A)Yes they should market the wax because it thaws before 55 seconds
C) Null hypothesis: mean <55 vs Alternative =55
D) The value is 51.33 moments
What is Hypothesis?
When The selected samples for the champion are: 57.9, 62.9, 50.6, 50.5,48.2,47.2,50.2 and 43.1
The mean is ;
Then, Sum =57.9 + 62.9 + 50.6 + 50.5+48.2+47.2+50.2+ 43.1 =410.6
After that, Mean is = 410.6/8 =51.33= mean
Considering Significant is =0.05, therefore, applying this level will give you 51.28-51.38
if particularly, the meantime to complete their standard test course should be less than 55 seconds for a former Olympic champion, then the nullified hypothesis is correct
After that; Null hypothesis: mean <55
Hence, Alternative hypothesis: mean =55
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To study the effectiveness of a certain adult reading program, researchers will select a random sample of adults who are eligible for the program. The selected adults will be given a pretest before beginning the program and a posttest after completing the program. The difference in the number of correct answers on the pretest and the number of correct answers on the posttest will be recorded for each adult in the sample.
Answer:
The matched pair t-interval for a population mean difference.
Step-by-step explanation:
The matched pair design (Used in the paired t-test or paired-samples t-test) compares the two means associated groups to conclude if there is a statistically significant difference amid these two means.
A (1 - α)% confidence interval for matched pair design can be used to determine whether there is any difference between the two groups of data.
We use the paired t-interval if we have two measurements on the same item, person or thing. We should also use this interval if we have two items that are being measured with a unique condition.
For instance, an experimenter tests the effect of a medicine on a group of patients before and after giving the doses.
In this case a study was conducted to determine the effectiveness of a certain adult reading program.
The selected adults will be given a pretest before beginning the program and a post-test after completing the program. Then the difference between the two scores would be computed for each adult.
This is an example of matched pair design.
To analyze this data, i.e. to determine whether the program is effective or not, compute the matched pair t-interval for a population mean difference.
The subject of the question is Statistics, which pertains to a before-and-after study design. It involves the use of a pretest and posttest to evaluate the effectiveness of an adult reading program. Proper random sample selection is key to avoid bias and increase study validity.
Explanation:The study mentioned in the question is under the subject of Statistics, particularly pertaining to Study Design and Data Analysis. This type of observational study is known as a before-and-after study or time series. It involves recording observations on variables of interest over a period of time before and after a treatment or intervention.
In this scenario, the intervention is the adult reading program. The observations are done in the form of a pretest (before the intervention) and a posttest (after the intervention). The difference between the results of the posttest and pretest is then calculated to determine if there has been an improvement, and if so, how significant that improvement is.
Such study designs have seen ample usage in education, health sciences, and various other fields. It's important to highlight that in this study design, it's critical to have a well-defined random sample to avoid selection bias and increase the generalizability of the study findings.
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A box contains 5 red and 5 blue marbles. Two marbles are withdrawn randomly. If they are the same color, then you win $1.10; if they are different colors, then you win -$1.00. (This is, you lose $1.00.) Calculate (a) the expected value of the amount you win; (b) the variance of the amount you win.
The expected value of the amount you would win from this game is $0.85 and the variance of the amount you win is $0.06.
Explanation:To calculate the expected value, we need to multiply each outcome with its probability and sum them up. The outcomes and corresponding probabilities are as follows:
Winning $1.10 (drawing two marbles of the same color): The probability is 4/9 (for blue) + 4/9 (for red) = 8/9Losing $1.00 (drawing two marbles of different colors): The probability is 1/9Then, the expected value is (8/9)*$1.10 + (1/9)*(-$1.00) = $0.96 - $0.11 = $0.85. So, (a) expected value of the amount you win is $0.85.
To calculate the variance, we need to subtract the expected value from each outcome, square the result, multiply by the probability of each outcome and sum it all up. Variance = (8/9)*($1.10-$0.85)^2 + (1/9)*(-$1.00-$0.85)^2 = $0.06. So, (b) the variance of the amount you win is $0.06.
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The expected value of the amount you win is approximately -$0.0667 and the variance of the amount you win is approximately $1.0888.
(a) To calculate the expected value of the amount you win, we first need to determine the probability of each outcome:
1. Probability of drawing two red marbles: [tex]\( P(\text{red-red}) = \frac{5}{10} \times \frac{4}{9} = \frac{20}{90} = \frac{2}{9} \)[/tex]
2. Probability of drawing two blue marbles: [tex]\( P(\text{blue-blue}) = \frac{5}{10} \times \frac{4}{9} = \frac{20}{90} = \frac{2}{9} \)[/tex]
3. Probability of drawing one red marble and one blue marble: [tex]\( P(\text{red-blue}) = 2 \times \frac{5}{10} \times \frac{5}{9} = \frac{50}{90} = \frac{5}{9} \)[/tex]
Now, let's calculate the expected value (\( E \)):
[tex]\[ E = (P(\text{red-red}) \times \$1.10) + (P(\text{blue-blue}) \times \$1.10) + (P(\text{red-blue}) \times -\$1.00) \]\[ E = \left(\frac{2}{9} \times \$1.10\right) + \left(\frac{2}{9} \times \$1.10\right) + \left(\frac{5}{9} \times -\$1.00\right) \]\[ E = \left(\frac{4}{9} \times \$1.10\right) + \left(\frac{5}{9} \times -\$1.00\right) \]\[ E = \$\left(\frac{4}{9} \times 1.10\right) + \$\left(\frac{5}{9} \times -1.00\right) \][/tex]
[tex]\[ E = \$\left(\frac{4.40}{9} - \frac{5.00}{9}\right) \]\[ E = \$\frac{-0.60}{9} \]\[ E = -\$0.0667 \][/tex]
So, the expected value of the amount you win is approximately -$0.0667.
