Here’s your next reading assignment. Read Section 5.3 in your textbook and answer the following questions by 8 a.m., Wednesday, April 4th. Be sure to login (using the link near the bottom of the sidebar) to the blog before leaving your answers in the comment section below.
- Which of p and “hat p” (seen in the margin on page 203) is a random variable?
- In a study designed to investigate whether certain detonators used with explosives in coal mining meet the requirement that more than 90% will ignite the explosive when charged, it is found that 185 of 200 detonators function properly. Is this convincing evidence that this type of detonators work as they should?
- If you want to determine the smallest sample size n so that the margin of error of a point estimate for a population proportion is no larger than m = 0.04, do you need to know the size of the population? Why or why not?
- What is one question you have about the reading?
1. Hat p because it is the sample and is subject to variation due to chance.
2. Po = .9
nPo = 200(0 .9) = 180 > 10
SE = sqrt((.9*.1)/200 = .0212
point estimate = 185/200 = .925
Z = (.925 – .9)/ .0212 = 1.18 -> p= .8810
Not convincing evidence.
3. Yes because ME = Z*SE and SE requires the population.
4. None.
1. P-hat (Which could maybe be a rap pseudonym, if the word “phat” is still cool).
2. Based on a 95% confidence interval of (.888, .962), the detonators work almost as well as they should, but there is enough uncertainty to perhaps warrant a more stringent process.
3. No – the margin of error for a point estimate is independent of the population size, depending only on the sample proportion and sample size.
4. Based on a 13% approval rating (http://www.langerresearch.com/uploads/1133a1CongressandObama.pdf), can a 95% confidence interval show that Congress is horrible at its job?
1. P hat is the random variable because it represents the sample proportion.
2. Ho: p = 0.9
Ha: p < 0.9
SE = 0.0212 as calculated using equation 5.15.
z = .925 – .9 / .0212 = 1.179
P (z < 1.179) = 0.8810 which is not less than 0.05.
Therefore we fail to reject the null hypothesis. This is not convincing evidence that the detonators do not work.
3. No. You're trying to find the smallest sample size n such that the margin of error for a point estimate, not the true proportion, is no larger than .04. For this calculation, all you care about is n.
4. None
1) hat p is a random variable.
2) No. A hypothesis test must be performed in order to get a convincing evidence.
3) We don’t need to know the size of the population. It is not needed in our calculation and it is very difficult to find an exact size of a population.
4) It looks like the sampling distribution of hat p can be approximated by binomial model also? (But the normal model looks easier to work with)
1) p-hat
2) hypothesis test:
H_0: p-hat = .9
H_a: p-hat p = .883
Fail to reject the null, strong evidence in favor of the detonators working properly.
3) No. the margin of error of a point estimate is based on the sample size, not on the population size
4) none
1) Hat p is the random variable because it varies based on the sample taken.
2) The Z-score in this case would be (185/200 – .9)/sqrt(.9*.1/200) = 1.1785. This makes the p-value 0.2386, which is not considered statistically significant. Therefore we do not reject the null hypothesis, which was that 90% of the detonators would work. Since we want more than this amount to work, we say that it is not functioning as expected.
3) You only need the size of the sample population. In order to calculate the margin of error you need the Z score for the confidence interval you want and standard error, which both use the sample population size but not the overall population size.
4) How does sample size affect margin of error?
1. “hat p” is a random variable.
2. No. H0: p = 0.90, HA: p > 0.90. The z-score of the sample is 1.19, so the p-value = 1-0.8830 = 0.117, which is greater than the significance value of 0.05 for a 95% confidence interval. Therefore we fail to to reject the null hypothesis, so the true proportion could be 90% or less.
3. No, because it does not require a value for the standard deviation of the population, just of the sample.
4. None.
1) p hat
2) SEphat = sqrt(p*(1-p)/n) = sqrt(.9 * .1/200) = .0212
185/200 – 1.96 * .0212 = .883
Not convincing evidence that they are functioning as they should.
