Question

The director of research and development is testing a new drug. She wants to know if there is evidence at the 0.05 level that the drug stays in the system for more than 393minutes. For a sample of 74 patients, the mean time the drug stayed in the system was 395 minutes. Assume the population variance is known to be 529. Find the P-value of the test statistic. Round your answer to four decimal places.

Answer 1

Answer:

= 393, =  529, = 23

n= 74, = 395

= 0.05

null and alternative hypothesis is

Ho:   393

H1:   > 393

formula for test statistics is

z = 0.748

test statistics: Z = 0.75

calculate P-Value

P-Value = 1 - P(z < 0.75)

using normal z table we get

P(z < 0.75) = 0.7734

P-Value = 1 - 0.7734

P-Value = 0.2266

Decision rule is

Reject Ho if ( P-Value) ( )

here, ( P-Value = 0.2266 ) > ( = 0.05 )

Hence,

Null hypothesis is NOT rejected.

There is enough not evidence to conclude that the drug stays in the system for more than 393 minutes.