(b) To calculate the variance of the amount you win, we'll use the formula for variance:
[tex]\[ \text{Var}(X) = E(X^2) - (E(X))^2 \][/tex]
We already know E(X) from part (a). Now, let's calculate E(X)²:
[tex]\[ E(X^2) = (P(\text{red-red}) \times (\$1.10)^2) + (P(\text{blue-blue}) \times (\$1.10)^2) + (P(\text{red-blue}) \times (-\$1.00)^2) \]\[ E(X^2) = \left(\frac{2}{9} \times (\$1.10)^2\right) + \left(\frac{2}{9} \times (\$1.10)^2\right) + \left(\frac{5}{9} \times (-\$1.00)^2\right) \]\[ E(X^2) = \left(\frac{4}{9} \times (\$1.21)\right) + \left(\frac{5}{9} \times \$1.00\right) \]\[ E(X^2) = \$\left(\frac{4 \times 1.21}{9} + \frac{5}{9}\right) \][/tex]
[tex]\[ E(X^2) = \$\left(\frac{4.84}{9} + \frac{5}{9}\right) \]\[ E(X^2) = \$\frac{9.84}{9} \]\[ E(X^2) = \$1.0933 \][/tex]
Now, we can calculate the variance:
[tex]\[ \text{Var}(X) = E(X^2) - (E(X))^2 \]\[ \text{Var}(X) = \$1.0933 - (-\$0.0667)^2 \]\[ \text{Var}(X) = \$1.0933 - \$0.0045 \]\[ \text{Var}(X) = \$1.0888 \][/tex]
So, the variance of the amount you win is approximately $1.0888.
Therefore, the solution is (a) Expected value: -$0.0667 and (b) Variance: $1.0888
2. Connecticut is located between New York and Boston, so baseball fans in the state typically root for
either the Yankees or Red Sox. In a recent poll of 803 randomly selected baseball fans in Connecticut,
55% said their favorite team was the Yankees Calculate and interpret a 90% confidence interval for
the proportion of all Connecticut baseball fans who say their favorite team is the Yankees
We are 90% confident that the true proportion of all Connecticut baseball fans who say their favorite team is the Yankees is between 0.511 and 0.589.
What is a confidence interval?A confidence interval is a range of values that is likely to contain the true value of an unknown parameter, such as a population means or proportion, based on a sample of data from that population.
It is a statistical measure of the degree of uncertainty or precision associated with a statistical estimate.
We have,
To calculate a 90% confidence interval for the proportion of all Connecticut baseball fans who say their favorite team is the Yankees, we can use the formula:
CI = p ± z √((p(1-p))/n)
where:
p is the sample proportion (0.55)
z* is the critical z-value for a 90% confidence level (1.645)
n is the sample size (803)
Plugging in the values,
CI = 0.55 ± 1.645 √((0.55(1-0.55))/803)
CI = 0.55 ± 0.042
Therefore,
The 90% confidence interval for the proportion of all Connecticut baseball fans who say their favorite team is the Yankees is (0.508, 0.592).
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To calculate a 90% confidence interval for the proportion of Connecticut baseball fans who favor the Yankees, the standard error was computed and multiplied by the Z-score for 90% confidence. The confidence interval was found to be between 52.11% and 57.89%, meaning there is a 90% chance the true proportion falls within this range.
Explanation:The question requires the calculation of a 90% confidence interval for the proportion of all Connecticut baseball fans who favor the New York Yankees. A sample proportion (p-hat) of 803 Connecticut baseball fans showed that 55% (or 0.55) favored the Yankees. To calculate the confidence interval, we use the formula for the confidence interval of a proportion:
CI = p-hat ± Z*sqrt[(p-hat)(1-p-hat)/n], where Z is the Z-score that corresponds to the level of confidence and n is the sample size. For a 90% confidence level, the Z-score is approximately 1.645.
Calculating the standard error (SE):
SE = sqrt[(0.55)(0.45)/803] = sqrt[0.2475/803] = sqrt[0.000308344] = 0.017562
Then, the margin of error (ME):
ME = Z*SE = 1.645 * 0.017562 = 0.028894
The 90% confidence interval (CI) is:
CI = 0.55 ± 0.028894
CI = [0.521106, 0.578894]
Interpreting this confidence interval, we are 90% confident that the true proportion of all Connecticut baseball fans who favor the Yankees falls between 52.11% and 57.89%.
Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.3. (Round your answers to four decimal places.)
(a) If the distribution is normal, what is the probability that the sample mean hardness for a random sample of 12 pins is at least 51?
The probability that the sample mean hardness for a random sample of 12 pins is at least 51 is approximately 0.0228.
Explanation:To determine the probability, we can use the central limit theorem. Since the distribution of the Rockwell hardness is normal, the distribution of the sample mean will also be normal. The formula for calculating the standard deviation of the sample mean (standard error) is given by the population standard deviation divided by the square root of the sample size. In this case, the standard error is calculated as 1.3 / sqrt(12), which is approximately 0.3746.
To find the z-score, we use the formula: z =[tex](X - μ) / σ[/tex], where X is the desired value (51), μ is the mean (50), and σ is the standard error (0.3746). Plugging in these values, we get a z-score of approximately 2.6744.
Using a standard normal distribution table, we find the probability that a z-score is greater than 2.6744 is approximately 0.0057. However, since we are interested in the probability that the sample mean hardness is at least 51, we need to consider the tail of the distribution on both sides. Therefore, the final probability is[tex]2 * 0.0057[/tex] = 0.0114.
To round the answer to four decimal places, the final probability is approximately 0.0228.
Probability that the sample mean hardness for 12 pins is at least 51 is approximately 0.0038.
To solve this problem, we'll use the Central Limit Theorem (CLT) and the properties of the normal distribution.
Given:
- Population mean[tex](\( \mu \))[/tex] = 50
- Population standard deviation [tex](\( \sigma \))[/tex] = 1.3
- Sample size (( n )) = 12
- Desired sample mean[tex](\( \bar{x} \)) = 51[/tex]
The CLT states that the distribution of sample means approaches a normal distribution as the sample size increases, regardless of the shape of the population distribution.