3) No, the error of a point estimate is given by z*sqrt(p*(1-p)/n), where z is the margin of error, p is the population proportion, and n is the sample size.
4) Is there a way to do this kind of analysis with lower sample sizes such that np or n(1-p) < 10?
1) “p hat” is the random variable, since it represents the sample proportion and not the actual population proportion. It will therefore vary between different samples.
2) Yes, there is a .09099 probability that the mean of the detonation rate is this high given that the mean is 0.9. This means that the actual mean is most likely higher than 0.9.
3)Yes, you need to verify that the data does, in fact, follow a normal distribution. One requirement of this is that your sample size must be less than 10% of the population. To know this, we must know the size of the population.
4) Can we do more checking of the initial conditions as practice in class? I realized that I rarely check to see if the population fits the requirements of a normal distribution before trying to answer the problem.
1. P^ is
2. Yes
3. No, you just need to know the size of the sample.
4. none
1. hat p
2. SE= square root (p*(1-p)/n). p= 0.9, n=200, so SE= 0.0212. then point estimate +/- z* SE = 185/200 – 1.96*.0212= 0.883. this is not enough evidence that the detonators work as they should.
3. You do not need to know the size of the population because the point estimate only uses the margin of error, the sample size, and the population proportion.
4. Is the success-failure condition only used to tell if a sample distribution is nearly normal?
1) hat p
2) sqrt(p*(1-p)/n) = sqrt(0.9 *(0 .1/200)) = .021
185/200 – 1.96 * .021 = .883
The evidence is not convincing that they are functioning correctly
3) No, you only need to know the population proportion, p, and the sample size, n
4) –
1. p-hat is a random variable.
2.) SE = sqrt(p(1-p)/n) = 0.019
Z = 0.9-0.925/0.019 = -1.32
This corresponds to a p-value of 1-0.0951 = 0.915, we can say that the device works, because there’s a 91.5% chance of getting such result if its 90% effective
3.) Yes, you need to know the size of the population before you can assume it is normal.
1. hat p
2. H0: p = .9
HA: p.05
Fail to reject HO. There is good evidence that the detonators are working properly.
3. Yes, to determine if n is less than 10% of the population.
1. p hat is a random variable because it is the proportion for the sample
2. H0: p = 0.90 Ha: p > 0.90
SE = 0.0212
Z = (0.925-0.9) / 0.0212 = 1.18
p = 1 – 0.8810 = 0.119
Because this is greater than 0.05, we cannot reject the null hypothesis meaning that the detonators do not meet the requirement that more than 90% will ignite.
3. You do not need to know the true size of the population but you do need to know an estimate of the proportion which can be gathered from a sample.
4. I really have no questions. It made complete sense.
1. p
2. No; using 90% as the null hypothesis and performing the hypothesis test set out in the reading we find that we should not reject the null hypothesis.
3. No. You need the sample size but not the population size.
4. In the example under the heading 5.3.2 the book uses p hat instead of p, as the formula given in the box just above the example specifies. I’m wondering why.
1) p hat is a random variable
2) This is not convincing evidence that the detonators will work as they should.
3) You do not need to know the size of the population because you can estimate the p that gives the worst case estimate if no other estimate is available. If there is an estimate from a recent poll then that would be used.
4) This section is a little confusing about the worst case estimate.
1) “hat p”
2) It is pretty convincing since p hat is contained in the 90% confidence interval.
3)No, because the formula only need the sample size.
4) No question.
1. p-hat
In a study designed to investigate whether certain detonators used with explosives in coal mining meet the requirement that more than 90% will ignite the explosive when charged, it is found that 185 of 200 detonators function properly. Is this convincing evidence that this type of detonators work as they should?
2.
H0: p = 0.9
HA: p > 0.9
SE = 0.0212
z = 1.17
p-value = 0.1210
fail to reject null hypothesis
unconvinced the detonators work as they should.
3. To actually calculate the sample size, no. But I think it would be necessary to have at least an estimate of the size of the population as a sanity check to make sure (1) there are at least that many members in the population, and (2) that the sample size represents less than 10% of the population in order to assume independence.
