The abc battery company claims that their batteries last 100 hours, on average. you decide to conduct a test to see if the compa
ny's claim is true. you believe that the mean life may be different from the 100 hours the company claims. you decide to collect data on the average battery life (in hours) of a random sample of n = 20 batteries. some of the information related to the hypothesis test is presented below.
1 answer:
Complete question is;
The abc battery company claims that their batteries last 100 hours, on average. You decide to conduct a test to see if the company's claim is true. You believe that the mean life may be different from the 100 hours the company claims. you decide to collect data on the average battery life (in hours) of a random sample of n = 20 batteries. some of the information related to the hypothesis test is presented below:
Test of H0: μ = 100 versus H1: μ ≠ 100
Sample mean: 98.5
Std error of mean: 0.777
Assuming the life length of batteries is normally distributed, what is the p-value associated with this test?
Answer:
p-value = 0.00001
Explanation:
We are given;
Null hypothesis; H0: μ = 100
Alternative Hypothesis; H1: μ ≠ 100
Sample mean: x = 98.5
Standard error of mean; s = 0.777
To find the test statistic, we will use the formula;
t = (x - μ)/(s/√n)
t = (98.5 - 100)/(0.777/√20)
t = -1.5/0.1737
t = -8.64
Now, from online p-value from t-score calculator attached, using t = -8.64; DF = n - 1 = 20 - 1 = 19; two tail distribution;significance level of 0.05; we have;
The p-value = 0.00001
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An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030 kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer.
What is the tension between her ears?
Would the astronaut find it difficult to keep from being torn apart by the gravitational forces?
Answer:
The tension between the ears = 2.07 KN
The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.
Explanation:
Given that:
Orbital radius of the spacecraft (R) = 120 Km = 120 × 10³ m
Mass of the black hole (m) =
where :
Then; we have:
Schwarzchild radius of the black hole
r - 15.0 km
Mass of each ear
Farther distance between one ear and the black hole (d) = 6.0 cm
= 0.06 m
Closer distance between the other ear and the black home is (d) 6.0 cm
= 0.6 cm
NOW, If we assume that the tension force should be T; then definitely the two ears will posses the same angular velocity .
The net force on the ear closer to the black hole will be:
The net force on the ear farther to the black hole is :
Equating equation (1) and (2) & therefore making (T) the subject of the formula; we have:
The tension between the ears = 2.07 KN
The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.