Rohan Mcpherson

2022-10-08

I have some practice homework questions. I did the first one I will go over the steps please tell me If I am doing it right.

a) As mentioned earlier, it is claimed that 70% of households in Ontario now own large-screen TVs. You would like to verify this statement for your class in mass communications. If you want your estimate to be within 5 percentage points, with 99 per cent level of confidence, how large of a sample must be acquired?

So for this i knew the targe parameter was p so the standard error is $\sqrt{\frac{pq}{n}}$ and that is equal to 0.05 and since we are finding the 99% C.I. our ${Z}_{\alpha}/2$ = 2.575

so my formula became

$0.05=2.575\cdot \sqrt{\frac{(0.7)(0.3)}{n}}\Rightarrow n=556.07=557$

b) You are to conduct a sample survey to determine the mean annual family income in a rural area. The question is, “How many families should be sampled?” In a pilot study of just 10 families, the standard deviation of the sample was $500. The sponsor of the survey wants to use the 95 per cent confidence level. The estimate is to be within $100. How many families’ should be interviewed?

For this I kinda got confused because it said how many family should be interviewed but we are given 10 families? but I was going to follow my procedure as in a but I didnt know where the n would come from because we are give the S.D. ?

or would it be this

$100=1.96\cdot \frac{500}{\sqrt{n}}$ and solve for n afterwards?

c) A student conducted a study and reported that an 80% confidence interval ranged from 48 to 52. He was sure the sample standard deviation was 16 and that the sample was at least 30, but could not remember the exact number (i.e., the size of the sample). Can you help him out?

for this since we are finding the 80% CI our area is 0.9 and our critical value now would be 1.285.

ans since we know it ranged from 48 to 52 our mean is $\frac{48+52}{2}=50$ and then that is equal to

$50=1.285\cdot \frac{16}{\sqrt{n}}$ and then solve for n only confusion was is that number 50 right?

a) As mentioned earlier, it is claimed that 70% of households in Ontario now own large-screen TVs. You would like to verify this statement for your class in mass communications. If you want your estimate to be within 5 percentage points, with 99 per cent level of confidence, how large of a sample must be acquired?

So for this i knew the targe parameter was p so the standard error is $\sqrt{\frac{pq}{n}}$ and that is equal to 0.05 and since we are finding the 99% C.I. our ${Z}_{\alpha}/2$ = 2.575

so my formula became

$0.05=2.575\cdot \sqrt{\frac{(0.7)(0.3)}{n}}\Rightarrow n=556.07=557$

b) You are to conduct a sample survey to determine the mean annual family income in a rural area. The question is, “How many families should be sampled?” In a pilot study of just 10 families, the standard deviation of the sample was $500. The sponsor of the survey wants to use the 95 per cent confidence level. The estimate is to be within $100. How many families’ should be interviewed?

For this I kinda got confused because it said how many family should be interviewed but we are given 10 families? but I was going to follow my procedure as in a but I didnt know where the n would come from because we are give the S.D. ?

or would it be this

$100=1.96\cdot \frac{500}{\sqrt{n}}$ and solve for n afterwards?

c) A student conducted a study and reported that an 80% confidence interval ranged from 48 to 52. He was sure the sample standard deviation was 16 and that the sample was at least 30, but could not remember the exact number (i.e., the size of the sample). Can you help him out?

for this since we are finding the 80% CI our area is 0.9 and our critical value now would be 1.285.

ans since we know it ranged from 48 to 52 our mean is $\frac{48+52}{2}=50$ and then that is equal to

$50=1.285\cdot \frac{16}{\sqrt{n}}$ and then solve for n only confusion was is that number 50 right?

Helena Bentley

Beginner2022-10-09Added 10 answers

a) is correct.

b) The problem is that the formula for the minimum required sample size is

$$n=\frac{{z}^{2}\cdot p\cdot (1-p)}{{d}^{2}}$$

where $z={z}_{\alpha /2}$ and d is half the length of the corresponding confidence interval. So in order to determine n you need to know the true value of p, which is what you want to estimate. Therefore, one often makes a pilot study (a much smaller test study) only to obtain knowledge of the parameter p so that one can decide a proper sample size. In your case you're told that the standard deviation of the sample $\sqrt{p(1-p)}$ is 500 and so your approach is correct.

c) I think your approach is not correct here. In your equation you have the average/mean on the left hand side, where in a) and b) you had half the length of the confidence interval d. In this case $d=(52-48)/2=2$ and so your minimum required sample size is

$$n=\frac{1.2815\cdot {16}^{2}}{{2}^{2}}=\mathrm{105.1.}$$

b) The problem is that the formula for the minimum required sample size is

$$n=\frac{{z}^{2}\cdot p\cdot (1-p)}{{d}^{2}}$$

where $z={z}_{\alpha /2}$ and d is half the length of the corresponding confidence interval. So in order to determine n you need to know the true value of p, which is what you want to estimate. Therefore, one often makes a pilot study (a much smaller test study) only to obtain knowledge of the parameter p so that one can decide a proper sample size. In your case you're told that the standard deviation of the sample $\sqrt{p(1-p)}$ is 500 and so your approach is correct.

c) I think your approach is not correct here. In your equation you have the average/mean on the left hand side, where in a) and b) you had half the length of the confidence interval d. In this case $d=(52-48)/2=2$ and so your minimum required sample size is

$$n=\frac{1.2815\cdot {16}^{2}}{{2}^{2}}=\mathrm{105.1.}$$

In a random survey 250 people participated. Out of 250 people who took part in the survey, 40 people listen to Pink Floyd. 30 people listen to Metallica and 20 people listen to John Denver. If 10 people listen to all three then find the no. of people who listen only Pink Floyd.

