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.
![H_0: \bar x =12.7\\H_a: \bar x >12,7](https://tex.z-dn.net/?f=H_0%3A%20%5Cbar%20x%20%3D12.7%5C%5CH_a%3A%20%5Cbar%20x%20%3E12%2C7)
(Right tailed test at 5% level)
Mean difference = 0.2
Std error = ![\frac{6}{\sqrt{1000} } \\=0.1897](https://tex.z-dn.net/?f=%5Cfrac%7B6%7D%7B%5Csqrt%7B1000%7D%20%7D%20%5C%5C%3D0.1897)
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:
A) 4 1/2
Step-by-step explanation:
8 - 4 1 / 2
4 1 / 2
= 4 1 / 2
Using the normal distribution, it is found that 44.43% of students are 21 or older and thus old enough to legally order alcohol at Griff's.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean
and standard deviation
is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The mean and the standard deviation are given, respectively, by:
![\mu = 20.66, \sigma = 2.4](https://tex.z-dn.net/?f=%5Cmu%20%3D%2020.66%2C%20%5Csigma%20%3D%202.4)
The proportion of students that are 21 or older is <u>one subtracted by the p-value of Z when X = 21</u>, hence:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{21 - 20.66}{2.4}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B21%20-%2020.66%7D%7B2.4%7D)
Z = 0.14
Z = 0.14 has a p-value of 0.5557.
1 - 0.5557 = 0.4443 = 44.43% of students are 21 or older and thus old enough to legally order alcohol at Griff's.
More can be learned about the normal distribution at brainly.com/question/4079902
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Answer:
5x^5y
Step-by-step explanation: