Answer:
Reject [tex]H_0[/tex] . The change is statistically significant. The software does appear to improve exam scores.
Step-by-step explanation:
We are given that a Statistics professor has observed that for several years students score an average of 114 points out of 150 on the semester exam. The software is expensive, and the salesman offers to let the professor use it for a semester to see if the scores on the final exam increase significantly.
In the trial course that used this software, 218 students scored an average of 117 points on the final with a standard deviation of 8.7 points.
Let [tex]\mu[/tex] = mean scores on the final exam.
SO, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \leq[/tex] 114 points {means that the mean scores on the final exam does not increases after using software}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] > 114 points {means that the mean scores on the final exam increase significantly after using software}
The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;
T.S. = [tex]\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample average points = 117 points
s = sample standard deviation = 8.7 points
n = sample of students = 218
So, test statistics = [tex]\frac{117-114}{\frac{8.7}{\sqrt{218} } }[/tex] ~ [tex]t_2_1_7[/tex]
= 5.091
Now, P-value of the test statistics is given by the following formula;
P-value = P( [tex]t_2_1_7[/tex] > 5.091) = Less than 0.05%
Since, in the question we are not given with the level of significance at which hypothesis can be tested, so we assume it to be 5%. Now at 5% significance level, the t table gives critical value of 1.645 at 217 degree of freedom for right-tailed test. Since our test statistics is higher than the critical value of t as 5.091 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.
Therefore, we conclude that the change is statistically significant. The software does appear to improve exam scores.
The test statistic is 3.448 and the P-value is 0.0014. The appropriate conclusion is that the software does appear to improve exam scores.
Explanation:The test statistic can be calculated using the formula:
test_statistic = (sample_mean - population_mean) / (sample_standard_deviation / sqrt(sample_size))
Substituting in the given values:
test_statistic = (117 - 114) / (8.7 / sqrt(218)) = 3.448
The P-value can be calculated using a statistical software or calculator. For this question, the P-value is 0.0014
Since the P-value is less than 0.05 (significance level), we can reject the null hypothesis. The appropriate conclusion is: iv. Reject H_0. The change is statistically significant. The software does appear to improve exam scores.
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A peach pie is divided into eight equal slices
with 280 calories in each slice. How many
calories are in the entire peach pie?
The entire peach pie contains 2240 calories.
Explanation:To find the number of calories in the entire peach pie, we need to multiply the number of calories in each slice by the total number of slices. In this case, there are 8 equal slices in the peach pie, and each slice has 280 calories. Therefore, the total number of calories in the entire peach pie is 8 multiplied by 280, which equals 2240 calories.
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During a recent drought, a water utility in a certain town sampled 100 residential water bills and found that 73 of the residences had reduced their water consumption over that of the previous year. f) If 95% confidence intervals are computed for 200 towns, what is the probability that more than 192 of the confidence intervals cover the true proportions
Final answer:
The probability of more than 192 confidence intervals containing the true proportion when constructing 95% confidence intervals across 200 towns, which follows a binomial distribution.
Explanation:
When constructing confidence intervals for a proportion, if we state that we have 95% confidence, this means that if we were to take many samples and build a confidence interval from each sample, we would expect 95% of those intervals to contain the population proportion. In the case with 200 towns, 95% confidence suggests that approximately 190 out of 200 intervals would contain the true population proportion, as 95% of 200 is 190.
To find the probability that more than 192 of these intervals contain the true proportion, we would use the binomial distribution, where each interval has a 0.95 probability of containing the true proportion (success), and we are looking for the sum of the probabilities of 193, 194, ..., 200 successes out of 200 trials.
A survey of 25 young professionals fond that they spend an average of $28 when dining out, with a standard deviation of $10. (a) Find and interpret the 95% confidence interval for the average spending. (b) Based on your answer in part (a), can you conclude statistically that the population mean is less than $30? Explain
Answer:
a) The 95% of confidence intervals for the average spending
(23.872 , 32.128)
b) The calculated value t= 1< 1.711( single tailed test) at 0.05 level of significance with 24 degrees of freedom.
The null hypothesis is accepted
A survey of 25 young professionals statistically that the population mean is less than $30
Step-by-step explanation:
Step:-(i)
Given data a survey of 25 young professionals fond that they spend an average of $28 when dining out, with a standard deviation of $10
The sample size 'n' = 25
The mean of the sample x⁻ = $28
The standard deviation of the sample (S) = $10.
Level of significance ∝=0.05
The degrees of freedom γ =n-1 =25-1=24
tabulated value t₀.₀₅ = 2.064
Step 2:-
The 95% of confidence intervals for the average spending
([tex](x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,x^{-} + t_{\alpha }\frac{S}{\sqrt{n} } )[/tex]
[tex](28 - 2.064 \frac{10}{\sqrt{25} } ,28 + 2.064\frac{10}{\sqrt{25} } )[/tex]
( 28 - 4.128 , 28 + 4.128)
(23.872 , 32.128)
a) The 95% of confidence intervals for the average spending
(23.872 , 32.128)
b)
Null hypothesis: H₀:μ<30
Alternative Hypothesis: H₁: μ>30
level of significance ∝ = 0.05
The test statistic
[tex]t = \frac{x^{-}-mean }{\frac{S}{\sqrt{n} } }[/tex]
[tex]t = \frac{28-30 }{\frac{10}{\sqrt{25} } }[/tex]
t = |-1|
The calculated value t= 1< 1.711( single tailed test) at 0.05 level of significance with 24 degrees of freedom.
The null hypothesis is accepted
Conclusion:-
The null hypothesis is accepted
A survey of 25 young professionals statistically that the population mean is less than $30
a data set has a lower quartile of 3 and an interquartile range of 5. Which box plot could represent this data set?
Answer:
Answer:
The first box plot.
Step-by-step explanation:
Step-by-step explanation:
In a box plot, the lower quartile is the right hand side of the box. In the first plot, this is 3.
The upper quartile is the left hand side of the box. In the first plot, this is 8.
The interquartile range is the difference between these two values:
8-3 = 5.
2. A spike train, commonly used to study neural activity, is a sequence of recorded times at which a neuron fires an action potential (spike). The time in between consecutive spikes is called the interspike interval (ISI). Answer the questions below for an experiment in which the firing rate for a neuron is 10 per second. (a) Let X represent a single interspike interval (ISI) having an exponential distribution. State the distribution of X and give its parameter value(s). (2 points) (b) Give the expected value and variance for an interspike interval (ISI). (4 points) (c) What is the probability that an ISI is less than 0.07 seconds
Answer:
a) X ~ exp ( 10 )
b) E(X) = 0.1 , Var (X) = 0.01
c) P ( X < 0.07 ) = 0.00698
Step-by-step explanation:
Solution:-
- The spike train, used to study neural activity, the given time in between consecutive spikes (ISI) where the firing rate = 10 neurons per seconds.
- Denote a random variable "X"represent a single interspike interval (ISI) having an exponential distribution.
- Where X follows exponential distribution defined by event rate parameter i.e λ.
X ~ Exp ( λ )
- The event rate (λ) is the number of times an event occurs per unit time. Since we are studying a single interspike interval (ISI) - which corresponds to the firing rate. So, event rate (λ) = firing rate = 10 neurons per second. Hence, the distribution is:
X ~ Exp ( 10 )
- The expected value E(X) denotes the amount of time in which a single an event occurs; hence, the time taken for a single neuron.
E(X) = 1 / λ
E(X) = 1 / 10
E(X) = 0.1 s per neuron.
- The variance is the variation in the time taken by a single neuron to be emitted. It is defined as:
Var (X) = 1 / λ^2
Var (X) = 1 / 10^2
Var (X) = 0.01 s^2
- The probability that ISI is less than t = 0.07 seconds: P ( X < t = 0.07 s):
- The cumulative distribution function for exponential variate "X" is:
P ( X < t ) = 1 - e^(-λ*t)
- Plug the values and the determine:
P ( X < 0.07 ) = 1 - e^(-0.1*0.07)
= 1 - 0.99302
= 0.00698
Suppose that A and B each randomly, and independently, choose 3 of 10 objects. Find the expected number of objects (a) chosen by both A and B; (b) not chosen by either A or B; (c) chosen by exactly one of A and B.
Answer:
a) N(A∩B) = 0.9
b) N(A∩B) = 4.9
c) N(A or B) = 4.2
Step-by-step explanation:
Given that A and B each randomly, and independently, choose 3 of 10 objects;
P(A) = P(B) = 3/10 = 0.3
P(A') = P(B') = 1 - 0.3 = 0.7
a) chosen by both;
Probability of being chosen by both;
P(A∩B) = 0.3 × 0.3 = 0.09
Expected Number of objects being chosen by both;
N(A∩B) = P(A∩B) × N(total) = 0.09×10
N(A∩B) = 0.9
b) not chosen by either A or B;
Probability of not being chosen by either A or B;
P(A'∩B') = 0.7 × 0.7 = 0.49
Expected Number of objects being chosen by both;
N(A'∩B') = P(A'∩B') × N(total) = 0.49×10
N(A∩B) = 4.9
c) chosen by exactly one of A and B.
Probability of being chosen by exactly one of A and B
P(A∩B') + P(A'∩B) = 0.3×0.7 + 0.7 × 0.3 = 0.42
Expected Number of objects being chosen by both;
N(A or B) = 0.42 × 10
N(A or B) = 4.2
The expected number of objects chosen by both A and B is 0.9. The expected number of objects not chosen by either A or B is 9.1. The expected number of objects chosen by exactly one of A and B is 4.2.
Explanation:To find the expected number of objects chosen by both A and B, we can use the multiplication rule. Since A and B each independently choose 3 objects from a set of 10, the probability of choosing a specific object is 3/10 for both A and B. Therefore, the expected number of objects chosen by both A and B is (3/10)(3/10)(10) = 0.9.
To find the expected number of objects not chosen by either A or B, we can subtract the expected number of objects chosen by both A and B from the total number of objects. So, the expected number of objects not chosen by either A or B is 10 - 0.9 = 9.1.
To find the expected number of objects chosen by exactly one of A and B, we can use the addition rule. Since A and B each independently choose 3 objects, the probability of choosing a specific object is 3/10 for both A and B. Therefore, the expected number of objects chosen by exactly one of A and B is 2(3/10)(7/10)(10) = 4.2.