1. p hat is the random variable.
2. The null proportion for this would be p0= 0.9. Z= 1.25 which correlates to a p-value of 0.106. This value is larger than 0.05 so I would say this is not significant evidence that this type of detonator works as it should.
3. Yes. In order to calculate margin of error you need to calculate the standard error and the size of the population is needed for this calculation.
4. I was a little confused on how to set up a confidence interval for proportions and also how to write the hypothesis for a hypothesis test using proportions.
1) hat p
2) Ho = .9 Ha > .9 z = 1.18 p = .119. So the null hypothesis cannot be ruled out. So this is not convincing evidence.
3) No, because it is a population proportion so the population isn’t needed.
4) Why does there have to be at least 10 successes or failures in order for hat p to be normal?
1) p hat is a random variable.
2)
SE(p hat) = sqrt(p*(1-p)/n) = sqrt(0.9 * 0.1/200) = 0.02121
P(Z>(0.925-0.9/0.02121)=P(Z>1.18)= 0.036
Reject the null at 5% significance level. Not convincing evidence that detonators work as they should.
3) No, the error of a point estimate = z*(sqrt(p*(1-p)/n)). Z is the margin of error, p is the population proportion, and n is the sample size and so the population size is not required.
4) No question.
1) p: true proportion p-hat: sample proportion
2) Yes it is sufficient data to ensure that the detonators function properly
3) Yes, in order to ensure that the data is independent, the sample size should not be more than 10% of the total population. Knowing the size of the total population ensures that the point value contains independent trials.
4) Nopers
1) hat p
2) SEhatp sqrt [(.9)* (.1/200)] = .0212; 185/200 – 1.96 *.0212 = .883
This is not enough evidence that they are functioning as they should.
3)Yes because the popoulation is included in the equation of the margin of error.
4) What are some applications of the margin of error?
1. “hat p” is a random variable. p represents the true population proportion.
2. Constructing a hypothesis test with H0: p=0.9 and HA: p>0.9 gives a p-value equal to 0.1190. This is not low enough to reject the null hypothesis therefore the study does not provide convincing evidence that the detonators work as they should.
3. No; You only need the z-score and the standard error which can be computed using only the estimate for p, and the number of observations n in the sample.
4. When would you use a two sided hypothesis test for a population proportion?
1. hat p because it is taken from a sample rather than a population.
2. x^= 185/200=0.925
SE=sqrt(p*(1-p)/n)= sqrt(0.925*0.075/200)=0.0186
H0: x = 0.90
HA: x > 0.90
p=P(Z>(.925-0.9)/0.0186)=P(Z>1.34)
p=1-0.9099= 0.0901
No there is not enough convincing evidence that the detonators work as they should because the data fails to reject the null hypothesis with an acceptance level of 0.05.
3. The size of the population is needed to ensure that the normal distribution can be used. This is checked by having at least 10 failures and 10 successes. If this condition is not met, then the margin of error calculated will be wrong because the normal distribution cannot be used to accurately portray the data.
4. Is there a way to guess the worst case estimate or do we always use a p of 0.5?
Which of p and “hat p” (seen in the margin on page 203) is a random variable?
hat p
In a study designed to investigate whether certain detonators used with explosives in coal mining meet the requirement that more than 90% will ignite the explosive when charged, it is found that 185 of 200 detonators function properly. Is this convincing evidence that this type of detonators work as they should?
h0: mu = .9
ha: mu > .9
p_hat = .925
SE = sqrt(.925 * .075 / 200)
=.0186
We can apply z score because the conditions necessary to model the normal distr are satisfied.
p(z > (.925 – .9)/.0186)
p-value = .0901.
So we do not reject H0 for Ha on sig level alpha = .05. i.e. this is NOT convincing evidence they work as they should..
If you want to determine the smallest sample size n so that the margin of error of a point estimate for a population proportion is no larger than m = 0.04, do you need to know the size of the population? Why or why not?