For a sample size of 12, we can use the following formula to calculate the standard error[tex](\( \text{SE} \))[/tex]:
[tex]\[ \text{SE} = \frac{\sigma}{\sqrt{n}} \][/tex]
Substitute the given values:
[tex]\[ \text{SE} = \frac{1.3}{\sqrt{12}} \][/tex]
[tex]\[ \text{SE} \approx \frac{1.3}{3.4641} \][/tex]
[tex]\[ \text{SE} \approx 0.3749 \][/tex]
Now, we need to find the z-score for the sample mean of 51 using the standard normal distribution formula:
[tex]\[ z = \frac{\bar{x} - \mu}{\text{SE}} \][/tex]
Substitute the given values:
[tex]\[ z = \frac{51 - 50}{0.3749} \][/tex]
[tex]\[ z = \frac{1}{0.3749} \][/tex]
[tex]\[ z \approx 2.6679 \][/tex]
Now, we look up the probability corresponding to this z-score in the standard normal distribution table.
The probability that the sample mean hardness for a random sample of 12 pins is at least 51 is the area under the standard normal curve to the right of ( z = 2.6679 ).
Consulting a standard normal distribution table or using statistical software, we find:
[ P(Z > 2.6679) ]
This probability represents the area to the right of ( z = 2.6679 ).
Now, if we're using a standard normal distribution table, we might find the value closest to 2.67 or 2.67 itself, then look up the corresponding probability.
This probability gives us the likelihood that the sample mean hardness for a random sample of 12 pins is at least 51.
Researchers are conducting a study in an attempt to establish a linear relationship between the number of online music video views and the number of guitar sales. A computer output for regression is shown and is based on a sample of seven observations. Predictor Coeff St. Dev t Ratio p-Value Constant 7.85671094 1.316226455 5.969118 0.001889 Music Video Views 0.094781123 0.027926367 3.393965 0.019378 What represents the 99% confidence interval for the slope of the regression line?
a. 0.0948 ± 3.499(1.3162)
b. 0.0948 ± 4.032(1.3162)
c. 0.0948 ± 3.707(0.0279)
d. 0.0948 ± 4.032(0.0279)
Answer:
d. 0.0948 ± 4.032(0.0279)
Step-by-step explanation:
A 99% confidence interval for the coefficient of promotional expenditures is, First, compute the t critical value then find confidence interval.
The t critical value for the 99% confidence interval is,
The sample size is small and two-tailed test. Look in the column headed es = 0.01 and the row headed in the t distribution table by using degree of freedom is here
for (n-2=5) degree of freedom and 99% confidence ; critical t =4.032
therefore 99% confidence interval for the slope =estimated slope -/+ t*Std error
= 0.094781123 -/+ 4.032* 0.027926367 = -0.017822 to 0.207384
What is the domain of f(x) = 3x – 22
Answer: 7.3333
Step-by-step explanation:
Which of the following is the point and slope of the equation y - 4 = 1/3(x - 5)?
Answer:
See answer below
Step-by-step explanation:
Hi there,
The formula above is not in point-slope form, which is of the form:
[tex]y = mx +b[/tex]
So, just simply put it in point-slope form by:
1. distributing the 1/3 term into the parentheses
[tex]y-4=\frac{1}{3}x -\frac{5}{3}[/tex]
2. add 4 to the right side of the equation to isolate y
[tex]y = \frac{1}{3} x-\frac{5}{3} +4 \\ = \frac{1}{3} x-\frac{5}{3}+\frac{12}{3} =\frac{1}{3} x+\frac{7}{3} \\ y =\frac{1}{3} x+\frac{7}{3}[/tex]
Adding 4 as 12/3 with common denominator, the resulting equation is now in point-slope form.
The "point" refers to y-intercept, which is 7/3, while the slope is the coefficient of x, which is 1/3.
thanks,
The point is (5, 4) and the slope is 1/3 in the given equation. The point-slope form of a linear equation is y - y1 = m(x - x1).
The point-slope form of a linear equation is y - y1 = m(x - x1).
Given the equation y - 4 = 1/3(x - 5), the point is (5, 4) and the slope is 1/3.
Therefore, the point is (5, 4) and the slope is 1/3.
Further, Boris must also take into consideration the chicken calorie specifications which, as you know, are imposed by customers and competitors. The target for the chicken calorie content is 475 calories---with a tolerance level of 50 calories in each direction. The upper control limit of the X-bar chart is: (Please round up) a) More than 520 calories but less than or equal to 540 calories. b) More than 540 calories but less than or equal to 560 calories. c) More than 560 calories but less than or equal to 580 calories. d) More than 580 calories but less than or equal to 600 calories. e) None of the above.The lower control limit of the X-bar chart is:a) More than 415 calories but less than or equal to 430 calories.b) More than 430 calories but less than or equal to 445 calories.c) More than 445 calories but less than or equal to 460 calories.d) More than 460 calories but less than or equal to 475 calories.The process capability index, Cp, is:a) Negative.b)Less than 1.c)More than 1 but less than 1.3d)More than 1 but less than 1.4.e)None of the above.The process capability index, Cpk, is:a) Negative.b) Less than 1.c) More than 1 but less than 1.05d) More than 1 but less than 1.08e) None of the above.
Answer:
a, a, b,b
Step-by-step explanation:
Upper bound: 475+50= 525
lower bound: 475-50= 425
Cp= (Upper bound-lower bound)/ 6σ
Cp= (525-525)/ (6× 50)
Cp= 0.33
Cpk is minimum of two following values: (Upper bound-mean)/ 3σ, (mean -lower bound)/ 3σ
Cpk in this case is: Min (0.33,0.33)
The Cp and Cpk values are calculated from the actual process values. Here target Cp and Cpk value is 0.33
Roster's Chicken can monitor their caloric content using an X-chart with control limits calculated at either three or four standard deviations from the mean. The analysis reveals that grilled chicken breast is more energy-dense than tortilla chips. After consuming 16 tortilla chips, significant portions of the daily values (DV) for fat, sodium, and fiber remain for other meals.
Control Limits for X-Chart
To design an X-chart for monitoring the caloric content of Roster's Chicken, we need to establish the upper and lower control limits based on the given parameters.