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what is the value of the 2 in 7,239,103
Answer:
(2x100,000) 200,000,00
Step-by-step explanation:
200,000
Answer: 2 hundred thousands
Explanation: To determine what the digit 2 means in 7,239,103, if we put 7,239,103 into the place value chart, we can recognize that the digit 2 is in the hundreds column of the thousands period.
So in this problem, 2 refers to 2 hundred thousands.
Place value chart is attached
in the image provided.
A racing car consumes a mean of 86 gallons of gas per race with a standard deviation of 7 gallons. If 41 racing cars are randomly selected, what is the probability that the sample mean would differ from the population mean by more than 3.1 gallons? Round your answer to four decimal places.
Answer:
0.0046 = 0.46% probability that the sample mean would differ from the population mean by more than 3.1 gallons
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this problem, we have that:
[tex]\mu = 86, \sigma = 7, n = 41, s = \frac{7}{\sqrt{41}} = 1.0932[/tex]
If 41 racing cars are randomly selected, what is the probability that the sample mean would differ from the population mean by more than 3.1 gallons?
Less than 86 - 3.1 = 82.9 or more than 86 + 3.1 = 89.1. Since the normal distribution is symmetric, these probabilities are equal, which means that we can find one of them and multiply by 2.
Less than 82.9
pvalue of Z when X = 82.9. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{82.9 - 86}{1.0932}[/tex]
[tex]Z = -2.84[/tex]
[tex]Z = -2.84[/tex] has a pvalue of 0.0023
2*0.0023 = 0.0046
0.0046 = 0.46% probability that the sample mean would differ from the population mean by more than 3.1 gallons
Final answer:
The probability that the sample mean of 41 racing cars would differ from the population mean by more than 3.1 gallons is 0.0046 or 0.46%.
Explanation:
To calculate the probability that the sample mean of 41 racing cars would differ from the population mean by more than 3.1 gallons, you use the Central Limit Theorem, which states that the sampling distribution of the sample mean will be normally distributed if the sample size is large enough. In this case, 41 is usually considered large enough. The formula to calculate the z-score for a given sample mean difference is z = (X - µ) / (σ/√n), where X is the sample mean, µ is the population mean, σ is the standard deviation, and n is the sample size.
First, we calculate the standard error of the mean (SEM): SEM = σ/√n = 7/√41 ≈ 1.0934. Next, we calculate the z-score for a difference of 3.1 gallons: z = 3.1 / 1.0934 ≈ 2.8354.
Now, we look up the z-score in standard normal distribution tables or use a calculator to find the probability corresponding to a z-score of 2.8354, which gives us the probability of a sample mean being 3.1 gallons or more away from the population mean on one side. However, since the question asks for the probability of the sample mean differing by more than 3.1 gallons on either side, we have to multiply this probability by 2 to get the final answer.
To obtain the final probability, we use statistical software or tables, which typically provide the probability for the region beyond a certain z-score. If P(z > 2.8354) is the probability of being beyond 2.8354 standard deviations from the mean on one side, the probability of being more than 3.1 gallons away from the mean on either side is 2 * P(z > 2.8354). Assuming P(z > 2.8354) = 0.0023 (from a standard normal distribution table), the final probability would be 2 * 0.0023 = 0.0046 or rounded to four decimal places, 0.0046.
5(- 3x - 2) - (x - 3) = -4(4x + 5) + 13 pls help
Answer:
infinite solutions
Step-by-step explanation:
5(- 3x - 2) - (x - 3) = -4(4x + 5) + 13
Ditribute
-15x -10 -x +3 = -16x - 20+13
Combine like term
-16x -7 = -16x -7
Add 16x to each side
-16x+16x -7 = -16x+16x-7
-7 =-7
This is always true so we have infinite solutions.
answers
All real numbers are solutions
step by step
5(−3x−2)−(x−3)=−4(4x+5)+13
Step 1: Simplify both sides of the equation.
5(−3x−2)−(x−3)=−4(4x+5)+13
5(−3x−2)+−1(x−3)=−4(4x+5)+13(Distribute the Negative Sign)
5(−3x−2)+−1x+(−1)(−3)=−4(4x+5)+13
5(−3x−2)+−x+3=−4(4x+5)+13
(5)(−3x)+(5)(−2)+−x+3=(−4)(4x)+(−4)(5)+13(Distribute)
−15x+−10+−x+3=−16x+−20+13
(−15x+−x)+(−10+3)=(−16x)+(−20+13)(Combine Like Terms)
−16x+−7=−16x+−7
−16x−7=−16x−7
Step 2: Add 16x to both sides.
−16x−7+16x=−16x−7+16x
−7=−7
Step 3: Add 7 to both sides.
−7+7=−7+7
0=0
Bottles of purified water are assumed to contain 250 milliliters of water. There is some variation from bottle to bottle because the filling machine is not perfectly precise. Usually, the distribution of the contents is approximately Normal. An inspector measures the contents of eight randomly selected bottles from one day of production. The results are 249.3, 250.2, 251.0, 248.4, 249.7, 247.3, 249.4, and 251.5 milliliters. Do these data provide convincing evidence at α = 0.05 that the mean amount of water in all the bottles filled that day differs from the target value of 250 milliliters?
