Question

 

1) The effectiveness of a blood-pressure drug is being investigated. An experimenter finds that, on average, the reduction in systolic blood pressure is 40.6 for a sample of size 593 and standard deviation 21.7.
Estimate how much the drug will lower a typical patient's systolic blood pressure (using a 80% confidence level).
Enter your answer as a tri-linear inequality accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).
____ < μμ < ____

2)You measure 42 textbooks' weights, and find they have a mean weight of 43 ounces. Assume the population standard deviation is 14.2 ounces. Based on this, construct a 90% confidence interval for the true population mean textbook weight .
Give your answers as decimals, to two places
____ < μμ < ____

Answer 1

Solution :

1)

When sample standard deviation is given then we use t-distribution.

sample size = n = 593

Degrees of freedom = df = n - 1 = 592

At 80% confidence level the t is ,

= 1 - 80% = 1 - 0.80= 0.20

/ 2 = 0.20 / 2 = 0.10

t _/2,df = t_0.10,592 = 1.283

Margin of error = E = t_/2,df * (s /√n)

= 1.283 * (21.7 / √ 593)

= 1.1

The 99% confidence interval estimate of the population mean is,

- E < < + E

40.6 - 1.1 < < 40.6 + 1.1

39.5 < < 41.7

2)

When population standard deviation is given then we use z-distribution.

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z_/2* ( /√n)

= 1.645 * (14.2 / √ 42 )

= 3.60

At 90% confidence interval estimate of the population mean is,

- E < < + E

43 - 3.60 < < 43 + 3.60

39.40 < < 46.60