.{denomγ = 21 (round down to be conservative), and φ ψ 3.5. Using those values, the approximate power is 0.99.

(c)  The estimated effect size (using sample values is): /σ∈ = (38 一 24)/′ 19.8424 = 3.1429027

(d)  Using the power table online, with a = 3, α = 0.10, 1 一 β = 0.95, and ∆/′MSE ψ 3, we find ni = 4 for all groups.

2. (a)

Yij = µ 2 + γi + eij , where i γi = 0, and the assumptions are:

i. A random sample of Yij was taken.

ii. The i groups are independent

iii. eij ζ N(0, variance = σ)

(b)

The estimated values of each γi are: i = i2 一 , or

1 = 38 一 31.3333333 = 6.6667, 2 = 32 一 31.3333333 = 0.6667 3 = 24 一 31.3333333 = 一7.3333

(c)

When a subject is in group II , the average value of Y tends to be 0.6667 higher than the overall mean.

(d)

i = (6.6667) + (0.6667) + (-7.3333) = 0

(e)

It is normal, with mean E{γˆi γ = µi 一

3. (a)

(b)

The confidence interval formula is: 1 ≥ t1 ′α/2,nT ′a ′MSE ( ) or 38 ≥ (2.08)( ′ 19.8424( ) or 38 ≥ (2.08)(1.5749) or (34.7242,41.2758) We are 95% confident that the true average time to complete physical therapy for the “below average” group is between 34.7242 and 41.2758 days.

The confidence interval formula is: I 一 III ≥ t1 ′α/2,nT ′a ′MSE ( + ) or (38 一 24) ≥ (2.831)( ′ 19.8424( + ) or (38 - 24 ) ≥ (2.831)(2.4057) or (7.1895,20.8105) We are 99% confident that the true average time to complete physical therapy for the “below average” group is longer than the “above average” group by between 7.1895 and 20.8105 days.

(c)

It does, since bound of the bounds for the confidence interval are above 30.

(d)

I expect the largest difference is 20.8105 days (with the above average group taking longer).








2022-05-25