Answer:
We accept H₀ we don´t have evidence of differences between the information from the sample and the population mean
Step-by-step explanation:
From data and excel (or any statistics calculator) we get:
X = 249,6 ml and s 1,26 ml
Sample mean and sample standard deviation respectively.
Population mean μ₀ = 250 ml
We have a normal distribution but we dont know the standard deviation of the population. Furthermore we have a two tails test since we are finding if the sample give us evidence of differences ( in both senses ) when we compare them with the amount of water spec ( 250 ml )
Our test hypothesis are: null hypothesis H₀ X = μ₀
Alternative Hypothesis Hₐ X ≠ μ₀
We also know that sample size is 8 therefore df = 8 - 1 df = 7 , with this value and the fact that we are required to test at α = 0,05 ( two tails test)
t = 2,365
Then we evaluate our interval:
X ± t* (s/√n) ⇒ 249,6 ± 2,365 * ( 1,26/√8 )
249,6 ± 2,365 * (1,26/2,83) ⇒ 249,6 ± 2,365 *0,45
249,6 ± 1,052
P [ 250,652 ; 248,548]
Then the population mean 250 is inside the interval, therefore we must accept that the bottles have being fill withing the spec. We accept H₀
Answer:
Because the p-value of 0.4304 is greater than the significance level of 0.05, we fail to reject the null hypothesis. We conclude the data provide convincing evidence that the mean amount of water in all the bottles filled that day does not differ from the target value of 250 milliliters.
Step-by-step explanation:
What is 6 yards 2 feet =
Answer:
20 feet
Step-by-step explanation:
An article titled "Teen Boys Forget Whatever It Was" appeared in the Australian newspaper The Mercury (April 21, 1997). It described a study of academic perfor- mance and attention span and reported that the mean time to distraction for teenage boys working on an inde- pendent task was 4 minutes. Although the sample size was not given in the article, suppose that this mean was based on a random sample of 50 teenage Australian boys and that the sample standard deviation was 1.4 minutes. Is there convincing evidence that the average attention span for teenage boys is less than 5 minutes? Test the relevant hypotheses using a .01.
Answer:
We conclude that the average attention span for teenage boys is less than 5 minutes.
Step-by-step explanation:
We are given that the mean time to distraction for teenage boys working on an independent task was 4 minutes.
Suppose that this mean was based on a random sample of 50 teenage Australian boys and that the sample standard deviation was 1.4 minutes.
Let [tex]\mu[/tex] = average attention span for teenage boys
SO, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \geq[/tex] 5 minutes {means that the average attention span for teenage boys is more than or equal to 5 minutes}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < 5 minutes {means that the average attention span for teenage boys is less than 5 minutes}
The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;
T.S. = [tex]\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean attention time span for teenage boys = 4 min
s = sample standard deviation = 1.4 min
n = sample of teenage boys = 50
So, the test statistics = [tex]\frac{4-5}{\frac{1.4}{\sqrt{50} } }[/tex] ~ [tex]t_4_9[/tex]
= -5.051
Now at 0.01 significance level, the t table gives critical value of -2.405 at 49 degree of freedom for left-tailed test. Since our test statistics is less than the critical value of t as -2.405 > -5.051, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.
Therefore, we conclude that the average attention span for teenage boys is less than 5 minutes.
Maria works as an electrician and earns $24.68/h. If she worked for 15 hours on one job, how much did she earn? *
Answer:
she would have $370.20
Step-by-step explanation:
24.68*15=370.2
The answer would be 320.2. 24.8 x 15= 320.5
Step-by-step explanation:
9. (12 pts) daniel is paying $600 for his auto insurance, and he is wondering if he is overpaying compared to his friends. he sent an email to all his friends in his contact list, and 9 of them replied with their paid amount. suppose the 9 friends who replied are a random sample, and the paid amount for auto insurance has approximately a normal distribution. (1) (2pts) what are the sample and population of the study? (2) (8 pts) use appropriate inference procedure to help daniel find out if he is overpaying compared to his friends. (3) (2pts) based on your conclusion in (2), what type of error is could you commit? explain the error using context words.
Answer:
Step-by-step explanation:
1) The sample and the population of this study is the friends who replied his email which includes in his contact list. then, the number of the replied to his email are 9 friends.
population: the whole friends include in his contact list.
3) Type I error occurs when one incorrectly rejects the null hypothesis
Here there is possibility of type I error
At a concert, 825 out of the 1500 audience are female.
What percentage of the audience are female?
Answer:
55%
Step-by-step explanation:
You could first find
10% of 1500=150
50%=750
25%=525
5%=75
750+75=825
The answer is 55% are female
The percentage of the audience that are female at the concert is 55%. This is calculated by dividing the number of females (825) by the total audience number (1500), and then multiplying that result by 100.
Explanation:To calculate the percentage of women in the audience, we need to divide the number of women by the total number of people in the audience, then multiply by 100 to convert the decimal into a percentage.
The number of women is 825.The total number of people is 1500.Thus, the calculation would be as follows:
(825 / 1500) * 100 = 55%
So, the percentage of the audience that are female is 55%.
