Answer: The lower bound of confidence interval would be 0.116.
Step-by-step explanation:
Since we have given that
p = 13.2%= 0.132
n = 1105
At 90% confidence,
z = 1.645
So, Margin of error would be
[tex]z\sqrt{\dfrac{p(1-p)}{n}}\\\\=1.645\times \sqrt{\dfrac{0.132\times 0.868}{1152}}}\\\\=0.0164[/tex]
So, the lower bound of the confidence interval would be
[tex]p-\text{margin of error}\\\\=0.132-0.0164\\\\=0.116[/tex]
Hence, the lower bound of confidence interval would be 0.116.
The confidence interval for percentage of U.S. adults who were victims of crime at the 90% confidence level is (0.114, 0.15). The lower bound of the confidence interval is 0.114.
Explanation:To estimate the percentage of U.S. adults who were victims of crime at the 90% confidence level, we can use the formula for a confidence interval:
[tex]CI = p +/- Z * \sqrt{((p * (1-p)) / n)[/tex]
where CI is the confidence interval, p is the sample proportion, Z is the Z-score corresponding to the desired confidence level, and n is the sample size. In this case, p = 0.132, Z = 1.645 (corresponding to a 90% confidence level), and n = 1105. Plugging in these values, we can calculate the confidence interval as:
[tex]CI = 0.132 +/- 1.645 * \sqrt((0.132 * (1-0.132)) / 1105)[/tex]
Simplifying the expression gives us:
CI = 0.132 ± 0.018
Therefore, the confidence interval for the percentage of U.S. adults who were victims of crime at the 90% confidence level is (0.114, 0.15). The lower bound of the confidence interval is 0.114.
Suppose a random sample of n = 16 observations is selected from a population that is normally distributed with mean equal to 101 and standard deviation equal to 12.
(a) Give the mean and the standard deviation of the sampling distribution of the sample mean x(with a line above). (Enter your standard deviation to two decimal places.)
mean =
standard deviation =
(b) Find the probability that x(line above it) exceeds 107. (Round your answer to four decimal places.)
Answer:
(c) Find the probability that the sample mean deviates from the population mean μ = 101 by no more than 4. (Round your answer to four decimal places.)
Answer:
101, 3, 0.025, 0.7416
Step-by-step explanation:
Given that a random sample of n = 16 observations is selected from a population that is normally distributed with mean equal to 101 and standard deviation equal to 12.
As per central limit theorem we have
a) Mean of sample mean = [tex]E(\bar x) =\\ \mu =101[/tex]
Std deviation of sample mean = [tex]\frac{\sigma}{\sqrt{n} } =3[/tex]
Mean = 101
Std dev =3.00
b) [tex]P(\bar x >107) = P(Z>\frac{107-101}{3} )\\= P(Z>2) = 0.025[/tex]
c) the probability that the sample mean deviates from the population mean μ = 101 by no more than 4.
=[tex]P(|\bar x-101|) \leq 4\\= P(|z|\leq 1.13)\\= 2(0.3708)\\=0.7416[/tex]
A statistical program is recommended.
A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specimen, resulting in the accompanying data.
1.12 5.09 0.97 1.59 4.60 0.32 0.55 1.45
0.18 4.47 1.20 3.50 5.02 4.67 5.22 2.69
3.96 3.17 3.03 2.21 0.69 4.47 3.31 1.17
0.79 1.17 1.57 2.62 1.66 2.05
The values of the sample mean, sample standard deviation, and (estimated) standard error of the mean are 2.484, 1.611, and 0.294, respectively. Does this data suggest that the true average percentage of organic matter in such soil is something other than 3%? Carry out a test of the appropriate hypotheses at significance level 0.10. [Note: A normal probability plot of the data shows an acceptable pattern in light of the reasonably large sample size.]
Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.)
t=
P-value=
Answer:
Which is the output of the formula =AND(12>6;6>3;3>9)?
A.
TRUE
B.
FALSE
C.
12
D.
9
Step-by-step explanation:
To test μ for an x distribution that is mound-shaped using sample size n ≥ 30, how do you decide whether to use the normal or Student's t distribution?
A.If σ is known, use the standard normal distribution. If σ is unknown, use the Student's t distribution with n – 1 degrees of freedom.
B. If σ is unknown, use the standard normal distribution. If σ is known, use the Student's t distribution with n – 1 degrees of freedom.
C. If σ is known, use the standard normal distribution. If σ is unknown, use the Student's t distribution with n degrees of freedom.
D. For large samples we always the standard normal distribution.
Answer:
The correct option is A) If σ is known, use the standard normal distribution. If σ is unknown, use the Student's t distribution with n – 1 degrees of freedom.
Step-by-step explanation:
Consider the provided information.
The t-distribution of the Student is a distribution of probability that is used when when the sample size is small and/or when the population variance is unknown to estimate population parameters.
The number of independent observations is equal to the sample size minus one when calculating a mean score or a proportion from a single sample.
Since µ and σ determine the shape of the distribution so we use standard normal distribution if σ is known.
Hence, the correct option is A) If σ is known, use the standard normal distribution. If σ is unknown, use the Student's t distribution with n – 1 degrees of freedom.
An ant is moving on a numbered, horizontal line every second. The number ranges from −[infinity] to [infinity] . It moves to the left integer with a probability of 1/4 and to the right integer with a probability of 3/4. Suppose initially it starts at 0, so what is the probability that after 3 seconds it will be at 1?
The probability that the ant will be at 1 after 3 seconds is 49/64.
Explanation:To find the probability that the ant will be at 1 after 3 seconds, we need to consider all possible paths it can take. After each second, the ant can either move left with a probability of 1/4 or move right with a probability of 3/4.