Part a) Control Limits with Four Standard Deviations
The average caloric content of the chicken breast is 420 calories with a standard deviation of 20 calories. For a sample size of 25, the standard error is:
Standard Error = σ / √n = 20 / √25 = 20 / 5 = 4
Therefore, the four standard deviation limits from the target mean of 420 calories are:
Upper Control Limit (UCL) = 420 + 4 * 4 = 420 + 16 = 436 calories
Lower Control Limit (LCL) = 420 - 4 * 4 = 420 - 16 = 404 calories
Part b) Control Limits with Three Standard Deviations
Using the same standard error, the three standard deviation limits are:
Upper Control Limit (UCL) = 420 + 3 * 4 = 420 + 12 = 432 calories
Energy Density Comparison
A serving of grilled chicken breast (4 oz) has about 190 calories. In comparison, 16 tortilla chips typically have around 140 calories. Chicken breast is more energy-dense because it has more calories per ounce compared to tortilla chips, which makes it a denser source of calories.
Percentage Daily Values (DV) in 16 Tortilla Chips
If 16 tortilla chips contribute 10% DV of fat, 8% DV of sodium, and 4% DV of dietary fiber, the remaining percentages for a healthy diet would be:
• Fat: 100% - 10% = 90% DV remaining
• Sodium: 100% - 8% = 92% DV remaining
• Fiber: 100% - 4% = 96% DV remaining
Complete Question:- Roster's Chicken advertises "lite" chicken with 30% fewer calories than standard chicken. When the process for "lite chicken breast production is in control, the average chicken breast contains 420 calories, and the standard deviation in caloric content of the chicken breast population is 20 calories. Roster's wants to design an X-chart to monitor the caloric content of chicken breasts, where 25 chicken breasts would be chosen at random to form each sample. a) What are the lower and upper control limits for this chart? These limits are chosen to be four standard deviations from the target. Upper Control Limit (UCL): 436 calories (enter your response as an integer) Lower Control Limit (LCL): 404 calories (enter your response as an integer) b) What are the limits with three standard deviations from the target? Upper Control Limit (UCL): calories (enter your response as an integer)
The mean height of an adult giraffe is 18 feet. Suppose that the distribution is normally distributed with standard deviation 0.8 feet. Let X be the height of a randomly selected adult giraffe. Round all answers to two decimal places.What is the probability that a randomly selected giraffe will be shorter than 17 feet tall?What is the probability that a randomly selected giraffe will be between 16 and 19 feet tall? The 90th percentile for the height of giraffes is?
Answer:
1. the probability that giraffe will be shorter than 17 feet tall is equal to 0.1056
2. the probability that a randomly selected giraffe will be between 16 and 19 feet tall is equal to 0.8882
3. the the 90th percentile for the height of giraffes is 19.024
Step-by-step explanation:
To calculate the probability that a giraffe will be shorter than 17 feet tall, we need to standardize 17 feet as:
[tex]z=\frac{x-m}{s}[/tex]
Where x is the height of the adult giraffe, m is the mean and s is the standard deviation, so 17 feet is equivalent to:
[tex]z=\frac{17-18}{0.8}=-1.25[/tex]
Now, the probability that giraffe will be shorter than 17 feet tall is equal to P(z<-1.25). Then, using the standard normal distribution table, we get that:
[tex]P(z<-1.25)=0.1056[/tex]
At the same way, 16 and 19 feet tall are equivalent to:
[tex]z=\frac{16-18}{0.8}=-2.5\\z=\frac{19-18}{0.8}=1.25[/tex]
So, the probability that a randomly selected giraffe will be between 16 and 19 feet tall is equal to:
[tex]P(-2.5<z<1.25)=P(z<1.25)-P(z<-2.5)\\P(-2.5<z<1.25)=0.8944-0.0062\\P(-2.5<z<1.25)=0.8882[/tex]
Finally, to find the the 90th percentile for the height of giraffes, we need to find the value z that satisfy:
[tex]P(Z<z)=0.9[/tex]
Now, using the standard normal distribution table we get that z is equal to 1.28. Therefore, the height x of the giraffes that is equivalent to 1.28 is:
[tex]z=\frac{x-m}{s} \\1.28=\frac{x-18}{0.8} \\x=(1.28*0.8)+18\\x=19.024[/tex]
it means that the the 90th percentile for the height of giraffes is 19.024
Using Z-scores, the probability that a randomly selected giraffe is less than 17 feet tall is approximately 0.106, the probability that it is between 16 and 19 feet tall is approximately 0.888. The 90th percentile in height for giraffes is about 19.02 feet.
Explanation:These questions are related to normal distribution. The formula of normal distribution (Z-score) is Z = (X - μ)/σ, where X is the specific value, μ is the mean, and σ is the standard deviation.
For the first question, we want to find the probability that a randomly selected giraffe is less than 17 feet tall. The Z-score is calculated as Z = (17 - 18) / 0.8 = -1.25. Using the Z-table, the probability that a giraffe is less than 17 feet tall is approximately 0.106.
For the second question, we need to find the probability that a randomly selected giraffe will be between 16 and 19 feet tall. We calculate two Z-scores: Z1 = (16 - 18) / 0.8 = -2.5 and Z2 = (19 - 18) / 0.8 = 1.25. Using the Z-table, the probability that a giraffe is between 16 and 19 feet tall is approximately 0.894-0.006 = 0.888.
For the third question, we need to find the 90th percentile for the height of giraffes. It corresponds to the Z-score of 1.282 on the Z-table. Hence, X = μ + Zσ = 18 + 1.282*0.8 = 19.02 feet.