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Amara was working on a report on Greek and Egyptian mathematicians. She decided to find a 98 percent confidence interval for the difference in mean age at the time of significant mathematics discoveries for Greek versus Egyptian mathematicians. She found the ages at the time of math discovery of all the members of both groups and found the 98 percent confidence interval based on a t-distribution using a calculator.
The procedure she used is not appropriate in this context becauseA. The sample sizes for the two groups are not equal.B. Age at the time of math discovery occurs at different intervals in the two countries, so the distribution of ages cannot be the same.C. Ages at the time of math discovery are likely to be skewed rather than bell shaped, so the assumptions for using this confidence interval formula are not valid.D. Age at the time of math discovery is likely to have a few large outliers, so the assumption for using this confidence interval formula is not valid.E. The entire population is measured in both cases, so the actual difference in means can be computed and a confidence interval should not be used.
Answer:
E. The entire population is measured in both cases, so the actual difference in means can be computed and a confidence interval should not be used.
Step-by-step explanation:
The entire population of Greek and Egyptian mathematicians was already measured by Amara. She can measure the actual difference in the mean. Therefore, she needn't use a confidence interval. It is mostly common for a researcher to be more interested in the difference between means than in the specific values of the means. The difference in sample means is used to compute the difference in population means.
Amara's sample is a random selection of Greek and Egyptian mathematicians. It is a smaller group drawn from the population that has the characteristics of the entire population. The observations and conclusions made against the sample data are attributed to the population. In this case, the entire position is measured.
A t-test is a type of inferential statistic used to determine if there is a significant difference between the means of two groups (Greek and Egyptian mathematicians) which may be related in certain features. It is mostly used when the data sets, like the data set recorded as the outcome from rolling a die 50 times, would follow a normal distribution and may have unknown variances. A t-test is used as a hypothesis testing tool, which allows testing of an assumption applicable to a population. Once the actual difference is known, a confidence interval should not be used.
The population of ground squirrels in a public park is measured to be 100. The growth of the population over time can be modeled by P(t)=600/(1+5base e ^-0.5t), with P representing the number of squirrels and t cu d. in months after the infuriating population measurement. According to this model, what is the maximum bed of squirrels the park can support?
Answer:
600
Step-by-step explanation:
As t gets very large, the exponential term goes to zero, and the expression nears the value ...
P(∞) = 600/(1 +5·0) = 600
The maximum number of squirrels the park can support is modeled as being 600.
Help me plz or I will fail
Answer:
D - Can't be determined
Step-by-step explanation:
10 * 2 *4 * 16 = 1280
1280/2 = 640
The data in NutritionStudy include information on nutrition and health habits of a sample of people. One of the variables is Smoke, indicating whether a person smokes or not (yes or no). Use technology to test whether the data provide evidence that the proportion of smokers is different from given that identify themselves as smokers.
Required:
Clearly state the null and alternative hypotheses. Your answer should be an expression.
Answer:
H₀: p = 0.20.
Hₐ: p ≠ 0.20.
Step-by-step explanation:
The question is:
The data in Nutrition Study include information on nutrition and health habits of a sample of 315 people. One of the variables is Smoke, indicating whether a person smokes or not (yes or no). Use technology to test whether the data provide evidence that the proportion of smokers is different from 20% given that 43 identify themselves as smokers. Clearly state the null and alternative hypotheses
In this case we need to test whether the proportion of smokers is different from 20%.
A one-proportion z-test can be used to determine the conclusion for this test.
The hypothesis defined as:
H₀: The proportion of smokers is 20%, i.e. p = 0.20.
Hₐ: The proportion of smokers is different from 20%, i.e. p ≠ 0.20.
The information provided is:
n = 315
X = number of people who identified themselves as smokers = 43
Compute the sample proportion of smokers as follows:
[tex]\hat p=\frac{X}{n}=\frac{43}{315}=0.137[/tex]
Compute the test statistic as follows:
[tex]z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{0.137-0.20}{\sqrt{\frac{0.20(1-0.20)}{315}}}=-2.80[/tex]
The test statistic is -2.80.
Compute the p-value as follows:
[tex]p-value=2\times P(Z<-2.80)\\=2\times [1-P(Z<2.80)]\\=2\times [1-0.99744]\\=0.00512[/tex]
*Use a z-table.
The p-value is 0.00512.
The p-value is quite small. So, the null hypothesis will be rejected at any significance level.
Thus, it can be concluded that the proportion of smokers is different from 20%.
The null hypothesis is that the proportion of smokers is 20% and the alternative hypothesis is that the proportion of smokers is not 20%.
What is null hypothesis?The null hypothesis simply means that there's no effect or relationship between the variables while the alternative hypothesis simply states that the prediction of the research has an effect
In this case, the null hypothesis is that the proportion of smokers is 20% and the alternative hypothesis is that the proportion of smokers is not 20%.
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At Cheng's Bike Rentals, it costs $36 to rent a bike for 9 hours.
How many hours of bike use does a customer get per dollar?
Answer:
4 hours
Step-by-step explanation:
Please help me I’ve included a picture!