Let's analyze all the possible paths:
The ant moves right in all three seconds - Probability = (3/4) * (3/4) * (3/4) = 27/64The ant moves right in the first two seconds and then moves left in the third second - Probability = (3/4) * (3/4) * (1/4) = 9/64The ant moves right in the first second, then moves left in the second second and right in the third second - Probability = (3/4) * (1/4) * (3/4) = 9/64The ant moves left in the first two seconds and then moves right in the third second - Probability = (1/4) * (1/4) * (3/4) = 3/64The ant moves left in all three seconds - Probability = (1/4) * (1/4) * (1/4) = 1/64Adding up the probabilities from each path, the total probability that the ant will be at 1 after 3 seconds is (27/64) + (9/64) + (9/64) + (3/64) + (1/64) = 49/64.
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A magazine currently has 8700 subscribers for its online web version. It is adding members at the rate of R(t) = 190e0.03t subscribers per month. If the proportion of members who remain subscribed t months from now is S(t) = e−0.06t, how many online subscribers will the magazine have three years from now? (Round your answer to the nearest integer.)
Number of subscriber the magazine will have after 3 years from now approximately be 8767
Solution:
Given that magazine currently has 8700 subscribers for its online web version
[tex]\begin{array}{l}{\mathrm{R}(\mathrm{t})=190 \mathrm{e}^{0.03 \mathrm{t}} \text { subscribers/month }} \\\\ {\mathrm{S}(\mathrm{t})=\mathrm{e}^{-0.06 \mathrm{t}}}\end{array}[/tex]
After 3 years, time(t) = 36 month
Total number of subscribers after 3 years from now :
Substitute "t" = 36
[tex]\begin{array}{l}{\mathrm{R}(36)=190 \mathrm{e}^{0.03 \times(36)}=190 \times(2.944)} \\\\ {\mathrm{R}(36) \approx 560} \\\\ {\mathrm{S}(36)=\mathrm{e}^{-0.06 \times(36)}=0.12}\end{array}[/tex]
Subscribers remaining = 0.12 x 560 = 67.2
The magazine currently has 8700 subscribers
Added Subscriber = 8700 + 560 = 9260
Remaining Subscriber = 8700 + 67.2 = 8767.2
Therefore number of subscriber the magazine will have after 3 years from now approximately be 8767
A sample of 20 cigarettes is tested to determine nicotine content and the average value observed was 1.2 mg. Compute a 99 percent two-sided confidence interval for the mean nicotine content of a cigarette if it is known that the standard deviation of a cigarette’s nicotine content is σ = .2 mg.
Answer: [tex]1.0848<\mu<1.3152[/tex]
Step-by-step explanation:
Confidence interval for population mean is given by :-
[tex]\overline{x}-z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}< mu< \overline{x}+ z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]
, where [tex]z_{\alpha/2}[/tex] = two -tailed z-value for [tex]{\alpha[/tex] (significance level)
n= sample size .
[tex]\sigma[/tex] = Population standard deviation.
[tex]\overline{x}[/tex] = Sample mean
By considering the given information , we have
[tex]\sigma=0.2\text{ mg}[/tex]
[tex]\overline{x}=1.2\text{ mg}[/tex]
n= 20
[tex]\alpha=1-0.99=0.01[/tex]
Using z-value table ,
Two-tailed Critical z-value : [tex]z_{\alpha/2}=z_{0.005}=2.576[/tex]
The 99 percent two-sided confidence interval for the mean nicotine content of a cigarette will be :-
[tex]1.2- (2.576)\dfrac{0.2}{\sqrt{20}}<\mu<1.2+ (2.576)\dfrac{0.2}{\sqrt{20}}\\\\=1.2- 0.1152<\mu<1.2+ 0.1152\\\\=1.0848<\mu<1.3152 [/tex]
Hence, the 99 percent two-sided confidence interval for the mean nicotine content of a cigarette: [tex]1.0848<\mu<1.3152[/tex]
An insurance company sets up a statistical test with a null hypothesis that the average time for processing a claim is 7 days, and an alternative hypothesis that the average time for processing a claim is greater than 7 days.
After completing the statistical test, it is concluded that the average time exceeds 7 days.
However, it is eventually learned that the mean process time is really 9 days.
What type of error occurred in the statistical test?
Answer:
There was no error
Step-by-step explanation:
There are two types of statistical errors, the type 1 error and the type 2 error. In this case we refute the null hypothesis when the hypothesis is, in fact, false, because the mean process is 9 days instead of 7. Therefore we made no errors.
If the null hypothesis were True, a type 1 error would have ocurred. If the null hypothesis were false and we didnt refute it, then a type 2 error would have ocurred.
You roll a fair die three times. What is the probability of each of the following?
a) You roll all 4's.
b) You roll all even numbers.
c) None of your rolls gets a number divisible by 2.
d) You roll at least one 2.
e) The numbers you roll are not all 2's.
This solution calculates the probability of different outcomes when rolling a fair die three times. The results are obtained by defining the successful outcomes versus the total possible outcomes for each specific event.
Explanation:This question is about probability. A fair die has 6 equally likely outcomes. Let's address each part:
You roll all 4's: There's 1 chance in 6 to roll a 4. Since you're rolling the die three times, the probability is (1/6) * (1/6) * (1/6) = 1/216. You roll all even numbers: There are 3 even numbers on a die (2, 4, 6), so the probability is (3/6) * (3/6) * (3/6) = 1/8. None of your rolls gets a number divisible by 2: This is the same as rolling all odd numbers. There are 3 odd numbers on a die (1, 3, 5), so the probability is (3/6) * (3/6) * (3/6) = 1/8. You roll at least one 2: The opposite of this are outcomes without any 2, i.e., combinations of 1, 3, 4, 5, and 6. Thus, subtract combinations without any 2 from total possible combinations: 1 - [(5/6) * (5/6) * (5/6)] = 91/216. The numbers you roll are not all 2's: The only case when this does not happen is when you roll a 2 three times. So, if we subtract the probability of rolling three 2's from 1, we get 1 - (1/6) * (1/6) * (1/6) = 215/216.Learn more about probability here:
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The probability of rolling all 4's is 1/64. The probability of rolling all even numbers is 7/64. The probability of not rolling all 2's is 63/64.
