The correct interpretation of the p-value is given by:
B If the average battery life really is 5 hours, then a sample of 10 observations having a sample mean of 4.25 hours or lower would only occur about 3.8% of the line.
<h3>How to find the p-value of a test?</h3>
It depends on the test statistic z, as follows:
- For a left-tailed test, it is the area under the normal curve to the left of z, which is the p-value of z.
- For a right-tailed test, it is the area under the normal curve to the right of z, which is 1 subtracted by the p-value of z.
- For a two-tailed test, it is the area under the normal curve to the left of -z combined with the area to the right of z, hence it is 2 multiplied by 1 subtracted by the p-value of z, which means that the p-value for a two-tailed test is twice the p-value of a one-tailed test.
In this problem, a left-tailed test is used, as we are testing if the mean is less than 5 hours.
The sample mean from the 10 times was of 4.25, and the p-value is of 0.038, which means that the area to the left of Z under the normal curve is of 0.038, that is, a sample mean of 4.25 hours or lower would only occur about 3.8% of the line, hence option B is correct.
You can learn more about p-values at brainly.com/question/13873630
Let Ch and C denote the events of a student receiving an A in <u>ch</u>emistry or <u>c</u>alculus, respectively. We're given that
P(Ch) = 88/520
P(C) = 76/520
P(Ch and C) = 31/520
and we want to find P(Ch or C).
Using the inclusion/exclusion principle, we have
P(Ch or C) = P(Ch) + P(C) - P(Ch and C)
P(Ch or C) = 88/520 + 76/520 - 31/520
P(Ch or C) = 133/520
Answer:
probably an update which sucks tell me about it lol
1. Pants=2, shirts= 3
# outcomes 2*3=6
2. Flavors=3, toppings= 4
# outcomes 3*4= 12
3. Type=3, color=3
# outcomes 3*3= 9