Answer:
257/2 degrees or 128.5 degrees
Step-by-step explanation:
radians to degrees is x radians * 180/π
[tex]\frac{257\pi }{360} * \frac{180}{\pi }= \frac{257}{2}[/tex]
it is 257/2 degrees or 128.5 degrees
The mean of a population is 74 and the standard deviation is 16. The shape of the population is unknown. Determine the probability of each of the following occurring from this population. Appendix A Statistical Tables a. A random sample of size 34 yielding a sample mean of 76 or more b. A random sample of size 120 yielding a sample mean of between 73 and 75 c. A random sample of size 218 yielding a sample mean of less than 74.8
Answer:
(a) P([tex]\bar X[/tex] [tex]\geq[/tex] 76) = 0.2327
(b) P(73 < [tex]\bar X[/tex] < 75) = 0.5035
(c) P([tex]\bar X[/tex] < 74.8) = 0.77035
Step-by-step explanation:
We are given that the mean of a population is 74 and the standard deviation is 16.
Assuming the data follows normal distribution.
Let [tex]\bar X[/tex] = sample mean
The z-score probability distribution for sample mean is given by;
Z = [tex]\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean = 74
[tex]\sigma[/tex] = standard deviation = 16
n = sample size
The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.
(a) Probability that a random sample of size 34 yielding a sample mean of 76 or more is given by = P([tex]\bar X[/tex] [tex]\geq[/tex] 76)
P([tex]\bar X[/tex] [tex]\geq[/tex] 76) = P( [tex]\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] [tex]\geq[/tex] [tex]\frac{76-74}{\frac{16}{\sqrt{34} } }[/tex] ) = P(Z [tex]\geq[/tex] 0.73) = 1 - P(Z < 0.73)
= 1 - 0.7673 = 0.2327
The above probability is calculated by looking at the value of x = 0.73 in the z table which has an area of 0.7673.
(b) Probability that a random sample of size 120 yielding a sample mean of between 73 and 75 is given by = P(73 < [tex]\bar X[/tex] < 75) = P([tex]\bar X[/tex] < 75) - P([tex]\bar X[/tex] [tex]\leq[/tex] 73)
P([tex]\bar X[/tex] < 75) = P( [tex]\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{75-74}{\frac{16}{\sqrt{120} } }[/tex] ) = P(Z < 0.68) = 0.75175
P([tex]\bar X[/tex] [tex]\leq[/tex] 73) = P( [tex]\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] [tex]\leq[/tex] [tex]\frac{73-74}{\frac{16}{\sqrt{120} } }[/tex] ) = P(Z [tex]\leq[/tex] -0.68) =1 - P(Z < 0.68)
= 1 - 0.75175 = 0.24825
Therefore, P(73 < [tex]\bar X[/tex] < 75) = 0.75175 - 0.24825 = 0.5035
The above probability is calculated by looking at the value of x = 0.68 in the z table which has an area of 0.75175.
(c) Probability that a random sample of size 218 yielding a sample mean of less than 74.8 is given by = P([tex]\bar X[/tex] < 74.8)
P([tex]\bar X[/tex] < 74.8) = P( [tex]\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{74.8-74}{\frac{16}{\sqrt{218} } }[/tex] ) = P(Z < 0.74) = 0.77035
The above probability is calculated by looking at the value of x = 0.74 in the z table which has an area of 0.77035.
Write a function that solves the matrix equation Ax = b using Gaussian Elimination (book section 6.2). Your function should accept as input a n-by-n matrix A and an n-by-1 vector b, and it should produce a n-by-1 vector x that satisfies Ax = b.
Answer:
See explaination
Step-by-step explanation:
public class GaussElim{
private static final double eps = 1e-10; % set epsilon value
public static doublic[] fun(double[][] A,double[] b){
int n=b.length; %calculate length of vector b.
for( int j=0;j<n;j++){
int max=j; %find and swap pivot row.
for (int i=j+1;i<n;i++){
if(Math.abs(A[i][j])>Math.abs(A[max][j])){
max=i;
}
}
double[] t1= A[j]; %swap
A[j]=A[max];
A[max]=t1;
double t= b[j]; %swap
b[j]=b[max];
b[max]=t;
if(Math.abs(A[j][j])<=eps){
throw new ArithmeticException("Matrix is singular."); % if matrix A is a singular matrix then throw error.
}
for(int i=j+1;i<n;i++){
double alpha= A[i][j]/A[j][j];
b[i]=b[i]-alpha*b[j];
for(int k=j;k<n;k++){
A[i][k]=A[i][k]-alpha*A[j][k];
}
}
}
double[] x=new double[n]; % back substitution starts here
for(int i=n-1;i>=0;i--){
double sum=0.0;
for(int j=i+1;j<n;j++){
sum=sum+A[i][j]*x[j];
}
x[i]=(b[i]-sum)/A[i][i];
}
return x;
}
public static void main(String[] args){
int n=3;
double[][] A={{1,2,1},{4,2,0},{-1,5,-3}};
double[] b={5,3,21};
double[] x=fun(A,b);
for(int i=0;i<n;i++){
StdOut.println(x[i]);
}
}
}
What is the slope of the line that’s contains the points (-2, 5)and (6,-3)?
Answer:
The slope is -1
Step-by-step explanation:
We can find the slope by using
m = (y2-y1)/(x2-x1)
= (-3-5)/(6 - -2)
= (-3-5)/(6+2)
= -8/8
-1
How much money does Bill need for his down payment if his house is $ 450,000? He has to put 10% down.