Explanation:a) The probability of rolling all 4's is ⅛3, or 1/64. There is only one way to roll a 4 on a fair die and a total of 6 possible outcomes on each roll, so the probability is 1/6. Since the rolls are independent, the probability of getting a 4 on all three rolls is (1/6)(1/6)(1/6) = 1/64.
b) The probability of rolling all even numbers is also 1/64. There are 3 even numbers on a die (2, 4, and 6), so the probability of rolling an even number on any single roll is 3/6 or 1/2. Since the rolls are independent, the probability of getting an even number on all three rolls is (1/2)(1/2)(1/2) = 1/8. However, we need to subtract the probability of rolling all 4's from this, which is also 1/64. So the final probability is 1/8 - 1/64 = 7/64.
c) The probability of none of the rolls getting a number divisible by 2 is 1 - (1/2)(1/2)(1/2) = 1 - 1/8 = 7/8. This is the complement of rolling all even numbers.
d) The probability of rolling at least one 2 is 1 - the probability of rolling no 2's. The probability of not rolling a 2 on any single roll is 5/6, so the probability of not rolling a 2 on all three rolls is (5/6)(5/6)(5/6) = 125/216. Therefore, the probability of rolling at least one 2 is 1 - 125/216 = 91/216.
e) The probability of rolling all 2's is 1/64. Therefore, the probability of not rolling all 2's is 1 - 1/64 = 63/64.
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Ms. Taylor's students give cards to each of
the 3 class parent helpers. There are
24 cards. How many cards will each helper
get if the students give an equal number of
cards to each helper?
To determine the number of cards each class parent helper will receive, divide the total number of cards (24) by the number of helpers (3), resulting in each helper receiving 8 cards.
Ms. Taylor's students are giving out cards to each of the 3 class parent helpers equally, and there are a total of 24 cards. To find out how many cards each helper gets, we can divide the total number of cards by the number of helpers.
Step 1: Determine the total number of cards: 24.
Step 2: Determine the number of class parent helpers: 3.
Step 3: Divide the total number of cards by the number of helpers: 24 ÷ 3.
Step 4: Calculate: 24 ÷ 3 = 8.
So, each class parent helper will receive 8 cards.
The 20 colleges of interest to a high school senior include 8 that are expensive ( tuition more than 20,000 per year), 8 that are far from home( more than 200 miles away), and 7 that are both expensive and far from home. If the student decides to select a college that is not expensive and within 200 miles from home, how many selections are possible?
Answer: 6
Step-by-step explanation:
Let S= Total colleges
A = colleges are expensive.
B= colleges are far from home( more than 200 miles away).
Given : n(S)= 20
n(A)=8
n(B)=8
n(A∩B) =2
Then, the number of college that are not expensive and within 200 miles from home :-
[tex]n(A'\cap B')=n(S)-n(A\cup B)\\\\=20-(n(A)+n(B)-N(A\cap B))\ \ [\because\ n(A\cup B)=n(A)+n(B)-N(A\cap B)]\\\\=20-(8+8-2)\\\\=20-14=6[/tex]
i.e. the number of college that are not expensive and within 200 miles from home=6
Hence, the number of possible selections are 6 .
An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed with a standard deviation of 45 hours. If a sample of 35 bulbs has an average life of 710 hours, find a 96% confidence interval for the population mean of all bulbs produced by this firm.
Answer:
[693,38:725,62]hs
Step-by-step explanation:
Hello!
Your study variable is X: the lifespan of a light bulb. This variable is said to have an approximately normal distribution.
X≈N(μ;σ²)
Were
μ= populatiom mean
σ= 45 hs standard deviation
A sample of n= 35
To estimate the population mean with a confidence interval you have to use the Z statistic:
X{bar} ± [tex]Z_{1-\alpha /2}[/tex]*(σ/√n)
[tex]Z_{1-\alpha /2} = Z_{0.98} = 2.054[/tex]
710 ± 2.054*(45/√35
[693,38:725,62]hs
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The police rounded up 35 people suspected of robbing a bank. Each person was assigned a number from 1 to 35. After a short investigation, three of them were arrested, numbers #14, #17 and #26. They made the following statements under intensive questioning: #14: I’m innocent. #17: I’m innocent. #26: #14 is guilty. If only one of these statements is true, who robbed the bank?
Answer:
If only one of these statements is true, #17 robbed the bank.
Step-by-step explanation:
1st Scenario: #26 tells the truth
If #26 is telling the truth, that means #14 is guilty and, consequently, lying. However, that would also mean that #17 is telling the truth and then more than one statement would be true.
2nd Scenario: #17 tells the truth
If #17 is telling the truth, #17 is innocent, but then again either #26 or #14 are lying and two statements would be true.
3rd Scenario: #14 tells the truth
If #14 is telling the truth, #14 is innocent and, consequently, #26 is lying. That leaves us with #17 claiming innocence, but since only #14 can be telling the truth, #17 is lying and robbed the bank.
I am graphing something with 29 points. I know that you have to label the points with a capital letter. However, there are only 26 letters in the English alphabet, so what do I name the remaining 3 points?
Answer:
a₁, a₂, a₃, . . .
Step-by-step explanation:
When we name something in Mathematics it is always advisable to number them as [tex]$ a_1, a_2, a_3, ...$[/tex].
Because the alphabets are only 26 in number we might run out of notations.
The simple logic behind the notion of [tex]$ a_1, a_2, a_3, ... $[/tex] is that the numbers have no end and we can number as many variables we want using this logic.
In this problem, you can continue your notation from [tex]$ a_{27}, a_{28}, a_{29} $[/tex] so that you don't have to make a change in your figure.
