Suppose a study estimated that 58% of the residents of a town (with an error range of ±8 percentage points at 95% confidence) favor expanding the park and recreation department. Which of the following percentages of the town's residents may favor expanding the park and recreation department? A:92% B:82% C:52% D:72%
@kirbykirby can you help me on this one too
If the confidence interval is \(58 \pm 8\), then your lower bound is \(58-8 = 50\) and the upper bound is \(58+8=66\). So, you should expect to find a value between 50 and 66 to be a value you need.
so it would be 52% which is C?
yes
thank you!
:)
is it okay if i ask you if i have any more questions? if not i completely understand i dont want to keep bothering you
you could, but I should be going in about ~20 mins or so
okay well i guess i will be using those twenty minutes wisely i have to take a test tomorrow so might as well get all the help i can, can you help me on this one
The owners of four companies competing for a contract are shown in the table below. If a report is released that advocates company B, which of the people having funded the report should result in the most skepticism? Company Owner of Company Company A Jake Adams Company B Debby Smith Company C Henry Rose Company D Rhonda Baker A. Jake Adams B. Debby Smith C. Rhonda Baker D. Henry Rose
Do you have at least an idea?
not really, i wouldnt know where to start. it wouldnt be debby smith for sure because she is the owner but idk theres 3 other options that could work
Well it should be Debby because she is biasing the report by self-reporting positive views, in which I would assume the report would have to be from some neutral source. (because she funded the report)
And there is no information anyway from the other companies to make any decision about them
OH makes sense, haha blonde moment
no worries :)
i have like 3 more questions then im done
sorry its alot
Suppose that a company claims that its coffee cans contain 9.4 ounces of coffee, on average. You take many large samples, and find, each time, that the mean weight of the sample is within the 95% confidence interval. Which of the following would be a reasonable conclusion? A. The company's coffee cans contain more than 9.4 ounces of coffee, on average. B. The company's coffee cans contain 9.4 ounces of coffee, on average. C. The company's coffee cans do not contain 9.4 ounces of coffee, on average. D. The company's coffee cans contain less than 9.4 ounces of coffee, on
average
I would go with B. The confidence interval is symmetric, so even if the values are not 9.4 exaclty, on average, they should measure out to be about 9.4 with a large sample
That was right! thanks 2 more... one second
:)
The Water Department checks the city water supply on a regular basis for contaminants such as lead. The Water Department takes 200 samples and estimates that the concentration of lead in your drinking water is 3 ppb (parts per billion), with a standard deviation of 0.3 ppb. Assuming the samples were random and unbiased, how much confidence can you have in this data? A. 95% confident the average concentration of PCBs in the water supply is between 2.4 ppb and 3.6 ppb. B. 97% confident the average concentration of PCBs in the water supply is between 2.6 ppb and 3.8 ppb. C. 99.7% confident the average concentration of PCBs in the water supply is between 2.7 ppb and 3.3 ppb.
D: 68% confident the average concentration of PCBs in the water supply is between 2.9 ppb and 3.1 ppb.
The confidence interval will be \(\Large \left( \bar{x}- z\frac{\sigma}{\sqrt{n}}, ~ \bar{x}+ z\frac{\sigma}{\sqrt{n}}\right) \)
im going to solve it, but would you mind solving it too so i can see if i got the right answer?
ok
thanks
i think its A
I didn't get A at all :o
.... then i will need to do this again
is it C? because thats the closest thing that is to my answer
I feel D gives me the closest answer
i believe you more than me.
i dont understand how you got it though im pretty sure i plugged everything in right?
Dang it! i was right
it was A
okay i have to get this last one right or i cant take the test tomorrow...
\[ 3 \pm 1.96 \frac{0.3}{\sqrt{200}}\] is what you did?
Which choice below is a boxplot for the following distribution? 58,50,48,46,44,42,40,38,36,34,32,30,28,26,24,22,14 A. Boxplot A B. Boxplot B C. Boxplot C D. Boxplot D E. Boxplot E
Those are the plots
and yeah thats what i did
idk why its saying that im still typing but im not lol
you still here?
dang
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