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Problem 15-05 (Algorithmic) Consider the following time series data. Week 1 2 3 4 5 6 Value 16 13 18 11 15 14 Choose the correct time series plot. (i) Time Series ValueWeek(t) (ii) Time Series ValueWeek(t) (iii) Time Series ValueWeek(t) (iv) Time Series ValueWeek(t) Plot (i) What type of pattern exists in the data? Horizontal Pattern Develop a three-week moving average for this time series. Compute MSE and a forecast for week 7. Round your answers to two decimal places. Week Time Series Value Forecast 1 16 2 13 3 18 4 11 15.67 5 15 14 6 14 14.67 MSE: 7.74 The forecast for week 7: 13.33 Use = 0.2 to compute the exponential smoothing values for the time series. Compute MSE and a forecast for week 7. Round your answers to two decimal places. Week Time Series Value Forecast 1 16 2 13 16 3 18 15.4 4 11 15.92 5 15 14.94 6 14 14.95 MSE: 8.17 The forecast for week 7: 14.76 Compare the three-week moving average forecast with the exponential smoothing forecast using = 0.2. Which appears to provide the better forecast based on MSE? Three-week moving average Explain. The input in the box below will not be graded, but may be reviewed and considered by your instructor. Use trial and error to find a value of the exponential smoothing coefficient that results in a smaller MSE than what you calculated for = 0.2. Find a value of for the smallest MSE. Round your answer to three decimal places. = 0.157
This answer involves time series analysis concepts, specifically focusing on three-week moving averages and exponential smoothing methods. It explains the results obtained from both methods for predicting week 7's forecast, comparing them using the Mean Squared Error (MSE). It suggests that the three-week moving average provides a better prediction due to a lower MSE.
Explanation:The question asked is a problem in time series analysis, a category of statistics that involves the use and interpretation of data points ordered in time. The method used to analyze this data includes calculating a three-week moving average, computing mean squared error (MSE), and conducting exponential smoothing.
In the scenario given, a three-week moving average tracks an average of data over the past three weeks. It showed a forecast for week 7 as 13.33 with an MSE value of 7.74. In contrast, the exponential smoothing method, which uses a weight assigned to historic data (referred to as alpha, =0.2 in this case) forecasted a higher value of 14.76 with a slightly greater MSE of 8.17.
Between the two methods, the three-week moving average forecast provided a better model due to its lower MSE, indicating a smaller average squared distance between predicted and actual values. Although the calculations for the best value are not included here, a lower MSE (compared to =0.2) was obtained with the value =0.157. This example showcases how using different methods and changing parameters within methods can provide different forecasting results in time series analysis.
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From a point along a straight road, the angle of elevation to the top of a hill is . From farther down the road, the angle of elevation to the top of the hill is . How high is the hill?
Complete Question:
From a point along a straight road, the angle of elevation to the top of a hill is 33° . A distance of 200 ft farther down the road, the angle of elevation to the top of the hill is 20°. How high is the hill?
Answer:
The hill is 165.87 ft high
Step-by-step explanation:
Check the file attached below for a pictorial understanding of the question
[tex]tan \theta = \frac{opposite}{Adjacent}[/tex]
From ΔABC
[tex]tan 33 = \frac{y}{x} \\[/tex]
[tex]y = x tan 33[/tex]..........(1)
From ΔABD
[tex]tan 20 = \frac{y}{x + 200} \\[/tex]
[tex]y = (x + 200) tan 20[/tex]............(2)
Equating (1) and (2)
[tex]x tan 33 = (x+200) tan20\\xtan33 = xtan20 + 200tan20\\0.649x = 0.364x + 72.794\\0.649x - 0.364x = 72.794\\0.285x = 72.794\\x = 72.794/0.285\\x = 255.42 ft[/tex]
Substitute the value of x into equation (1)
[tex]y = 255.42 tan 33[/tex]
y = 165.87 ft
The height of the hill is 550 feet.
Given:
- [tex]\(\theta_1 = 31^\circ\)[/tex]
- [tex]\(\theta_2 = 24^\circ\)[/tex]
- [tex]\(d = 320\) feet[/tex]
We use the formula for the height of the hill \(h\):
[tex]\[h = \frac{d \tan(\theta_2) \tan(\theta_1)}{\tan(\theta_1) - \tan(\theta_2)}\][/tex]
First, we calculate [tex]\(\tan(31^\circ)\)[/tex] and [tex]\(\tan(24^\circ)\)[/tex]:
[tex]\[\tan(31^\circ) \approx 0.6009\][/tex]
[tex]\[\tan(24^\circ) \approx 0.4452\][/tex]
Now, substitute these values into the formula:
[tex]\[h = \frac{320 \times 0.4452 \times 0.6009}{0.6009 - 0.4452}\][/tex]
Calculate the numerator:
[tex]\[320 \times 0.4452 \times 0.6009 \approx 85.676256\][/tex]
Calculate the denominator:
[tex]\[0.6009 - 0.4452 \approx 0.1557\][/tex]
Now, divide the numerator by the denominator:
[tex]\[h = \frac{85.676256}{0.1557} \approx 550.41\][/tex]
Rounding to the nearest foot, the height of the hill is:
550 feet.
Complete question is here
From a point along a straight road, the angle of elevation to the top of a hill is 31º. From 320 n farther down the road, the angle of elevation to the top of the hill is 24º. How high is the hill? Round to the nearest foot.
The line AB has midpoint (2,5).
A has coordinates (1, 2).
Find the coordinates of B.
Answer:
[tex] X_m = \frac{A_x +B_x}{2}= \frac{1+B_x}{2}= 2[/tex]
And we can solve for [tex] B_x[/tex] and we got:
[tex] 1+B_x = 4[/tex]
[tex]B_x = 3[/tex]
[tex] Y_m = \frac{A_y +B_y}{2}= \frac{2+B_y}{2}= 5[/tex]
And we can solve for [tex] B_x[/tex] and we got:
[tex] 2+B_y = 10[/tex]
[tex]B_y = 8[/tex]
So then the coordinates for B are (3,8)
Step-by-step explanation:
For this case we know that the midpoint for the segment AB is (2,5)
And we know that the coordinates of A are (1,2)
We know that for a given segment the formulas in order to find the midpoint are given by:
[tex] X_m = \frac{A_x +B_x}{2}= \frac{1+B_x}{2}= 2[/tex]
And we can solve for [tex] B_x[/tex] and we got:
[tex] 1+B_x = 4[/tex]
[tex]B_x = 3[/tex]
[tex] Y_m = \frac{A_y +B_y}{2}= \frac{2+B_y}{2}= 5[/tex]
And we can solve for [tex] B_x[/tex] and we got:
[tex] 2+B_y = 10[/tex]
[tex]B_y = 8[/tex]
So then the coordinates for B are (3,8)
The coordinates of point B are (4, 8).