A study of 20 worldwide financial institutions showed the correlation between their assets and pretax profit to be 0.81.State the decision rule for 0.025 significance level: H0: rho ≤ 0; H1: rho > 0. (Round your answer to 3 decimal places.)Compute the value of the test statistic. (Round your answer to 2 decimal places.)Can we conclude that the correlation in the population is greater than zero? Use the 0.025 significance level.
Answer:
t=5.86
We can conclude that the population correlation between their assets and pretax profit is higher than 0 at the significance level provided.
Step-by-step explanation:
n= 20 random sample taken
r=0.81 correlation coeffcient obtained
[tex]\alpha=0.025[/tex] significance level obtained
1) System of hypothesis
The system of hypothesis given are:
Null hypothesis :[tex]\rho \leq 0[/tex]
Alternative hypothesis: [tex] \rho >0[/tex]
2) Calculate the statistic
The statistic in order to test an hypothesis for the correlation coefficient is given by:
[tex]t =\frac{r\sqrt{n-2}}{\sqrt{1-r^2}}[/tex]
This statistic follows a t distribution with n-2 degrees of freedom
If we replace the values given we got:
[tex]t =\frac{0.81\sqrt{20-2}}{\sqrt{1-(0.81^2)}}=5.86[/tex]
3) P value
For this case w eneed to calculate first the degrees of freedom
[tex]df=n-2=20-2=18[/tex]
And then analyzing the alternative hypothesis we can calculate the p value on this way:
[tex]p_v =P(t_{18} >5.86) =1-P(t_{18} <5.86)=1-0.99999=7.51x10^{-6}[/tex]
Since the P-value is smaller than the significance level, we have enough evidence to reject the null hypothesis in favor of the alternative. We conclude "there is sufficient evidence at the significance level to conclude that there is a linear relationship in the population between the two variables analyzed."
The decision rule for a 0.025 significance level intends to reject the null hypothesis if the p-value from the test statistic is less than 0.025. Using a standard Z test to compute the test statistic, we substitute the given values to find the result. As the p-value is 0.026, which is greater than the 0.025 significance level, we do not reject the null hypothesis, indicating insufficient evidence to conclude that the population correlation is greater than zero.
Explanation:To answer your question, let's first state the decision rule for a 0.025 significance level. Considering your null hypothesis (H0: rho ≤ 0) and the alternative hypothesis (H1: rho > 0), we want to reject the null hypothesis if the p-value from the test statistic is less than 0.025.
Since we aren't given a specific formula or context about how the test statistic should be computed, I'll assume that it's via a standard Z test using the formula:
[tex]Z = (r\times \sqrt(n-2))/(\sqrt(1-r^2)),[/tex]where r is the sample correlation (0.81 in this case), and n is the number of data points (20 in this case). Let's substitute these values into the formula to find the test statistic.
Based upon the p-value from your computing software as 0.026 and the significance level being 0.025, we do not reject the null hypothesis because the p-value is greater than the significance level. In other words, there is insufficient evidence at the 0.025 significance level to conclude that the population correlation is greater than zero.
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The Greater Pittsburgh Area Chamber of Commerce wants to estimate the mean time workers who are employed in the downtown area spend getting to work. A sample of the 15 workers reveals the following number of minutes spent traveling. 29 38 38 33 38 21 45 34 40 37 37 42 30 29 35Develop a 98 percent confidence interval for the population mean. Interpret the results.
Answer:
98% Confidence interval: (31.74,38.4)
Step-by-step explanation:
We are given the following data set:
29, 38, 38, 33, 38, 21, 45, 34, 40, 37, 37, 42, 30, 29, 35
Formula:
[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]
where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{526}{15} = 35.07[/tex]
Sum of squares of differences = 506.93
[tex]S.D = \sqrt{\displaystyle\frac{506.93}{14}} = 6.02[/tex]
98% Confidence interval:
[tex]\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}[/tex]
Putting the values, we get,
[tex]t_{critical}\text{ at degree of freedom 14 and}~\alpha_{0.02} = \pm 2.145[/tex]
[tex]35.07 \pm 2.145(\frac{6.02}{\sqrt{15}} ) = 35.07 \pm 3.33 = (31.74,38.4)[/tex]
Thus, there is 98% confidence that the population mean number of minutes spent traveling by workers is between 31.74 mins and 38.40
Let X equal the thickness of spearmint gum manufactured for vending machines. Assume that the distribution of X is N(mu, sigma^2). The target thickness is 7.5 hundreds of an inch. We shall test the null hypothesis H_0: mu=7.5 against a two-sided alternative hypothesis, using 10 observations. Let m denote the sample mean and S^2 denote the sample variance.
Define the test statistic in terms of m and s.
Answer:
[tex]t=\frac{m-7.5}{\frac{s}{\sqrt{10}}}=\sqrt{10} (\frac{m-7.5}{s})[/tex]
Step-by-step explanation:
1) Notation
n=10 represent the sample size
[tex]\bar X=m[/tex] represent the sample mean
[tex]s[/tex] represent the sample standard deviation
m represent the margin of error
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".
The null hypothesis attempts "to show that no variation exists between variables or that a single variable is no different than its mean"
The alternative hypothesis "is the hypothesis used in hypothesis testing that is contrary to the null hypothesis"
2) State the null and alternative hypotheses.
We need to conduct a hypothesis in order to determine if the mean for the population is 7.5 or no, the system of hypothesis would be:
Null hypothesis:[tex]\mu =7.5[/tex]
Alternative hypothesis:[tex]\mu \neq 7.5[/tex]
We don't know the population deviation, so for this case we can use the t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
3) Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{m-7.5}{\frac{s}{\sqrt{10}}}=(\sqrt{10})\frac{m-7.5}{s}[/tex]
and we have our statistic in terms of m (mean) and the sample standard deviation s.
According to the record of the registrar's office at a state university, 35% of the students arefreshman, 25% are sophomore, 16% are junior, and the rest are senior. Among the freshmen,sophomores, juniors, and seniors, the portion of students who live in the dormitory are,respectively, 80%, 60%, 30%, and 20%. What is the probability that a randomly selected studentis a freshman who lives in a dormitory?
