1 a) t = (34 - JustAnswer



1 a) t = (34.43 – 34) / (2.53 / √43) = 1.115

b) t = (33.57 – 34) / (2.53 / √43) = –1.115

c) t = (33.2 – 34) / ( 2.05 / √60) = – 3.023

d) t = (34.38 – 34) / (1.8 / √100) = 2.111

2. Ho: u = 37, Ha: u > 37

Use T-statistic with 34 degrees of freedom. Reject Ho if t > 1.692 (from T table or calculator)

t = ( 37.5 – 37) / (1 / √35) = 2.958 so reject Ho. Conclusion is that viral infection elevates body temperature.

3. Ho: u1 = u2, Ha: u1 > u2 where u1 is the true drug mean and u2 is the true placebo mean. Then use the 2 sample T statistic which is

T = (SM1 – SM2) / SQRT{S^2*[(n1 + n2) / (n1*n2)]}

with n1 + n2 – 2 = 68 degrees of freedom. Reject Ho if T > 2.382. SM1 is sample mean on drug, SM2 is sample mean on placebo, S^2 is the pooled estimate of variance, n1 and n2 are the sample sizes, and SQRT is the square root. First, compute the pooled estimate of variance (assuming variances of two groups are equal). It is

S^2 = [(n1–1)*s1^2 + (n2–1)^s2^2] / (n1 + n2 – 2)

where s1^2 is the sample variance of drug and s2^2 is the sample variance of placebo (assuming values given in question are standard deviations).

S^2 = [34*12.05^2 + 34*13.78^2] / (35 + 35 – 2)

= 167.55

T = (25.32 – 31.96) / SQRT{167.55 [70 / 1225]}

= – 6.64 / 3.094

= – 2.146

T is much less than 2.382 so Ho can not be rejected. The conclusion is that the drug did not reduce uterine cramping pain.

4. Ho: P1 = P2, Ha: P1 > P2 where P1 is the true dementia rate in the hormone group and P2 is the dementia rate in the placebo group. The

Z-statistic can be used to test the hypothesis. It is

Z = (p1 – p2) / SQRT[p*(1-p)*( 1/n1 + 1/n2)]

where p1 and p2 are the observed rates for the 2 groups and p is the combined rate. Reject Ho if Z > 2.327 (from Z table or calculator for alpha = 0.01)

Z = (40/2266 – 21/2266) / SQRT[ (61/4532)*(1-61/4532)*(1/2266 + 1/2266)]

= .00838/ SQRT[ .01328 * .000883]

= .00838 / .00342

= 2.448

Since Z = 2.448 > 2.327 Ho can be rejected. The conclusion is that dementia has a higher rate in those treated with Prempro.

5. a) Ho: p = 2/3

b) Ha: p > 2/3

c) This is a one sided test only wanting to reject Ho if the observed p is much greater than 2/3. Using the standard normal approximation, Z, reject if Z > 1.645.

Z = {(Sample p – p) / SQRT[p*(1–p)] / n)

= {( 164/200 – 2/3) / SQRT(2/3*1/3 / 100)

= {.82 – 2/3} / .0471

= .1533 / .0471

= 3.255

Can reject Ho since Z = 3.255 > 1.645. The pratical conclusion is that the screening program is effective and therefore should be used.

d) P-value is P( Z > 3.255) = .0006 from Z calculator. The interpretation is that if there is no effect of the screening program then this or more extreme outcome is expected to occur in only 0.0006 of all trials.

6. No question

7. a) The 99% confidence interval, CI, of a difference in 2 means is

(Mean1 – Mean2) – z(.01)*SQRT{s1^2/n1 + s2^2/n2)}

< u1 – u2 < (Mean1 – Mean2) + z(.01)*SQRT{(s1^2/n1 + s2^2/n2)}

where SQRT is the square root. The CI is

(9017 – 5853) – 2.575*SQRT(394571 + 48069) < u1 – u2 < (9017 – 5853) + 2.575* SQRT(394571 + 48069)

3164 – 2.575*665.3 < u1 – u2 < 3164 – 2.575*665.3

1451 < u1 – u2 ≤ 4877

b) Breaststroke swimmers practice from 1451 to 4877 meters more per week on the breaststroke than individual medley swimmers. The reason for the difference is that individual medley swimmers have to practice other strokes.

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