Yes and No. Yes, you need to know that your sample size <= population size. But whether you need to know more depends on what you want to do with this sample size. If you want to conduct an experiment, you need to know that the sample size is no larger than 10% of the population size in order to ensure independence of a random variable. So, you need some information about the population size, but it depends on what statistics you are doing.
What is one question you have about the reading?
Please explain what a distribution of 1's and 0's would look like and how it can be normal?
1) P Hat
2) SE of P Hat = root(p(1-p)/n) = root(.9(1-.9)/200) = .0212
(.925-.9)/.0212 = Z = 1.18, P = .881, which is >.05
not convincing evidence at all
3) No: N the error of a point estimate is z*sqrt(p*(1-p)/n), where P is the population proportion. The actual population count is not used.
4) none; pretty simple
1. Which of p and “hat p” (seen in the margin on page 203) is a random variable?
Hat p is a random variable that is normally distributed
2. In a study designed to investigate whether certain detonators used with explosives in coal mining meet the requirement that more than 90% will ignite the explosive when charged, it is found that 185 of 200 detonators function properly. Is this convincing evidence that this type of detonators work as they should?
Sprt(po(1-p0)/n) = 0.0212132
Z = (.92 – .9)/0.0212132 = 0.942809
P(Z=.94) = .8264
1-.8264 = .1736
larger than .05, so we fail to reject the null hypothesis
3. If you want to determine the smallest sample size n so that the margin of error of a point estimate for a population proportion is no larger than m = 0.04, do you need to know the size of the population? Why or why not?
No, because the population is 600 and a population of over 50 is sufficient to acquire statistics
4. What is one question you have about the reading?
Why doesn’t margin of error change with the size of the population?
1. hat p is a random variable
2. 185/200 = 0.925, null proportion = 0.9
SE = sqrt(.9(1-.9)/200) = 0.02121
Z-score = (0.925-0.9)/0.02121 = 1.1785
From table, P(Z) = 1 – 0.879 = 0.121
Because this is bigger than 0.05, we fail to reject the null hypothesis. Therefore we conclude that the detonators are functioning as they should.
3. I think we do, though I am not sure why…
4. The explanations in the book was not too clear to me. Can you go over them in class?
Hat p is a random variable
no this is not convincing evidence the detonators work
no we don’t need the population size to calculate m
“hat p” is a random variable, because it is the representation of a set out outcomes that are assigned numbers.
Given that the sample observations are independent, and that we expect to see at least 10 successes and failures (180 and 20 respectively in a 200 sample set), we can assume this is nearly normal data. The standard error in this case is 0.0186. Our point estimate is 0.925, and our null value is 0.9. Ho is that greater than or equal to 90% of the detonators go off, Ha is that 1.34) = 0.0901. We fail to reject the null hypothesis in this case.
No, because the margin of error relies on the point estimate “hat p”, not the entire population.
How do you come up with these equations?
1. “Hat p” is the random variable because the population proportion is always the same number, while the sample proportion can vary based on the individual data in the sample.
2. This is not convincing evidence that the detonators function properly because a quick hypothesis test shows that the null hypothesis (90% ignition) is not rejected, so the detonators may have a lot of variance between different samples.
3. Knowing the sample size is necessary to determine the margin of error of a point estimate for a population proportion, however knowing the population size is not. The standard error of the proportion is computed using the sample size, and then the confidence interval will the give the estimated proportion of the population without knowing the actual size of the population.
4. Why do I need to spend my time searching through each minute detail of the reading and confusing myself about said reading material just to come up with a question to post?
1. Hat p is a random variable.
2. SE * Hat P = .0212
185/200 – 1.96*.0212 = .883
Since this value is too small, it is not convincing evidence that the detonators should be accepted as working as they should.
3. No, you only need to know the margin of error, the population proportion, and the sample size.
4. I’m still slightly confused on the definition of a random variable.
1.
“hat p”
2.
SE p-hat = sqrt[p(1-p)/n] = sqrt[(.9(1-.9)/200] = 0.0212
95% confidence = 185/200 +/- 1.96*.0212 = (.883, .966)
Since this range has parts below 0.9 or 90%, it is not convincing evidence
3.