A) 0.28
B) 0.32
C) 0.52
D) 0.38
Answer:
the probability is 0.28
Step-by-step explanation:
using Bayes's theorem
P(A|B)=P(A∩B)/P(B)
where
P(A∩B) = probability that events A and B happen
P(A|B) = probability that event A happen if B already happened
P(B)= probability of event B
therefore
P(A∩B)=P(A|B)*P(B)
if event A= selection of a student that lives in a dormitory and event B = selection of a freshmen student
P(A|B) = 0.8 (live in a dormitory knowing that is a freshmen student )
P(B) = 0.35 (freshmen student)
P(A∩B)=P(A|B)*P(B) = 0.8* 0.35 =0.28
The probability that a randomly selected student is a freshman who lives in a dormitory is 0.28. So the correct option is A.
Explanation:To find the probability that a randomly selected student is a freshman who lives in a dormitory, we need to multiply the probability that a student is a freshman by the probability that a freshman lives in the dormitory. According to the given information, 35% of the students are freshmen, and among them, 80% live in the dormitory.
We calculate this probability using the formula:
Probability of (Freshman who lives in dorm) = Probability of (Freshman) × Probability of (Dorm | Freshman)
Probability of (Freshman who lives in dorm) = 0.35 × 0.80
Probability of (Freshman who lives in a dorm) = 0.28
Thus, the correct answer is A) 0.28.
In a random sample of 150 customers of a high-speed Internetprovider, 63 said that their service had been interrupted one ormore times in the past month.A) Find a 95% confidence interval for the proportion of customerswhose service was interrupted one or more times in the pastmonth.B) Find a 99% confidence interval for the proportion of customerswhose service was interrupted one or more times in the pastmonth.C) Find the sample size needed for a 95% confidence interval tospecify the proportion to within -+ 0.05.D) Find the sample size needed for a 99% confidence interval tospecify the proportion to within -+ 0.05.
Answer:
a) The 95% confidence interval would be given by (0.341;0.499)
b) The 99% confidence interval would be given by (0.316;0.524)
c) n=335
d)n=649
Step-by-step explanation:
1) Notation and definitions
[tex]X_{IS}=63[/tex] number of high speed internet users that had been interrupted one or more times in the past month.
[tex]n=150[/tex] random sample taken
[tex]\hat p_{IS}=\frac{63}{150}=0.42[/tex] estimated proportion of high speed internet users that had been interrupted one or more times in the past month.
[tex]p_{IS}[/tex] true population proportion of high speed internet users that had been interrupted one or more times in the past month.
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".
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})[/tex]
1) Part a
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:
[tex]t_{\alpha/2}=-1.96, t_{1-\alpha/2}=1.96[/tex]
The confidence interval for the mean is given by the following formula:
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
If we replace the values obtained we got:
[tex]0.42 - 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.341[/tex]
[tex]0.42 + 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.499[/tex]
The 95% confidence interval would be given by (0.341;0.499)
2) Part b
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2 =0.005[/tex]. And the critical value would be given by:
[tex]t_{\alpha/2}=-2.58, t_{1-\alpha/2}=2.58[/tex]
The confidence interval for the mean is given by the following formula:
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
If we replace the values obtained we got:
[tex]0.42 - 2.58\sqrt{\frac{0.42(1-0.42)}{150}}=0.316[/tex]
[tex]0.42 + 2.58\sqrt{\frac{0.42(1-0.42)}{150}}=0.524[/tex]
The 99% confidence interval would be given by (0.316;0.524)
3) Part c
The margin of error for the proportion interval is given by this formula:
[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] (a)
And on this case we have that [tex]ME =\pm 0.05[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:
[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] (b)
And replacing into equation (b) the values from part a we got:
[tex]n=\frac{0.42(1-0.42)}{(\frac{0.05}{1.96})^2}=374.32[/tex]
And rounded up we have that n=335
4) Part d
The margin of error for the proportion interval is given by this formula:
[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] (a)
And on this case we have that [tex]ME =\pm 0.05[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:
[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] (b)
And replacing into equation (b) the values from part a we got:
[tex]n=\frac{0.42(1-0.42)}{(\frac{0.05}{2.58})^2}=648.599[/tex]
And rounded up we have that n=649
Wire electrical-discharge machining (WEDM) is a process used to manufacture conductive hard metal components. It uses a continuously moving wire that serves as an electrode. Coating on the wire electrode allows for cooling of the wire electrode core and provides an improved cutting performance. An article gave the following sample observations on total coating layer thickness (in µm) of eight wire electrodes used for WEDM. 21 16 29 36 42 25 24 25
Calculate a 99% CI for the standard deviation of the coating layer thickness distribution. (Round your answers to two decimal places.) , Is this interval valid whatever the nature of the distribution? Explain.
Yes, there are enough data points for this interval to be valid.
No, validity of this interval requires that coating layer thickness be, at least approximately, normally distributed.
To calculate a 99% confidence interval for the standard deviation of the coating layer thickness distribution, use the sample variance, sample size, and chi-square distribution. The coating layer thickness should be approximately normally distributed for the confidence interval to be valid.
Explanation:To calculate a 99% confidence interval for the standard deviation of the coating layer thickness distribution, we can use the chi-square distribution. First, we need to calculate the sample variance and sample size. Then, we can use the chi-square distribution table to find the critical values for a 99% confidence interval with (n-1) degrees of freedom. Finally, we can calculate the confidence interval using the formula:
CI = sqrt((n - 1) * s^2 / X^2)
where CI is the confidence interval, n is the sample size, s^2 is the sample variance, and X^2 is the critical value from the chi-square distribution.
In this case, the coating layer thickness should be approximately normally distributed for the confidence interval to be valid.
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In the given case, the answer No, validity of this interval requires that coating layer thickness be, at least approximately, normally distributed.
To calculate the 99% confidence interval (CI) for the standard deviation of the coating layer thickness distribution, we can use the chi-square distribution.