Explanation:To find the coordinates of point B, we can use the midpoint formula. The midpoint formula states that the coordinates of the midpoint M(xm, ym) of a line segment with endpoints A(x1, y1) and B(x2, y2) are given by:
xm = (x1 + x2) / 2
ym = (y1 + y2) / 2
In this case, we are given that the midpoint M is (2, 5) and A is (1, 2). We can substitute these values into the formula:
2 = (1 + x2) / 2
5 = (2 + y2) / 2
Now, we can solve for x2 and y2:
x2 = 4
y2 = 8
Therefore, the coordinates of point B are (4, 8).
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Point A is at (-2, -7) and point M is at (2.5, -1.5).
Point M is the midpoint of point A and point B.
What are the coordinates of point B?
Answer:
B = (7, 4)
Step-by-step explanation:
B = 2M -A = 2(2.5, -1.5) -(-2, -7)
B = (5+2, -3+7)
B = (7, 4)
__
Derivation
You know the midpoint is calculated from ...
M = (A+B)/2
Solving for B, we get ...
2M = A+B
B = 2M-A . . . . the formula we used above
Answer:
Give more points
Step-by-step explanation:
is The polar form of a complex number is unique?
Answer:
no
Step-by-step explanation:
The angle in the polar form of a complex number can have any multiple of 2π radians added to it, and the number will be the same number. That is, there are an infinite number of representations of a complex number in polar form.
Final answer:
The polar form of a complex number is not unique due to the angle component, which can vary by multiples of full rotations (2π). However, when considering the principal value of the angle, the polar form becomes unique.
Explanation:
The polar form of a complex number is not entirely unique, the reason being that the angle (θ) in the polar form can be expressed as θ + 2πk, where k is any integer.
This is because complex numbers are represented in the complex plane, which is similar to a circle, where angles that differ by full rotations (2π radians) represent the same point.
However, the magnitude (or modulus) and the principal value of the angle (typically between -π and π) combine to give a unique representation of a complex number in polar form.
Toxaphene is an insecticide that has been identified as a pollutant in the Great Lakes ecosystem. To investigate the effect of toxaphene exposure on animals, groups of rats were given toxaphene in their diet. An article reports weight gains(in grams) for rats given a low dose (4ppm) and for control rats whose diet did not include the insecticide. The sample standard deviation for 23 female control rats was 32g and for 20 female low-dose rats was 54g. Does this data suggest that there is more variability in low-dose weight gains than in control weight gains? Assuming normality, carry out a test of hypotheses at significance level .05.
Answer:
Check the explanation
Step-by-step explanation:
The details of the given samples are as follows:
Low dose. Number numbers of rats 20
The standard deviation . [tex]S_1[/tex]=54g
Control:
Number of rats [tex]N_2[/tex]=23
The standard deviation [tex]S_1[/tex]=32g
[tex]H_0[/tex]: there is no more variability in low — dose weight gains than in the control weight gins
[tex]H_1[/tex]: There is mere In-viability in low — dose weight gains than in control weight gains
>4 The level of significance.
a= 0.05 The test statistic to test the above hypothesis is Larger variance Smaller variance
The critical value of F (19.22) dr and at 0.05
significance level is 2.0837
The P-Value is 0.01 The test significance value is greater the the critical level is 20837.
Also. the 8-value is less then the significance level Hence, we reject the null hypothesis.
Therefore, we conclude that there is sufficient evidence to conclude that there is more variability in low — dose weight gains than in control weight gains .
A university administrator was interested in determining if there was a difference in the distance students travel to get from class from their current residence(in miles). Men and women at UF were randomly selected. The Minitab output is below. What is the best interpretation for the output? Difference = mu (F) - mu (M) T-Test of difference = 0 (vs not =): T-Value = -1.05 P-Value = 0.305 DF = 21
The Minitab output from the t-test signifies that there is no statistically significant difference in the distances traveled by men and women at UF to get to class. The t-value and p-value obtained don't give enough evidence to reject the null hypothesis. The degrees of freedom (DF) indicate the number of independent observations in the sample.
Explanation:The output from Minitab that you've shared is the result of a paired t-test comparing the mean distances traveled by men and women to get to class at UF. The null hypothesis in this context is that there is no difference in the average distances traveled by men and women (Difference = mu (F) - mu (M)). The t-value of -1.05 and the p-value of 0.305 do not provide enough evidence to reject the null hypothesis at the conventional 0.05 level of significance. Therefore, we could interpret the output as not detecting a statistically significant difference between the mean distances men and women travel to get to class at UF.
The 'DF' or degrees of freedom, indicates the number of independent observations in your sample that are free to vary once certain constraints (like the sample mean) are calculated. In this case, DF = 21, which is the sample size (pairs of men and women) minus 1.
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Michael's bank contains only nickels, dimes, and quarters. There are 57 coins in all, valued at $4.55. The number of nickels is 7 short of being three times the sum of the number of dimes and quarters together. How many dimes are in the bank
Answer:
10 dimes.
Step-by-step explanation:
Let n represent number of nickels, d represent number of dimes and q represent number of quarters.
We have been given that there are 57 coins in all. We can represent this information in an equation as:
[tex]n+d+q=57...(1)[/tex]
We are also told that the value of all coins is $4.55. We can represent this information in an equation as:
[tex]0.05n+0.10d+0.25q=4.55...(2)[/tex]
The number of nickels is 7 short of being three times the sum of the number of dimes and quarters together. We can represent this information in an equation as:
[tex]n+7=3(d+q)...(3)[/tex]
From equation (1), we will get:
[tex]d+q=57-n[/tex]
Substituting this value in equation (3), we will get:
[tex]n+7=3(57-n)[/tex]
[tex]n+7=171-3n[/tex]
[tex]n+3n+7-7=171-7-3n+3n[/tex]
[tex]4n=164[/tex]
[tex]\frac{4n}{4}=\frac{164}{4}[/tex]
[tex]n=41[/tex]
Therefore, there are 41 nickels in the bank.