The volume of a sphere is increasing at a rate of 3 cubic centimetres per second. How fast is the surface area increasing when the radius is 2 centimetres ?

A)$3c{m}^{2}/s$

B)$6c{m}^{2}/s$

C)$12c{m}^{2}/s$

D)$1c{m}^{2}/s$Determine the average value of function $y=Asi{n}^{2}x$ in the range x=0 to $x=\pi $.

Find the average angular speed of the minute hand of a normal clock in 30 minutes.

Find the average value ${f}_{ave}$ of the function f on the given interval.

$f(x)=3{x}^{2}+8x,[-1,2]$A negative potential energy is possible. Explain

Find the average value fave of the function f on the given interval.

$f(t)=e\mathrm{sin}(t),\mathrm{cos}(t),[0,\frac{\pi}{2}]$What is key in user research?

On the generalized Sierpinski space

In Sierpiński topology the open sets are linearly ordered by set inclusion, i.e. If $S=\{0,1\}$, then the Sierpiński topology on S is the collection $\{\varphi ,\{1\},\{0,1\}\}$ such that $\varphi \subset \{0\}\subset \{0,1\}$ we can generalize it by defining a topology analogous to Sierpiński topology with nested open sets on any arbitrary non-empty set as follows: Let X be a non-empty set and I a collection of some nested subsets of X indexed by a linearly ordered set $(\mathrm{\Lambda},\le )$ such that I always contains the void set $\varphi $ and the whole set X, i.e.

$I=\{\mathrm{\varnothing},{A}_{\lambda},X:{A}_{\lambda}\subset X,\lambda \in \mathrm{\Lambda}\}$

such that ${A}_{\mu}\subset {A}_{\nu}$ whenever $\mu \le \nu $.

Then it is easy to show that I qualifies as a topology on X.

My questions are:

(1) Is there a name for such a topology in general topology literature?

(2) Is there any research paper studying such type of compact, non-Hausdorff and connected chain topologies?Is the invariant subspace problem open for invertible maps?

Let $T:H\to H$ be a bounded linear operator with bounded inverse on the separable complex Hilbert space. Does T preserve a closed proper non-trival invariant subspace?

I'm aware the question is (famously) open for bounded linear maps, and of partial results, but no survey (or Tao's blog, etc) seem to address the invertible case.

If it is open, does a positive or negative answer imply the answer in the non-invertible case?Finite Math (Probability/Venn Diagram)

13) A survey revealed that 25% of people are entertained by reading books, 39% are entertained by watching TV, and 36% are entertained by both books and TV. What is the probability that a person will be entertained by either books or TV? Express the answer as a percentage.

Is this problem stated correctly? How can 36% of the people be entertained by both books and TV, when only 25% of the people are entertained by reading books?

EDIT

Here are two other questions from the exam, that the instructor said followed the same logic as the question above.

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15) A survey of senior citizens at a doctor's office shows that 42% take blood pressure-lowering medication, 45% take cholesterol-lowering medication, and 13% take both medications. What is the probability that senior citizen takes either blood pressure-lowering or cholesterol-lowering medication? Express the answer as a percentage.Homotopy equivalence between two mapping tori of compositions

For any maps $s:X\to K$ there is defined a homotopy equivalence

$T(d\circ s:X\to X)\to T(s\circ d:K\to K);\phantom{\rule{1em}{0ex}}(x,t)\mapsto (s(x),t).$

Here, T(f) denotes the mapping torus of a self-map $f:Z\to Z$ (not necessarily a homeomorphism). It is very surprising to me that this holds with no extra conditions on d and s. I'm guessing that the homotopy inverse is the map:

$T(s\circ d)\to T(d\circ s),\phantom{\rule{1em}{0ex}}(k,t)\mapsto (d(k),t).$

If the above is a genuine homotopy inverse, then the map:

$(x,t)\mapsto (d(s(x)),t)$

would have to be homotopic to the identity somehow. However, after banging my head against the wall on this for a while I can't come up with a valid homotopy. So my questions are:

Is the map $T(s\circ d)\to T(d\circ s)$ I've defined above actually a homotopy inverse? If so, what is the homotopy from the composition I wrote down above to the identity map?

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A. Response bias.

The wording of the survey may be confusing or provoke a certain response.

B. Undercoverage.

The entire population is not reached,

C. Nonresponse bias.

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Calculate an average or typical time. If I would just use the median (or other "normal" types of calculating an average), for example $23:59$ and $00:01$ would yield $12:00$ when it should $00:00$. Is there a better method?