Given that the sample size is small (n = 8), we'll use the chi-square distribution with n - 1 = 7 degrees of freedom.
The formula for the confidence interval is:
[tex]\[ \left( \sqrt{\frac{(n-1)S^2}{\chi^2_{\alpha/2, n-1}}}, \sqrt{\frac{(n-1)S^2}{\chi^2_{1-\alpha/2, n-1}}} \right) \][/tex]
Given the sample observations: 21, 16, 29, 36, 42, 25, 24, 25, we first need to calculate the sample standard deviation:
[tex]\[ S = \sqrt{\frac{\sum_{i=1}^{n}(X_i - \bar{X})^2}{n-1}} \][/tex]
Let's perform the calculations:
[tex]\[ S = \sqrt{\frac{(21-27.625)^2 + (16-27.625)^2 + \ldots + (25-27.625)^2}{7}} \][/tex]
[tex]\[ S \approx 9.38 \][/tex]
Now, plug the values into the formula:
[tex]\[ \left( \sqrt{\frac{7 \times 9.38^2}{18.48}}, \sqrt{\frac{7 \times 9.38^2}{2.17}} \right) \][/tex]
[tex]\[ \left( \sqrt{26.97}, \sqrt{229.78} \right) \][/tex]
[tex]\[ \left( 5.19, 15.15 \right) \][/tex]
Therefore, the 99% confidence interval for the standard deviation of the coating layer thickness distribution is approximately [tex]\( (5.19, 15.15) \)[/tex] micrometers.
Therefore, the answer is: No, validity of this interval requires that coating layer thickness be, at least approximately, normally distributed.
An agricultural field trial compares the yield of two varieties of tomatoes for commercial use – one is the common variety used by farmers and the other is a new hybrid. The researchers plant each type of tomato on one half of each of 10 plots. At harvest time the difference in the yield of the hybrid tomato minus the common variety in pounds per plant (ppp) is calculated for each plot. The average difference is 1.05 ppp with a standard deviation of 1.23 ppp. In testing whether the hybrid tomato has a higher yield with a significance level of α = 0.05, what is the p-value? You may assume the conditions to carry out the relevant hypothesis test are satisfied.
Answer:
p-value 0.0124
Step-by-step explanation:
Hello!
In this experiment, 10 plots were randomly sampled and each plot was divided in a half, one half was planted with the common variety of tomato and the other half was planted with the new hybrid variety. At harvest time it was measured the pounds per plant harvested and the variable difference was established. This study variable is defined as the difference between the pounds per plant of the hybrid variety minus the pound per plant of the common variety. Symbolized: Xd: Xnew - Xcommon.
The purpose of this experiment is to test whether there is a difference between the yield of both species. This is a classic example of a paired test, in which you want to put two dependent samples to test. The easiest way to recognize this type of test, if it is not specified in the problem, is that both variables are measured in the same sampling unit. In this case, the sampling unit is "one plot" that was divided and both verities of tomatoes were planted on it.
This test is also called "paired samples T-test" and as a statistic, you have to use the Student t.
The data given is:
sample: 10 plots
sample mean: Xd[bar]: 1.05ppp
sample standard deviation Sd: 1.23ppp
If the hybrid has a higher yield, that would mean that the difference between them would be positive if the difference is positive, its safe to assume that the population mean of the difference will be positive as well. Symbolically: μd>0
With this, you can state the hypothesis as:
H₀:μd≤0
H₁:μd>0
α: 0.05
t= Xd[bar] - μd ~ [tex]t_{n-1}[/tex]
Sd/n
t[tex]t_{obs}[/tex]= (1.05 - 0) / (1.23/√10) = 2.69
This is a one-tailed test with critical value [tex]t_{9;0.95} = 1.83[/tex]
the p-value for this test is also one tailed.
p([tex]t_{9}[/tex]≥2.69) = 1-P([tex]t_{9}[/tex]<2.69) = 1 - 0.9876 = 0.0124
The decision is to reject the null hypothesis.
I hope it helps!
Please help if so thank you And explain
Answer:
x = 34°
Step-by-step explanation:
Given AC and BD are perpendicular bisectors, we can say that at point E, there are 4 right angles [perpendicular bisectors intersect to create 4 90 degree angles].
Now, if we look at the triangle AED, we know that it is a right triangle, meaning that angle E is a right angle.
Also,
We know sum of 3 angles in a triangle is 180 degrees. Thus, we can write:
∠A + ∠E + ∠D = 180
Note: Angle A and Angle D are just the half part of the diagram. More exactly we can write:
∠EAD + ∠ADE + ∠DEA = 180
Given,
∠EAD = 56
∠DEA = 90
We now solve:
∠EAD + ∠ADE + ∠DEA = 180
56 + ∠ADE + 90 = 180
146 + ∠ADE = 180
146 + x = 180
x = 180 - 146
x = 34°
In a recent poll, 600 people were asked if they liked soccer, and 72% said they did. Based on this, construct a 99% confidence interval for the true population proportion of people who like soccer.
As in the reading, in your calculations:
--Use z = 1.645 for a 90% confidence interval
--Use z = 2 for a 95% confidence interval
--Use z = 2.576 for a 99% confidence interval.
Give your answers as decimals, to 4 decimal places.
Out of 500 people sampled, 80 had kids. Based on this, construct a 90% confidence interval for the true population proportion of people with kids.
As in the reading, in your calculations:
--Use z = 1.645 for a 90% confidence interval
--Use z = 2 for a 95% confidence interval
--Use z = 2.576 for a 99% confidence interval.
Give your answers to three decimals
Give your answers as decimals, to three decimal places.
Answer:
(0.6728, 0.7672)
(0.1330, 0.1870)
Step-by-step explanation:
Given that in a recent poll, 600 people were asked if they liked soccer, and 72% said they did.