Bill has a program that gives him 80 percent of that down payment, the rest he has to come up with. How much money does Bill need?
PLEASE HELP ME
Answer:
He would need 45,000 for his down payment. He would only have to come up with 9,000.
Step-by-step explanation:
10% times 450,000 is 45,000.
80% times 45,000 is 36,000. 45,000 minus 3600 is 9,000.
Answer:
$9,000
Step-by-step explanation:
10% of 450000
10/100 × 450000
= 45000
He has 80%, needs to arrange for 100-80 = 20% of the down payment
20/100 × 45000 = 9,000
1
Find the surface area of the triangular
3 cm
4 cm
10 cm
5 cm
Step-by-step explanation:
you are going to have to be more specific... please let us know which ones are the legs
Two different types of polishing solutions are being evaluated for possible use in a tumble-polish operation for manufacturing interocular lenses used in the human eye following cataract surgery. Three hundred lenses were tumble polished using the first polishing solution, and of this number, 253 had no polishing-induced defects. Another 300 lenses were tumble-polished using the second polishing solution, and 196 lenses were satisfactory upon completion.
Is there any reason to believe that the two polishing solutions differ? Use α = 0.05. What is the P-value for this test?
Answer:
[tex]z=\frac{0.843-0.653}{\sqrt{0.748(1-0.748)(\frac{1}{300}+\frac{1}{300})}}=5.358[/tex]
[tex]p_v =2*P(Z>5.358) = 4.2x10^{-8}[/tex]
Comparing the p value with the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that we have singificantly differences between the two proportions.
Step-by-step explanation:
Data given and notation
[tex]X_{1}=253[/tex] represent the number with no defects in sample 1
[tex]X_{2}=196[/tex] represent the number with no defects in sample 1
[tex]n_{1}=300[/tex] sample 1
[tex]n_{2}=300[/tex] sample 2
[tex]p_{1}=\frac{253}{300}=0.843[/tex] represent the proportion of number with no defects in sample 1
[tex]p_{2}=\frac{196}{300}=0.653[/tex] represent the proportion of number with no defects in sample 2
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the value for the test (variable of interest)
[tex]\alpha=0.05[/tex] significance level given
Concepts and formulas to use
We need to conduct a hypothesis in order to check if is there is a difference in the the two proportions, the system of hypothesis would be:
Null hypothesis:[tex]p_{1} - p_2}=0[/tex]
Alternative hypothesis:[tex]p_{1} - p_{2} \neq 0[/tex]
We need to apply a z test to compare proportions, and the statistic is given by:
[tex]z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}[/tex] (1)
Where [tex]\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{253+196}{300+300}=0.748[/tex]
z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.
Calculate the statistic
Replacing in formula (1) the values obtained we got this:
[tex]z=\frac{0.843-0.653}{\sqrt{0.748(1-0.748)(\frac{1}{300}+\frac{1}{300})}}=5.358[/tex]
Statistical decision
Since is a two sided test the p value would be:
[tex]p_v =2*P(Z>5.358) = 4.2x10^{-8}[/tex]
Comparing the p value with the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that we have singificantly differences between the two proportions.
A potato chip company produces a large number of potato chip bags each day and wants to investigate whether a new packaging machine will lower the proportion of bags that are damaged. The company selected a random sample of 150 bags from the old machine and found that 15 percent of the bags were damaged, then selected a random sample of 200 bags from the new machine and found that 8 percent were damaged. Let pˆO represent the sample proportion of bags packaged on the old machine that are damaged, pˆN represent the sample proportion of bags packaged on the new machine that are damaged, pˆC represent the combined proportion of damaged bags from both machines, and nO and nN represent the respective sample sizes for the old machine and new machine. Have the conditions for statistical inference for testing a difference in population proportions been met?
A No, the condition for independence has not been met, because random samples were not selected.
B No, the condition for independence has not been met, because the sample sizes are too large when compared to the corresponding population sizes.
C No, the condition that the distribution of pˆO−pˆN is approximately normal has not been met, because nN(pˆC) is not greater than or equal to 10.
D No, the condition that the distribution of pˆO−pˆN is approximately normal has not been met, because nO(1−pˆC) is not greater than or equal to 10.
E All conditions for making statistical inference have been met.
If the conditions for statistical inference for testing a difference in population proportions been met then we can say that -(E )All conditions for making statistical inference have been met.
Step-by-step explanation:
We can test the claim and the assumptions about the population proportion under the following conditions
The method of random sampling should be adopted by the company so as to ensure that the observation conducted is independent and not biased The outcome data of the sampled unit must give rise to two outputs -On that is said to be successful and the other that is said to be a failure
Thus we can say that by studying the question we can say that the above mentioned condition have been met.Hence
If the conditions for statistical inference for testing a difference in population proportions been met then we can say that -(E )All conditions for making statistical inference have been met.
The conditions for statistical inference for testing a difference in population proportions have not been met.
Explanation:In order to test for a difference in population proportions, certain conditions need to be met. Firstly, the condition for independence must be met, which means that random samples need to be selected. Secondly, the distribution of the difference in sample proportions needs to be approximately normal. This is determined by checking whether nO(pC) is greater than or equal to 10 and nO(1-pC) is greater than or equal to 10. In this case, the conditions for statistical inference have not been met because random samples were not selected (option A) and nN(pC) is not greater than or equal to 10 (option C).