My guess is no from the example on page 205. You can use the worst case scenario of p = 0.5, and then solve for the least amount of participants needed n.
4.
I am confused about question number three. To get the smallest n, would you need to know the p value that leads to the smallest n value?
1. hat P is a random variable.
2. NO it is not.
3. NO you don’t. You can simply estimate p to be the worst case scenario (highest margin of error).
4. The reading is straightforward.
1. p hat
2. SEphat = sqrt(.9 * .1/200) = .0212
(185/200) – 1.96 (.0212) = .883
evidence is not very convincing
3. No. The equation for the error of a point estimate is a function of z, the margin of error, p, the population proportion, and n, the sample size.
4. The last bit about estimating population sizes and the standard deviations was a little confusing.
1. p is a random variable
2. p=185/200=.925
n=200;
z=1.96 (95%)
Point Estimate=.9;
SE=sqrt(p*(1-p)/n);
low=PE-z*SE
high=PE+z*SE
95% Confidence Interval: ( 0.8635, 0.9365)
Yes, this is convincing evidence that the detonators work as they should.
3. Yes, you have to know the size of the population. The margin of error is equal to the z score times the standard error, and the standard error depends on n.
1. Hat p
2. SE of hatP = (p*(1-p)/n)^.5 =(.9*(.1)/200))^.5 = 0.0213
185/200 – 1.65*0.0213 = .8899
It’s not convincing enough
3. No, z*(p*(1-p)/n)^.5 is the error of a point estimate, where p is the population *proportion* not the size
4. None
1. p hat
2. SEphat = 0.0212
185/200-1.96*SEphat = 0.883
So not convincing evidence
3. No, only the margin of error, population proportion, sample size are needed
4. Whether this kind of analysis is available when sample sizes is small?
1) p hat
2) yes
3) no
4) none
1) p hat
2) SEphat =0.0212
185/200 – 1.96 * 0.0212 = 0.883
Not enough evidence shows they are functioning as they should.
3) No, the error = z*sqrt(p*(1-p)/n) n is sample size.
1. p-hat
2. SE=sqrt(.925*.075/200)=.0186, .0186*1.65=ME=.03069 Not super convincing, as margin of erro will fall bellow .90.
3. No, because the absolute population size doesn’t enter the ME equation, only p, n, and z do. Thus in the worst case(p=.5) for a 95% percent confidence (z=1.96) interval, we need at most 601 observations to determine the margin of error to within .04
4. None came up
1. p hat
2. SEphat = sqrt(p*(1-p)/n) = .0212
185/200 – 1.96 * .0212 = .883
It’s not convincing evidence.
3. No, the error of a point estimate is z*sqrt(p*(1-p)/n), where z is the margin of error, p is the population proportion, and n is the sample size. Therefore the population size is not needed.
4. Does the conclusion of problem 3 means the error is independent of the population size?
1. p hat, as it’s never a fixed constant like p could be.
2. The null hypothesis is the detonators work 90% of the time. The alternative is that they work less than 90% of the time. phat = 185/200 = 0.925. SE for phat = sqrt((pnot*(1-pnot))/n) = sqrt((.9(1-.9))/200) = 0.0212. Z-score = (0.925-0.9)/0.0212 = 1.179. Since the Z-score is well above one, the area to the left in a normal probability plot will be well above .05 (let alone 0.5), the usual significance level.
3. No, because you only need to know the z-score and standard error, which in total only use the values of the sample proportion, population proportion, and sample size; the population size is unnecessary.
4. Why did they choose ten to be a minimum number of successes and failures?
1. p-hat
2. SE = sqrt ((0.9 * 0.1)/200) = 0.0212
185/200 -1.96*SE = 0.883
not convincing enough
3. no, because the equation for calculating the error does not require population.
1. The “hat p” is a random variable, as it is taken from the sample distribution.
2. This is not necessarily convincing evidence that this type of detonator works properly. This margin of error is only 0.025, which is relatively small, and depending on the confidence interval, this wouldn’t necessarily hold up.