Std error = [tex]\sqrt{\frac{pq}{n} } \\=\sqrt{\frac{0.72(0.28)}{600} } \\=0.0183[/tex]
Margin of error for 99% we would use the value [tex]z = 2.576[/tex]
Margin of error = [tex]2.576*SE\\=2.576*0.0183\\=0.0472[/tex]
Confidence interval lower bound = [tex]0.72-0.0472=0.6728[/tex]
Upper bound = [tex]0.72+0.0472=0.7672[/tex]
99% confidence interval for the true population proportion of people who like soccer.=(0.6728, 0.7672)
b) n =500
Sample proportion p=[tex]\frac{80}{500} =0.16[/tex]
Margin of error for 90% = [tex]1.645*\sqrt{\frac{0.16*0.84}{500} } \\\\=0.0270[/tex]
90% confidence interval for the true population proportion of people with kids. =[tex](0.16-0.0270, 0.16+0.0270)\\=(0.1330, 0.1870)[/tex]
Which of the following equations could be the result of using the comparison method to solve the system shown? x + y = 5 2x + y = 7
Answer:
so (x,y) = (2,3)
Step-by-step explanation:
we have
x +y= 5 ------ equation 1
2x+y= 7-------equation 2
Multiplying equation--1 by 2, we will have equation---3
2x +2y= 10-----equation 3
now subtracting equation 2 from equation 3
2x + 2y= 10
2x + y = 7
y = 3
adding value of y in equation 1
x + y = 5
x= 5 -y
x= 5 - 3
x =2
so (x,y) = (2,3)
Answer: the correct answer is
5 - x = 7 - 2x
Step-by-step explanation:
Let the number of chocolate chips in a certain type of cookie have a Poisson distribution. We want the probability that a cookie of this type contains at least two chocolate chips to be greater than 0.99. Find the smallest value of the mean that the distribution can take.
Answer:
[tex]\lambda \geq 6.63835[/tex]
Step-by-step explanation:
The Poisson Distribution is "a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event".
Let X the random variable that represent the number of chocolate chips in a certain type of cookie. We know that [tex]X \sim Poisson(\lambda)[/tex]
The probability mass function for the random variable is given by:
[tex]f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...[/tex]
And f(x)=0 for other case.
For this distribution the expected value is the same parameter [tex]\lambda[/tex]
[tex]E(X)=\mu =\lambda[/tex]
On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:
[tex]P(X\geq 2)=1-P(X<2)=1-P(X\leq 1)=1-[P(X=0)+P(X=1)][/tex]
Using the pmf we can find the individual probabilities like this:
[tex]P(X=0)=\frac{e^{-\lambda} \lambda^0}{0!}=e^{-\lambda}[/tex]
[tex]P(X=1)=\frac{e^{-\lambda} \lambda^1}{1!}=\lambda e^{-\lambda}[/tex]
And replacing we have this:
[tex]P(X\geq 2)=1-[P(X=0)+P(X=1)]=1-[e^{-\lambda} +\lambda e^{-\lambda}[][/tex]
[tex]P(X\geq 2)=1-e^{-\lambda}(1+\lambda)[/tex]
And we want this probability that at least of 99%, so we can set upt the following inequality:
[tex]P(X\geq 2)=1-e^{-\lambda}(1+\lambda)\geq 0.99[/tex]
And now we can solve for [tex]\lambda[/tex]
[tex]0.01 \geq e^{-\lambda}(1+\lambda)[/tex]
Applying natural log on both sides we have:
[tex]ln(0.01) \geq ln(e^{-\lambda}+ln(1+\lambda)[/tex]
[tex]ln(0.01) \geq -\lambda+ln(1+\lambda)[/tex]
[tex]\lambda-ln(1+\lambda)+ln(0.01) \geq 0[/tex]
Thats a no linear equation but if we use a numerical method like the Newthon raphson Method or the Jacobi method we find a good point of estimate for the solution.
Using the Newthon Raphson method, we apply this formula:
[tex]x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}[/tex]
Where :
[tex]f(x_n)=\lambda -ln(1+\lambda)+ln(0.01)[/tex]
[tex]f'(x_n)=1-\frac{1}{1+\lambda}[/tex]
Iterating as shown on the figure attached we find a final solution given by:
[tex]\lambda \geq 6.63835[/tex]
The problem pertains to Poisson Distribution in probability theory, focusing on finding the smallest mean (λ) such that the probability of having at least two chocolate chips in a cookie is more than 0.99. This involves solving an inequality using the formula for Poisson Distribution.
Explanation:This problem pertains to the Poisson Distribution, often used in probability theory. In particular, we're looking at the number of events (in this case, the number of chocolate chips) that occur within a fixed interval. Here, the interval under study is a single cookie. The question requires us to find the smallest value of λ (the mean value of the distribution) such that the probability of getting at least two chocolate chips in a cookie is more than 0.99.
Using the formula for Poisson Distribution, the probability of finding k copies of an event is given by:
P(X=k) = λ^k * exp(-λ) / k!
The condition here is that the probability of finding at least 2 copies is more than 0.99. Therefore, you formally need to solve the inequality:
P(X>=2) = 1 - P(X=0) - P(X=1) > 0.99
Substituting the values of P(X=0) and P(X=1) from our standard formula, you will need to calculate and find the smallest value of λ that satisfies this inequality.
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Peter makes six cakes of 5 lb each three chocolate cakes and three vanilla cakes to serve at a party for 120 people. How many ounces of cake will each guest have? If the guest list increases to 150 guests, how many more cakes does Peter need to make?
Answer:
4 oz2 cakesStep-by-step explanation:
Since 6 cakes serve 120 people, each cake serves 120/6 = 20 people. Each cake weighs (5 lb)(16 oz/lb) = 80 oz. Then each person gets ...
(80 oz)/(20 persons) = 4 oz/person
__
150 servings will require 150/20 = 7.5 cakes. Peter already has 6 cakes, so needs to make 2 more.