[tex]d+q=57-n...(1)[/tex]
[tex]d+q=57-41...(1)[/tex]
[tex]d+q=16...(1)[/tex]
[tex]q=16-d...(1)[/tex]
Upon substituting equation (1) and value of n in equation (2), we will get:
[tex]0.05(41)+0.10d+0.25(16-d)=4.55[/tex]
[tex]2.05+0.10d+4-0.25d=4.55[/tex]
[tex]6.05-0.15d=4.55[/tex]
[tex]6.05-6.05-0.15d=4.55-6.05[/tex]
[tex]-0.15d=-1.5[/tex]
[tex]\frac{-0.15d}{-0.15}=\frac{-1.5}{-0.15}[/tex]
[tex]d=10[/tex]
Therefore, there are 10 dimes in the bank.
In a large lecture class, the professor announced that the scores on a recent exam were normally distributed with a range from 51 to 87. Using the Empirical Rule μ ± 3σ to estimate σ, how many students would you need to sample to estimate the true mean score for the class with 90 percent confidence and an error of ±2?
Answer:
The standard deviation of the scores on a recent exam is 6.
The sample size required is 25.
Step-by-step explanation:
Let X = scores of students on a recent exam.
It is provided that the random variable X is normally distributed.
According to the Empirical rule, 99.7% of the normal distribution is contained in the range, μ ± 3σ.
That is, P (μ - 3σ < X < μ + 3σ) = 0.997.
It is provided that the scores on a recent exam were normally distributed with a range from 51 to 87.
This implies that:
P (51 < X < 87) = 0.997
So,
μ - 3σ = 51...(i)
μ + 3σ = 87...(ii)
Subtract (i) and (ii) to compute the value of σ as follows:
μ - 3σ = 51
(-)μ + (-)3σ = (-)87
______________
-6σ = -36
σ = 6
Thus, the standard deviation of the scores on a recent exam is 6.
The (1 - α)% confidence interval for population mean is given by:
[tex]CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]
The margin of error of this interval is:
[tex]MOE = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]
Given:
MOE = 2
σ = 6
Confidence level = 90%
Compute the z-score for 90% confidence level as follows:
[tex]z_{\alpha/2}=z_{0.10/2}=z_{0.05}=1.645[/tex]
*Use a z-table.
Compute the sample required as follows:
[tex]MOE = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\\2=1.645\times \frac{6}{\sqrt{n}}\\n=(\frac{1.645\times 6}{2})^{2}\\n=24.354225\\n\approx 25[/tex]
Thus, the sample size required is 25.
The Empirical Rule guides estimating sample sizes with specific confidence levels and margins of error.
The Empirical Rule states that about 95 percent of the values lie within two standard deviations of the mean in a normal distribution.
To estimate the sample size needed to estimate the true mean score with 90 percent confidence and an error of ±2, we can use the formula for the margin of error.
By rearranging this formula and plugging in the required values, we can determine the sample size needed to achieve the desired confidence level and margin of error.
which is greater, 7 or 24/4
Answer:
7
Step-by-step explanation:
24/4=6. 7>6
You roll a six-sided die twice. What is the probability of rolling a 3 and then a 5?
My guess is that it is 2/12 chance to get it, or 16.6
Answer:
1/36
Step-by-step explanation:
Multiply the two independent probabilities to find the compound probability.
P(3 and 5) =
1
6
×
1
6
=
1
36
9. A contractor is considering a sale that promises a profit of $27,000 with a probability of 0.7 or a loss (due to bad weather, strikes, and such) of $12,000 with a probability of 0.3. What is the expected value?
Answer:
Hence the expected value for the contractor for sales is 15,300 $.
Step-by-step explanation:
Given:
Winning $27000 is 0.7 and losing 12000 $ of it about 0.3
To find :
Expected value for the contractor for sales.
Solution:
Th expected value is the average occurred of the event.
{Suppose
a series of number like 10,30,30,30,30,60,78.
for this expected value will be
(10+30+30+30+30+60+78+78) / 8
=10(1/8)+30(4/8)+60(1/8)+78(2/8).
78 ,10 ,30 and 60 are just like cost and 1/8 ,4/8,2/8 are probabilities of respective cost.
}
Similar for given values
Expected value with probability is =
Winning probability *cost of winning +(-losing probability * losing cost)
losing means negative impact on value so it is negative
=27000*0.7-12000*0.3
=18900-3600.
=15300 $.
Which expression is equivalent to
2(5m) + m?
a) 11m
b) 12m
c) 5m + 2
d) 7m + 2m
Answer:
choice a.) 11m
Step-by-step explanation:
2(5m) + m = 10m + m = 11m
Find an equation in standard form for the ellipse with vertical major axis of length 18 and minor axis length of 6
Answer:
The equation of the ellipse is standard form is expressed as:
x²/81 + y²/9 = 1
Step-by-step explanation:
General equation of an ellipse is expressed as shown below:
x²/a² + y²/b² = 1
Length of the major axis = 2a
Length of the Minor axis = 2b
Given length of major axis = 18
18 = 2a
a = 9
Similarly, if the length of the minor axis is 6, then
6 = 2b
b = 3
The equation of the ellipse becomes;
x²/9² + y²/3² = 1
x²/81 + y²/9 = 1
Finding the LCM
(x²+9y²)/81 = 1
x²+9y² = 81
the plane that delta airlines uses to travel from New York to Los Angeles has a total of 80 seats. On the most recent flight, only 4/5 of the seats were filled with passengers. Of those filled with passagers 3/4 were people traveling for business. How many people on the flight were traveling for business?
Answer:
48
Step-by-step explanation:
If only 4/5 of the 80 seats were filled, then 4/5*80=64 of the seats were filled. Of those filled, if 3/4 were travelling for business, then 3/4*64=48 of the people were travelling for business. Hope this helps!
you have 5 red crayons, 3 yellow crayons, 4 blue crayons and purple crayons. what multiplication would you do to find the probability of picking a blue then a red crayon.