Use the value of the first integral I to evaluate the two given integrals. IequalsIntegral from 0 to 1 (x cubed minus 5 x )dxequalsnegative nine fourths a. Integral from 0 to 1 (10 x minus 2 x cubed )dx b. Integral from 1 to 0 (5 x minus x cubed )dx
Answer:
a) (9/2)
b) (9/4)
Step-by-step explanation:
I = ∫¹₀ (x³ - 5x) dx = -(9/4)
a) ∫¹₀ (10x - 2x³) dx = -2 ∫¹₀ (x³ - 5x) dx
∫¹₀ (x³ - 5x) dx = -(9/4) from the given value for I
-2 ∫¹₀ (x³ - 5x) dx = -2 × (-9/4) = (9/2)
b) ∫¹₀ (5x - x³) dx = -1 ∫¹₀ (x³ - 5x) dx
∫¹₀ (x³ - 5x) dx = -(9/4) from the given value for I
-1 ∫¹₀ (x³ - 5x) dx = -1 × (-9/4) = (9/4)
Hope this Helps!!!
Answer:
a.
[tex]\int\limits_{0}^{1} 10x - 2x^3 \,dx = -2(\int\limits_{0}^{1} x^3 -5x\,dx) = -2*(-9/4) = 9/2[/tex]
b.
[tex]\int\limits_{0}^{1} 5x - x^3 \,dx = -1*(\int\limits_{0}^{1} x^3 -5x\,dx) = -1*(-9/4) = 9/4[/tex]
Step-by-step explanation:
According to the information given.
[tex]\int\limits_{0}^{1} x^3 - 5x \,dx = -9/4\\[/tex]
Now.
a.
[tex]\int\limits_{0}^{1} 10x - 2x^3 \,dx = -2(\int\limits_{0}^{1} x^3 -5x\,dx) = -2*(-9/4) = 9/2[/tex]
b.
[tex]\int\limits_{0}^{1} 5x - x^3 \,dx = -1*(\int\limits_{0}^{1} x^3 -5x\,dx) = -1*(-9/4) = 9/4[/tex]
The population p of a small community on the outskirts of a city grows rapidly over a 20-year period: t05101520p1002004509502000 As an engineer working for a utility company, you must forecast the population 5 years into the future in order to anticipate the demand for power. Employ an exponential model and linear regression to make this prediction. (Round the final answer to the nearest whole number.)
Answer:
The population of the small community, 5 years into the future, after the initial 20-year period = 4268.
Step-by-step explanation:
t | 0 | 5 | 10 | 15 | 20
p | 100 | 200 | 450 | 950 | 2000
The exponential function will look like
p = aeᵏᵗ
where a and k are constants.
Take the natural logarithms of both sides
In p = In aeᵏᵗ
In p = In a + In eᵏᵗ
In p = In a + kt
In p = kt + In a.
We then use linear regression to fit the data of In p against t to obtain k and In a.
t | 0 | 5 | 10 | 15 | 20
p | 100 | 200 | 450 | 950 | 2000
In p | 4.605 | 5.298 | 6.109 | 6.856 | 7.601
In p = kt + In a.
y = mx + b
m = k and b = In a
Performing a linear regression analysis on the now-linear relationship between In p and t and also plotting a graph of the variables.
The regression equation obtained is
y = 0.151x + 4.584
The first attached image shows the equations necessary for the estimation of the linear regression parameters.
The second attached image shows the use of regression calculator and the plot of the function In p versus t.
Comparing
y = 0.151x + 4.584
With
In p = kt + In a.
y = In p
k = 0.151
x = t
In a = 4.584
a = 97.905
The exponential function relating p and t,
p = aeᵏᵗ now becomes
p = 97.905 e⁰•¹⁵¹ᵗ
So, to predict the population 5 years into the future, that is 5 years after the 20 year period.
we need p at t=25 years.
0.151 × 25 = 3.775
p(t=25) = 97.905 e³•⁷⁷⁵ = 4268.41 = 4268.
Hope this Helps!!!
To forecast another 5 years, an exponential growth model can be used. The growth rate 'r' can be estimated using linear regression on the natural log of population figures against time, and this rate can be used to compute the predicted population. However, such a model may not account for influences like resource depletion.
Explanation:To predict the population 5 years into the future, we can use an exponential model and linear regression. The population growth described suggests it follows a sort of exponential process where population counts increase more rapidly as time progresses.
Exponential growth can be modeled using the equation P(t) = P0 * e^(rt), where P(t) is the population at time t, P0 is the initial population, r is the growth rate, and e is the base of natural logarithms. To find 'r', we can plot the natural log of the population against time and apply linear regression. The slope of the regression line estimates the growth rate 'r'. Once we've estimated 'r', we can plug the estimated 'r', the current population P0, and the time (t=25, for 5 years into the future) into the formula to calculate the predicted population.
Though this approach gives an estimate, it's important to note that real-life population dynamics can be influenced by various factors not accounted for in a simple exponential model, such as carrying capacity and resource depletion. Thus, it's more of an optimistic estimate, assuming ideal conditions for continued growth.
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