5lb per cake *6 cakes= 30lb
30lb / 120 people = 0.25lb per person
150 people - 120 people = 30 people
0.25lb per person * 30 people = 7.5lb
5lb___1 cake
7.5lb___X=1.5 cakes
(7.5lb * 1 cake)/5lb = 1.5 cakes
The area of a rectangular plot 32 feet long and 25 feet wide will be doubled by adding an equal width to each side of the plot. Which equation can be used to find this added width?
(x + 32)(x + 25) = 1600
(2x + 32)(2x + 25) = 1600
(2x + 32)(x + 25) = 800
(x + 32)(x + 25) = 800
Final answer:
To find the added width of the rectangular plot, set up the equation (32 + x)(25 + x) = 2(32)(25) based on the fact that the area is doubled. Solving this equation will give the added width.
Explanation:
To find the added width of the rectangular plot, we need to determine the increase in length and width that will double the area. The original dimensions of the plot are 32 feet long and 25 feet wide. Since equal width is added to each side, the new length and width will be 32 + x and 25 + x, respectively. To find x, we can set up the equation (32 + x)(25 + x) = 2(32)(25) using the fact that the area is doubled. Solving this equation will give us the added width.
Expanding (32 + x)(25 + x), we get 800 + 57x + x^2 = 1600. Rearranging and simplifying, we get x^2 + 57x - 800 = 0. We can then solve this quadratic equation using factoring or the quadratic formula to find the value of x.
The correct equation to use in order to find the added width is (x + 32)(x + 25) = 800.
In the triangle above, which angle is smallest in measure?
A. ∠A
B. ∠B
C. ∠C
D. All three angles are congruent.
Answer:
A. ∠A
Step-by-step explanation:
The shortest side is BC. The smallest angle is opposite the shortest side, so is angle A.
∠A is the smallest in measure.
In a poll conducted by the Gallup organization in April 2013, 48% of a random sample of 1022 adults in the U.S. responded that they felt that economic growth is more important than protecting the environment. Calculate and interpret a 95% confidence interval for the proportion of all U.S. adults in 5 April 2013 who felt that economic growth is more important than protecting the environment. Make sure to include all steps.
Answer: 95% confidence interval would be (0.449, 0.511)
Step-by-step explanation:
Since we have given that
n = 1022
p = 48% = 0.48
We need to find the 95% confidence interval first.
z = 1.96
Margin of error would be
[tex]z\sqrt{\dfrac{p(1-p)}{n}}\\\\=1.96\times \sqrt{\dfrac{0.48\times 0.52}{1022}}\\\\=0.031[/tex]
95% confidence interval would be
[tex]p\pm 0.031\\\\=(0.48-0.031,0.48+0.031)\\\\=(0.449,0.511)[/tex]
It means true proportion who felt that economic growth is more important than protecting the environment is within 0.449 and 0.511 using 95% confidence.
At 95% confidence, how large a sample should be taken to obtain a margin of error of .03 for the estimation of the population proportion? Assume past data are not available for developing a planning value for p*.
Answer:
The sample should be 1,068.
Step-by-step explanation:
Consider the provided information.
Confidence level is 95% and margin of error is 0.03.
Thus,
1-α=0.95
α=0.05, E=0.03 and planning value [tex]\hat p=0.5[/tex]
Formula to calculate sample size is: [tex]n=\frac{\hat p(1-\hat p)(z_{\alpha/2})^2}{E^2}[/tex]
From the table we can find:
[tex]z_{\alpha/2}=z_{0.05/2}\\z_{0.025}=1.96[/tex]
Substitute the respective values in the above formula we get:
[tex]n=\frac{0.5(0.5)(1.96)^2}{(0.03)^2}[/tex]
[tex]n=\frac{0.25(1.96)^2}{(0.03)^2}\approx 1067.111[/tex]
Hence, the sample should be 1,068.
The function y = 1.29x + 3 represents the price y a website store charges for shipping x items. Which is a reasonable range for this function?
F- {…1.71, 3, 4.29, 5.58, 6.87, …}
G- {4.29, 5.58, 6.87, …}
H- {1, 2, 3, …}
J- all positive real numbers
Final answer:
The function y = 1.29x + 3 represents the price y a website store charges for shipping x items. The reasonable range for this function would be all positive real numbers.
Explanation:
The function y = 1.29x + 3 represents the price y a website store charges for shipping x items. The reasonable range for this function depends on the context. Since the function represents the price of shipping, the range should be positive, as shipping cannot have a negative price.
Therefore, the reasonable range for this function would be J- all positive real numbers.
Final answer:
The reasonable range for the shipping cost function starts from 3 and includes values that increase by 1.29 for each additional item, corresponding to whole numbers of items shipped. Therefore, the answer is G - {4.29, 5.58, 6.87, ...}.
Explanation:
The function given is y = 1.29x + 3, which represents the price y that a website store charges for shipping x items. Since shipping cannot have a negative cost and the minimum number of items shipped is either zero or a positive integer, the reasonable range for this function would begin at the point where x is zero. Therefore, we start our range by calculating the shipping cost for zero items: 1.29(0) + 3 = 3. As x increases, the cost will also increase linearly according to the function. Hence, all subsequent shipping prices will be greater than 3.
Now, let's consider what the reasonable range for shipping items would be. It would be abnormal to have a fractional number of items shipped because items are discrete entities. Thus, the list of shipping prices should only include charges for whole numbers of items. So, the range should only include prices that correspond to whole numbers of items shipped.
As a result, the reasonable range would include values starting from y = 3 onwards at intervals of 1.29 times an integer value. Option G, which starts from 4.29 and increases at a constant rate of 1.29, represents these intervals since 4.29 is the price for shipping one item (1.29*1 + 3). Any positive number of items shipped will result in a corresponding shipping price that is greater than 3, and since it is discreetly incremented, the prices will form a sequence of specific numbers, not all positive real numbers. Therefore, option G - {4.29, 5.58, 6.87, ...} - is the most appropriate answer.