Answer:
3/46
Step-by-step explanation:
To illustrate the effects of driving under the influence of alcohol, a police officer brought a DUI simulator to a local high school. Student reaction time in an emergency was measured with unimpaired vision and also while wearing a pair of goggles to simulate the effects of alcohol on vision. For a random sample of nine teenagers, the time (in seconds) required to bring the vehicle to a stop of 60 miles per hour was recorded. Complete parts (a) and (b).Note: A normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers.Data table:Subj. 1 2 3 4 5 6 7 8 9Normal 4.49 4.24 4.58 4.56 4.31 4.80 4.59 5.00 4.79XiImpaired 5.86 5.85 5.45 5.32 5.90 5.49 5.23 5.63 5.63XiA. (pick one below)a. This is a good idea in designing the experiment because the sample size is not large enough.b. This is a good idea in designing the experiment because reaction times are different.c. This is a good idea in designing the experiment because it controls for any "learning" that may occur in using the simulator.B. Use 95% confidence interval to test if there is a difference in braking time with impaired vision and normal vision where the differences are computed as "impaired minus normal."The 95% confidence interval is (_, _). (Round to the nearest thousandth as needed.)State the appropriate conclusion. Choose the correct answer below.a. There is sufficient evidence to conclude there is a difference in braking time with impaired vision and normal vision.b. There is insufficient evidence to conclude there is a difference in braking time with impaired vision and normal vision.
Answer:
Check the explanation
Step-by-step explanation:
Here we have to first of all carry out dependent sample t test. consequently wore goggles first was selected at random for the reason that the reaction time in an emergency taken with goggles would be greater than the amount of reaction time in an emergency taken with not so weakened vision. So that we will get the positive differences d = impaired - normal
b)
To find 95% confidence interval first we need to find sample mean and sample sd for difference d = impaired minus normal.
We can find it using excel that is in the first attached image below,
Therefore sample mean [tex]( \bar{X}_{d} )[/tex] = 0.98
Sample sd [tex]( \bar{S}_{d} )[/tex] = 0.3788
To find 95% Confidence interval we can use TI-84 calculator,
Press STAT ----> Scroll to TESTS ---- > Scroll down to 8: T Interval and hit enter.
Kindly check the attached image below.
Therefore we are 95% confident that mean difference in braking time with impaired vision and normal vision is between ( 0.6888 , 1.2712)
Conclusion : As both values in the interval are greater than 0 , mean difference impaired minus normal is not equal to 0
There is significant evidence that there is a difference in braking time with impaired vision and normal vision at 95% confidence level .
Bringing a DUI simulator to a high school is a good idea as it controls for any "learning" that may occur. The 95% confidence interval shows that there is a difference in braking time with impaired vision and normal vision.
In designing the experiment, bringing a DUI simulator to a high school is a good idea because it controls for any "learning" that may occur in using the simulator. It allows the students to experience the effects of impaired vision due to alcohol and measure their reaction times under both normal and impaired conditions.
To test if there is a difference in braking time with impaired vision and normal vision, a 95% confidence interval is used. The 95% confidence interval is calculated by determining the mean difference in braking time and finding the range within which the true mean difference lies.
The 95% confidence interval for the difference in braking time with impaired vision and normal vision is (0.254, 1.088) seconds. Based on this interval, there is sufficient evidence to conclude that there is a difference in braking time with impaired vision and normal vision.
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DEFINE PARALLELLOGRAM
Answer:
Parallelograms are shapes that have four sides with two sides that are parallel. For example, some shapes that are parallelograms are squares, rectangle, rhombus and rhomboid.
Step-by-step explanation:
No explanation :)
Final answer:
A parallelogram is a quadrilateral with parallel and congruent opposite sides, congruent opposite angles, and diagonals that bisect each other.
Explanation:
A parallelogram is a four-sided polygon, also known as a quadrilateral, where opposite sides are parallel and equal in length. There are several properties that define a parallelogram:
Opposite sides are congruent (equal in length).Opposite angles are congruent (equal in measure).Consecutive angles are supplementary (their measures add up to 180 degrees).The diagonals of a parallelogram bisect each other.Each diagonal divides the parallelogram into two congruent triangles.An example of a parallelogram is a rectangle, which also has all angles equal to 90 degrees, making it a special type of parallelogram. Another example is a rhombus, which has all sides of equal length. The term parallelogram comes from the Greek 'parallelos' and 'gramme', meaning 'parallel lines' and 'line' respectively.
Work out the length X.
Answer:
17.67 cm
Step-by-step explanation:
[tex]\sqrt{a^2+b^2}[/tex] = x
16^2 = 256
7.5^2 = 56.25
Add.
256 + 56.25 = 312.25
[tex]\sqrt{312.25}= 17.67[/tex]
The missing side of the triangle is x = 17.67 cm
What is Pythagoras Theorem?If ABC is a triangle with AC as the hypotenuse and angle B with 90 degrees then we have:
|AC|^2 = |AB|^2 + |BC|^2
where |AB| = length of line segment AB. (AB and BC are rest of the two sides of that triangle ABC, AC being the hypotenuse).
The Pythagorean theorem formula:
a² + b² = c²
Replacing by the values we have;
16² + 7.5² = x²
x² = 256 + 56.25
x² = 312.25
x = √312.25 cm
x = 17.67 cm
Therefore, the correct answer is x = 17.67 cm
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What is true about the function graphed below?
see image v v v
A. The coordinates of the vertex are (3, 0).
B. The function graphed is look at the picture.
C. The axis of symmetry is the x-axis.
D. The axis of symmetry is the y-axis.
Answer:
D. The axis of symmetry is the y-axis.
Step-by-step explanation:
The vertex is at coordinates (0, 3). Since this parabola opens vertically, its axis of symmetry is the x=coordinate of the vertex: x = 0. That is the equation of the y-axis.
The axis of symmetry is the y-axis.
Jeanette grew 12 flowers with 2 seeds packets how many seed packets does jeanntte need to have a total of 18 flowers in the graden
Answer: 3 packets
12/2 = 6 seeds per packets
18/6 = 3 packets
Answer:
3
Step-by-step explanation:
If Jeanette grew 12 flowers with 2 seeds packets, that means that each seed packet grows 6 flower. 12/2=6
If each seed packet grows 6 flowers, 3 seed packets will give 18 flowers. Hope this helps!