Will you please help me with the following problem? Two pain relief drugs are be

Question

Will you please help me with the following problem? Two pain relief drugs are being considered. A random sample of 8 doses of the first drug showed that the average amount of time required before the drug was absorbed into the bloodstream was x1 = 24 minutes with population standard deviation = 4 minutes. For the second drug, a random sample of 10 doses showed the average time required for absorption was x2 = 29 minutes with population standard deviation = 3.9 minutes.The absorption times follow an approximately normal distribution. a) Should you use the critical value (zc) or (tc) in determining a confidence interval in this problem? Explain. (If I have cigma I use z, if I don't have cigma I use t....right?) b) Find a 90% confidence interval for the difference in average absorption time for the two drugs. c) Does it appear that one drug is absorbed faster than the other (at the 90% level)? Explain. So for this problem would I use this formula?: (x1-x2) - E < u1-u2 <(x1-x2) + E where E = z sqrt (o 2/1)/n1) + (o 2/2)/n2) totally don't know how to type that! Please help! So confusing.

Explanation / Answer

Two pain-relief drugs are being considered. A random sample
of 8 doses of the first drug showed that the average amount of
time required before the drug was absorbed into the bloodstream was 24 minutes with standard deviation s1 = 3 minutes. For the second drug, a random sample of 10 doses showed the
average time required for absorption was 29 minutes with
standard deviation s2 = 4.9 minutes. Assume that the absorption
times follow a normal distribution. Find a 90% confidence
interval for the difference in average absorption times for the
two drugs. Does it appear that one drug is absorbed faster
than the other (at the 90% level)?

Assuming that the time populations have the same variance then a pooled
estimate of this variance is (7X3^2+9X4.9^2)/(7+9)=17.443
For the critical value a t value with 16 degrees of freedom, using p=0.975
t=2.120 so the interval is
((29-24) - 2.120sqrt[(17.443)(1/8+1/10), (29-24) + 2.120sqrt[(17.443)(1/8+1/10))
=(0.800,9.200)
This is for 2-1 and provides evidence that 2>1 but not at the 90%
level since the interval is two-sided and a test for 2>1 is one-sided.
My calculation assumes that the standard deviations are based on unbiased
estimates of the two variances. and according to the choices it is not the case.
You will have to recalculate the pooled estimate again

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