# Basic Computation:hat{p} Distribution Suppose we have a binomial experiment in which success is defined to be a particular quality or attribute that interests us. (a) Suppose n = 33 and p = 0.21. Can we approximate the hat{p} distribution by a normal distribution? Why? What are the values of mu_{hat{p}} and sigma_ {hat{p}}.?

Question
Normal distributions
Basic Computation:$$\hat{p}$$ Distribution Suppose we have a binomial experiment in which success is defined to be a particular quality or attribute that interests us.
(a) Suppose $$n = 33$$ and $$p = 0.21$$. Can we approximate the $$\hat{p}$$
distribution by a normal distribution? Why? What are the values of $$\mu_{hat{p}}$$ and $$\sigma_ {\hat{p}}$$.?

2021-02-14
We have binomial experiment with $$n = 33$$ and $$p = 0.21$$
$$np = 33(0.21)$$
$$np = 6.93$$
$$nq = 33(1 — 0.21)$$
$$nq = 26.07$$
Since both the values np and ng are greater than 5, hence, we can approximate the $$\hat{p}$$ distribution by a normal distribution.
The formula for the mean of the hat p distribution is $$\mu_{hat{p}} = \hat{p}$$.
$$\mu_{\hat{p}} = 0.21$$
The formula for the standard error of the normal approximation to the $$\hat{p}$$ distribution is
$$\sigma_{hat{p}} = \sqrt{\frac{pq}{n}}$$
$$\sigma_{hat{p}} = \sqrt{\frac{0.21(1-0.21)}{33}}$$
$$\sigma_{hat{p}} = 0.071$$

### Relevant Questions

Basic Computation:$$\hat{p}$$ Distribution Suppose we have a binomial experiment in which success is defined to be a particular quality or attribute that interests us.
(c) Suppose $$n = 48$$ and $$p= 0.15$$. Can we approximate the $$\hat{p}$$ distribution by a normal distribution? Why? What are the values of $$\mu_{hat{p}}$$ and $$\sigma_{p}$$.?
Distribution Suppose we have a binomial experiment in which success is defined to be a particular quality or attribute that interests us.
(a) Suppose $$n = 100$$ and $$p= 0.23$$. Can we safely approximate the \hat{p} distribution by a normal distribution? Why?
Compute $$\mu_{hat{p}}$$ and $$\sigma_{hat{p}}$$.
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The table below shows the number of people for three different race groups who were shot by police that were either armed or unarmed. These values are very close to the exact numbers. They have been changed slightly for each student to get a unique problem.
Suspect was Armed:
Black - 543
White - 1176
Hispanic - 378
Total - 2097
Suspect was unarmed:
Black - 60
White - 67
Hispanic - 38
Total - 165
Total:
Black - 603
White - 1243
Hispanic - 416
Total - 2262
Give your answer as a decimal to at least three decimal places.
a) What percent are Black?
b) What percent are Unarmed?
c) In order for two variables to be Independent of each other, the P $$(A and B) = P(A) \cdot P(B) P(A and B) = P(A) \cdot P(B).$$
This just means that the percentage of times that both things happen equals the individual percentages multiplied together (Only if they are Independent of each other).
Therefore, if a person's race is independent of whether they were killed being unarmed then the percentage of black people that are killed while being unarmed should equal the percentage of blacks times the percentage of Unarmed. Let's check this. Multiply your answer to part a (percentage of blacks) by your answer to part b (percentage of unarmed).
Remember, the previous answer is only correct if the variables are Independent.
d) Now let's get the real percent that are Black and Unarmed by using the table?
If answer c is "significantly different" than answer d, then that means that there could be a different percentage of unarmed people being shot based on race. We will check this out later in the course.
Let's compare the percentage of unarmed shot for each race.
e) What percent are White and Unarmed?
f) What percent are Hispanic and Unarmed?
If you compare answers d, e and f it shows the highest percentage of unarmed people being shot is most likely white.
Why is that?
This is because there are more white people in the United States than any other race and therefore there are likely to be more white people in the table. Since there are more white people in the table, there most likely would be more white and unarmed people shot by police than any other race. This pulls the percentage of white and unarmed up. In addition, there most likely would be more white and armed shot by police. All the percentages for white people would be higher, because there are more white people. For example, the table contains very few Hispanic people, and the percentage of people in the table that were Hispanic and unarmed is the lowest percentage.
Think of it this way. If you went to a college that was 90% female and 10% male, then females would most likely have the highest percentage of A grades. They would also most likely have the highest percentage of B, C, D and F grades
The correct way to compare is "conditional probability". Conditional probability is getting the probability of something happening, given we are dealing with just the people in a particular group.
g) What percent of blacks shot and killed by police were unarmed?
h) What percent of whites shot and killed by police were unarmed?
i) What percent of Hispanics shot and killed by police were unarmed?
You can see by the answers to part g and h, that the percentage of blacks that were unarmed and killed by police is approximately twice that of whites that were unarmed and killed by police.
j) Why do you believe this is happening?
Do a search on the internet for reasons why blacks are more likely to be killed by police. Read a few articles on the topic. Write your response using the articles as references. Give the websites used in your response. Your answer should be several sentences long with at least one website listed. This part of this problem will be graded after the due date.
Select all that apply. We show that our sample statistics have (at minimum) a somewhat normal distribution because
this allows us to use t and z critical values.
this allows us to use t and z tables for probabilities.
this tells us that our sampling is appropriate.
normal distributions are cool and that's all we talk about in this class.
A new thermostat has been engineered for the frozen food cases in large supermarkets. Both the old and new thermostats hold temperatures at an average of $$25^{\circ}F$$. However, it is hoped that the new thermostat might be more dependable in the sense that it will hold temperatures closer to $$25^{\circ}F$$. One frozen food case was equipped with the new thermostat, and a random sample of 21 temperature readings gave a sample variance of 5.1. Another similar frozen food case was equipped with the old thermostat, and a random sample of 19 temperature readings gave a sample variance of 12.8. Test the claim that the population variance of the old thermostat temperature readings is larger than that for the new thermostat. Use a $$5\%$$ level of significance. How could your test conclusion relate to the question regarding the dependability of the temperature readings? (Let population 1 refer to data from the old thermostat.)
(a) What is the level of significance?
State the null and alternate hypotheses.
$$H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}>?_{2}^{2}H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}\neq?_{2}^{2}H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}?_{2}^{2},H1:?_{1}^{2}=?_{2}^{2}$$
(b) Find the value of the sample F statistic. (Round your answer to two decimal places.)
What are the degrees of freedom?
$$df_{N} = ?$$
$$df_{D} = ?$$
What assumptions are you making about the original distribution?
The populations follow independent normal distributions. We have random samples from each population.The populations follow dependent normal distributions. We have random samples from each population.The populations follow independent normal distributions.The populations follow independent chi-square distributions. We have random samples from each population.
(c) Find or estimate the P-value of the sample test statistic. (Round your answer to four decimal places.)
(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?
At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the ? = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.
(e) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings.Fail to reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings. Fail to reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings.Reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings.
(d)$$n=1000$$,$$p=.001$$
(с)$$n=500$$,$$p=.001$$