3. Yes, you would need to know the sample size. Example 5.20 deals directly with this question. You would also need to know what confidence interval you were trying to construct to properly answer this question.
4. I think it all made sense. Probably will have questions later, though.
1) Which of p and “hat p” (seen in the margin on page 203) is a random variable?
Answer: “hat p”
2) In a study designed to investigate whether certain detonators used with explosives in coal mining meet the requirement that more than 90% will ignite the explosive when charged, it is found that 185 of 200 detonators function properly. Is this convincing evidence that this type of detonators work as they should?
Answer: Yes.
3) If you want to determine the smallest sample size n so that the margin of error of a point estimate for a population proportion is no larger than m = 0.04, do you need to know the size of the population? Why or why not?
Answer: No. Population size isn’t in the formula to determine ME.
ME=(z*)(SE) = (z*)sqrt((p(1-p))/n)
Where ME is margin of error, z* is the confidence level, p is population proportion (or an estimate of it), and n is the sample size.
4) What is one question you have about the reading?
Answer: Why is p_hat used for the sample proportion, as opposed to p_bar or something else? Is there a mathematical convention for variable usage in things like this or is it just traditional?
1. “hat p”
2. SE = sqrt((0.9 * 0.1)/200) = 0.0212. Z = (0.925 – 0.90)/0.0212 = 1.18. So, the p-value is 0.8810 > 0.05, so we fail to reject the null hypothesis, meaning that there is no convincing evidence that the detonators work as they should.
3. The formula to calculate the margin of error is ME = z*SE <= m. SE relates to the size of the sample, but the calculation has nothing to do with the size of the population.
4. I have seen this material before, and I therefore had no questions about the reading.
1. “Hat p” is the random variable
2. SE of “hat p” = sqrt(p*(1-p)/n) = sqrt(.9* (.1/200))= .0212
(185/200) – 1.96 * .0212 = .883
This is convincing evidence that they are functioning correctly since .88 is within p of 90%.
3. No you do not need to know the size of the population since p is a proportion of the population and n is the sample size. You can use these two values to find z the margin of error.
4. How does the hypothesis testing and proportions correlate?
1. “P hat” is the random variable because we can describe it as a random process or variable with a numerical outcome.
2. 185/200 = 0.9285. The proportion of “success” meets the requirement that more than 90% will ignite, so I believe this is convincing evidence that this type of detonator will work as stated. However, we would need more information to be certain.
3. No, you do not need to know the size of the population (look at the expression at the bottom of page 204). You obtain a worst case estimate and plug that value into the equation to solve for n.
4. Are we expected to check whether the conditions are normal for each and every one of these problems?
1) p hat
2) No, this is not sufficient evidence. The 95% confidence interval for this study is 88.8% to 96.1%.
3) No, you just need to know p and n
4) I don’t understand why p or p hat are random
1. Hat p is a random variable.
2. Meets the two conditions; they are all independent and with greater than 10 of each successes and failures. This means that the distribution is nearly normal.
SE = sqrt((hat p)(1-hat p)/n) = sqrt(0.925*0.075/200) = 0.0186
0.925 +- 1.96*0.0186 = (0.8885, 0.9615) = 95% confidence interval.
Therefore, this is not convincing evidence because numbers below 90% fall into the 95% confidence interval, so there is a good chance that the mean might be below 90% success rate.
If you want to determine the smallest sample size n so that the margin of error of a point estimate for a population proportion is no larger than m = 0.04, do you need to know the size of the population? Why or why not?
3. No you don’t need to know the size of the population. All you need to know is the sample size and the results of each experiment. The whole population is just implied to be large.
4. I guess I don’t understand how the book got from point estimate +- z*SE to Z = (point estimate – null value)/SE. They seem so similar, but if they were derived from each other, why did (for example) the “+-” in the first equation turn into just “-” in the second?
1) p hat
2) No. There could be a high low probability that 185/200 work if the actual mean is 90%
3) No, because z*SE <= m, and SE = sqrt(p(1-p)/n). We just need to know the proportions of success for a given sample.