the following data is the amount of time (in minutes) for a headache to be gone

ID: 3360976 • Letter: T

Question

the following data is the amount of time (in minutes) for a headache to be gone after taking medication 14,14,16,19,20,23

is the amount of time until the headache is gone from the population where this sample came from symmetric?

If so, answer the following:

a) what is the best single estimate for the amount of time for a headache to be gone after taking the medication?

b) How long will it take for the headache to be gone with 95% confidence? with 99.7% confidence?

c) Out of 2000 individuals who use the medicine, how many would you expect tohave the headache gone in 28.5 minutes or more?

d) Our of 2000 people who take the medicine, how many would you expect to have the headache gone in 10.5 to 21.3 minutes?

A) single estimate = (14 + 14 + 16 + 19 + 20 + 23)/ 6 = 17.67

B) standard deviation = sqrt((14 - 17.67)^2 + (14 - 17.67)^2 + (16 - 17.67)^2 + (19 - 17.67)^2 + (20 - 17.67)^2 + (23 - 17.67)^2)/6) = 3.3

95% cinfidence interval the critical value is 1.96

The confidence interval is

Mean +/- Zc * sd/Sqrt (n)

17.67 +/- 1.96 * 3.3/sqrt(6)

= 17.67 +/- 2.64

= 15.03, 20.31

At 99.7% confidence interval the critical value is Zc = 2.96

The cinfidence interval is

17.67 +/- 2.96 * 3.3/sqrt(6)

= 17.67 +/- 3.99

= 13.68, 21.66

C) P(X > 28.5) = P((x - mean)/(SD/sqrt(n) > (28.5 - 17.67)/(3.3/sqrt(6))

= P(Z > 8.04)

= 1 - P(Z < 8.04)

= 1 - 1 = 0

Expected value = 2000 * 0 = 0

D) P(10.5 < x < 21.3)

= P((10.5 - 17.67)/(3.3/sqrt(6) < (x - mean)/(SD/sqrt(n)) < (21.3 - 17.67)/(3.3/sqrt(6))

= P(-5.32 < Z < 2.69)

= P(Z < 2.69) - P(Z < -5.32)

= 0.9964 - 0

= 0.9964

Expected value = 2000 * 0.9964 = 1992.8 = 1993

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