Data was collected for a sample of organic snacks. The amount of sugar (in mg) in each snack is summarized in the histogram belo
w. Which statement best describes the meaning of one of the bars in the histogram?
2
4
6
8
10
12
amount of sugar (mg)
220
224
228
232
236
240
244
248
Frequency
2 snacks have about 226 mg of sugar.
226 snacks have 2 mg of sugar.
226 snacks have between 224 and 228 mg of sugar.
The largest number of snacks have 12 mg of sugar
2
4
6
8
10
12
amount of sugar (mg)
220
224
228
232
236
240
244
248
Frequency
2 snacks have about 226 mg of sugar.
226 snacks have 2 mg of sugar.
226 snacks have between 224 and 228 mg of sugar.
The largest number of snacks have 12 mg of sugar
1 answer:
Answer:
Step-by-step explanation:
Hello!
The histogram summarizes the amount of sugar in organic snacks. (mg)
Y-axis shows the number of snacks
X-axis shows the amount of sugar per snack
The first column of the histogram show that 2 snacks contain sugar between 220 and 224 mg of sugar.
The second column shows that about 3 snacks have between 224 and 228 mg of sugar.
The third column shows that 8 snacks have between 228 and 232 mg of sugar.
The fourth column shows that 12 snacks have between 232 and 236 mg of sugar.
The fifth column shows that 11 snacks have between 236 and 240 mg of sugar.
The sixth column shows that 3 snacks have between 240 and 244 mg of sugar.
The seventh column shows that 2 snacks have between 244 and 248 mg of sugar.
The total of observations is 2+3+8+12+11+3+2= 41 snacks.
I hope this helps!
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Answer:
B. 10000 years old, with a margin of error of 400 years
Step-by-step explanation:
Assuming this complete problem: "12. An archaeologist uses an accelerator mass spectrometer to find the age of a buried branch. At the 68% confidence level, the spectrometer estimates that the branch was 10,000 years old with a following could the spectrometer estimate as the age of the branch at the 95% confidence level?"
The possible options are:
A. 9500 years old, with a margin of error of 500 years
B. 10000 years old, with a margin of error of 400 years
C. 9500 years olds, with a margin of error of 50 years
D. 10000 years old, with a margin of error of 40 year
Solution to the problem
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
represent the sample mean for the sample
population mean (variable of interest)
s represent the sample standard deviation
n represent the sample size
The confidence interval for the mean is given by the following formula:
(1)
The margin of error for this case is given by this formula:
For the first case the confidence was 68% so then and
we can find a critical value on the normal standard distribution that accumulates 0.16 of the area on each tail and this value is
, because P(z<-0.944) = 0.16 and P(Z>.944) = 0.16
The margin of error can be expressed like this:
We can solve for the standard error on this case and we got:
And then for the new confidence interval we need to calculate the new . For the first case the confidence was 95% so then
and
we can find a critical value on the normal standard distribution that accumulates 0.025 of the area on each tail and this value is
, because P(z<-1.96) = 0.025 and P(Z>1.96) = 0.025. So then the new margin of error would be:
The estimation for the mean not changes and from the before and new case is 1000 years. So then the best option for this case would be:
B. 10000 years old, with a margin of error of 400 years
And makes sense since a larger confidence interval means a wider interval.