An observational study determined that there is a strong correlation between eating breakfast and weight loss. Can it be determi
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To start this question, we should know what is the atomic number of cobalt. The atomic number (the number of protons) of Cobalt is Z =27.
Now, we know that a Cobalt 60 isotope means an isotope of Cobalt whose Atomic Mass is 60.
Thus, in a Cobalt 60 isotope, the number of neutrons in the nucleus are
From the question we know that the given nuclear mass is 59.933820 u.
Now, the mass defect of Cobalt 60 can be easily calculated by adding the masses of the protons and the neutrons as per our calculations and subtracting the given nuclear mass from it.
Thus,
Mass Defect = (Number of Protons Mass of Proton given in the question) + (Number of Neutrons Mass of Neutron given in the question)-59.933820 u
Thus, the required Mass Defect is 0.5634u.
In eV, the Mass Defect is
Answer:
We conclude that the population proportion is equal to 0.70.
Step-by-step explanation:
We are given that a random sample of 100 observations from a binomial population gives a value of pˆ = 0.63 and you wish to test the null hypothesis that the population parameter p is equal to 0.70 against the alternative hypothesis that p is less than 0.70.
Let p = <u><em>population proportion.</em></u>
(1) The intuition tells us that the population parameter p may be less than 0.70 as the sample proportion comes out to be less than 0.70 and also the sample is large enough.
(2) So, Null Hypothesis, : p = 0.70 {means that the population proportion is equal to 0.70}
Alternate Hypothesis, : p < 0.70 {means that the population proportion is less than 0.70}
The test statistics that would be used here <u>One-sample z-test</u> for proportions;
T.S. = ~ N(0,1)
where, = sample proportion = 0.63
n = sample of observations = 100
So, <u><em>the test statistics</em></u> =
= -1.528
The value of z-test statistics is -1.528.
<u>Now at 0.05 level of significance, the z table gives a critical value of -1.645 for the left-tailed test.</u>
Since our test statistics is more than the critical value of z as -1.528 > -1.645, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u><em>we fail to reject our null hypothesis</em></u>.
Therefore, we conclude that the population proportion is equal to 0.70.
(c) The observed level of significance in part B is 0.05 on the basis of which we find our critical value of z.
Answer:
The null hypothesis:
The alternative hypothesis:
Test statistics : z = −2.30
P-value: = 0.02144
Decision Rule: Since the p-value is lesser than the level of significance; then we reject the null hypothesis.
Conclusion: We accept the alternative hypothesis and conclude that under the same circumstances the proportion of offspring peas will be yellow is not equal to 0.27
Step-by-step explanation:
From the given information:
Let's state the null and the alternative hypothesis;
Since The claim is that 27% of the offspring peas will be yellow.
The null hypothesis state that the proportion of offspring peas will be yellow is equal to 0.27.
i.e
The alternative hypothesis state that the proportion of offspring peas will be yellow is not equal to 0.27
<u>The test statistics:</u>
we are given 437 green peas and 129 yellow apples;
Hence;
where ;
= sample proportion
x = number of success
n = total number of the sample size
Now; the test statistics can be computed as :
z = −2.30
C. P-value
P-value = P(Z < z)
P-value = P(Z< -2.30)
By using the P-value method and the normal distribution as an approximation to the binomial distribution.
from the table of standard normal distribution
move left until the first column is reached. Note the value as –2.0
move upward until the top row is reached. Note the value as 0.30
find the probability value as 0.010724 by the intersection of the row and column values gives the area to the left of
z = -2.30
P- value = 2P(z ≤ -2.30)
P-value = 2 × 0.01072
P - value = 0.02144
Decision Rule: Since the p-value is lesser than the level of significance; then we reject the null hypothesis.
Conclusion: We accept the alternative hypothesis and conclude that under the same circumstances the proportion of offspring peas will be yellow is not equal to 0.27