How do I figure out1/2(35)(12)
1 answer:
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Answer:
b. Do not reject H0. We do not have convincing evidence that the mean weekly time spent using the Internet by Canadians is greater than 12.7 hours.
Step-by-step explanation:
Given that in a study of computer use, 1000 randomly selected Canadian Internet users were asked how much time they spend using the Internet in a typical week. The mean of the sample observations was 12.9 hours.
(Right tailed test at 5% level)
Mean difference = 0.2
Std error =
Z statistic = 1.0540
p value = 0.145941
since p >alpha we do not reject H0.
b. Do not reject H0. We do not have convincing evidence that the mean weekly time spent using the Internet by Canadians is greater than 12.7 hours.
Answer:
Hi your question is incomplete attached below is the complete question
answer : p ( X < 725 ) = 0.0116
Step-by-step explanation:
Given data:
Average life of bulbs (μ ) = 750 hours
standard deviation (б ) = 55 hours
n ( sample size ) = 25
X = 725
<u>Probability that the mean life of a random sample of 25 bulbs will be less than725 hours </u>
p ( X < 725 ) = p (( X - μ )/ б √n < 725 - 750 / 55√25 )
= P ( Z > - 2.27 )
Hence P ( X < 725 ) = 0.0116 ( using Z-table )
