Answer:
[tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-0.714 -0}{\frac{1.38}{\sqrt{7}}}=-1.369[/tex]
[tex]p_v =2*P(t_{(6)}<-1.369) =0.220[/tex]
So the p value is higher than any significance level given, so then we can conclude that we FAIL to reject the null hypothesis that the difference mean between Drug A and Drug B is equal to 0.
Step-by-step explanation:
Data given
1 2 3 4 5 6 7
Drug A 6 3 4 5 7 1 4
Drug B 5 1 5 5 5 2 2
A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.
Let put some notation
x=test value for A , y = test value for B
The system of hypothesis for this case are:
Null hypothesis: [tex]\mu_y- \mu_x = 0[/tex]
Alternative hypothesis: [tex]\mu_y -\mu_x \neq 0[/tex]
The first step is calculate the difference [tex]d_i=y_i-x_i[/tex] and we obtain this:
d: -1, -2, 1, 0, -2, 1, -2
The second step is calculate the mean difference
[tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}=-0.714[/tex]
The third step would be calculate the standard deviation for the differences, and we got:
[tex]s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =1.38[/tex]
The 4 step is calculate the statistic given by :
[tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-0.714 -0}{\frac{1.38}{\sqrt{7}}}=-1.369[/tex]
The next step is calculate the degrees of freedom given by:
[tex]df=n-1=7-1=6[/tex]
Now we can calculate the p value, since we have a two tailed test the p value is given by:
[tex]p_v =2*P(t_{(6)}<-1.369) =0.220[/tex]
So the p value is higher than any significance level given, so then we can conclude that we FAIL to reject the null hypothesis that the difference mean between Drug A and Drug B is equal to 0.
find the number of subsets of s = 1 2 3 . . . 10 that contain exactly 4 elements including 3 or 4 but not both.
Answer:
112
Step-by-step explanation:
Let A be a subset of S that satisfies such condition.
If 3∈A, then the other three elements of A must be chosen from the set B={1,2,5,6,7,8,9,10} (because 3 cannot be chosen again and 4 can't be alongside 3). B has eight elements, then there are [tex]\binom{8}{3}=56[/tex] ways to select the remaining elements of A (the binomial coefficient counts this). The remaining elements determine A uniquely, then there are 56 subsets A.
If 4∉A, we have to choose the remaining elements of A from the set B={1,2,5,6,7,8,9,10}. B has eight elements, then there are [tex]\binom{8}{3}=56[/tex] ways to select the remaining elements of A. Thus, there are 56 choices for A.
By the sum rule, the total number of subsets is 56+56=112
A researcher asks participants to taste each of three meals and to choose the one they like best. The same foods are in each meal, however the calorie total of each meal is different. One is low in calories, one is moderate in calories and one is high in calories. Based on the observed frequencies given below, what is an appropriate conclusion for this test at a .05 level of significance?
Type of Meal
Low Calorie Moderate Calorie High Calorie
fo 6 7 17
fe 10 10 10
A. Participants liked the high calorie meal more than the low calorie meal.
B. Participants liked the low calorie meal less than the moderate calorie meal.
C. Participants liked the high calorie meal more than was expected.
D. All of the above
Answer:
D. All of the above
Step-by-step explanation:
By the given data, some persons were in the experiment and each of them was given with three meals, Low Calorie, Moderate Calorie and High Calorie. They results for the meal they liked is as below:
Low Calorie Meal got likes = 6
Moderate Calorie Meal got likes = 7
High Calorie Meal got likes = 17
so,
Option A is correct as High Calorie meal got 17 likes while Low Calorie meal got 6 likes.
Option B is also correct as Low Calorie meal got 6 likes while Moderate Calorie meal got 7 likes.
Option C is correct too as High Calorie meal got largest number of likes even more than the double of Low calorie and Moderate calorie meal so it was more than expected.
Final answer:
Based on the data provided, without performing an actual chi-squared test due to a lack of p-value or statistical data, the most supported conclusion is that participants liked the high calorie meal more than was expected (option C). This is inferred from the observed preference for the high calorie meal, which was chosen significantly more often (17 times) compared to its expected frequency (10 times).
Explanation:
The question presents a scenario where participants are asked to taste three types of meals with different calorie counts and choose the one they like best. The observed frequencies (fo) for each type of meal (low calorie, moderate calorie, and high calorie) are given as 6, 7, and 17, respectively. The expected frequencies (fe) are all 10, assuming no preference among the meals. Using a significance level of .05, we must employ a chi-squared test to determine if the observed preferences significantly differ from what was expected given no effect of calorie count.
The chi-squared test result for this would involve calculating the sum of squared differences between observed and expected frequencies, divided by the expected frequencies for each category and summing those values. However, since the data on the actual chi-squared statistic or the p-value are not provided, we cannot perform the exact calculation. Nonetheless, we can make a conclusion based on the observed and expected frequencies:
There is a notable preference for the high calorie meal, as it was chosen substantially more often than expected (17 vs. 10).
The low calorie meal was chosen less often than expected (6 vs. 10).
The moderate calorie meal was also chosen less than expected, but to a lesser extent (7 vs. 10).
Without a specific p-value or chi-squared statistic, we can't formally conclude that the participants liked the high calorie meal significantly more than the other meals, but the observed data suggest that might be the case. Also, we cannot definitively reject the possibility that the preference for the high-calorie meal is due to chance.
Therefore, the most reasonable conclusion based on the given information without formal test statistics would be option C: Participants liked the high calorie meal more than was expected.
This is in line with the general principle that a higher calorie meal tends to be preferred potentially due to higher fat content, which has been associated with better taste and satisfaction, particularly in fast-food meals. At a .05 level of significance and without the actual test statistic, option C is the most supported by the provided data.
region R is revolved about the y-axis. The volume of the resulting solid could (in principle) be found by using the disk/washer method and integrating with respect to
Answer:
A region R is revolved about the y-axis. The volume of the resulting solid could (in principle) be found using the disk/washer method and integrating with respect to y or using the shell method and integrating with respect to x.
Step-by-step explanation:
We assume this question :
Fill in the blanks: A region R is revolved about the y-axis. The volume of the resulting solid could (in principle) be found using the disk/washer method and integrating with respect to _ or using the shell method and integrating with respect to _____.
We can calculate the volume of a region when revolved about y-axis with two common methods: washer method and shell method. We need to take in count, that the general formula for these methods are different respect to the variable used to calculate the volume.
Since the region R is revolved about y-axis, disk/washer method needs to calculate the integral respect to y, and by the other hand the shell method will calculate the integral respect to x.
Washer method
[tex]V\approx \sum_{k=1}^n \pi r^2 h = \sum_{k=1}^n \pi f(y_k)^2 dy[/tex]
[tex]V= \pi \int_{c}^d f(y)^2 dy[/tex]
Shell method
[tex]V = \lim_{n\to\infty} 2\pi x_i f(x_i)\Delta x =\int_{a}^b 2\pi x f(x) dx[/tex]
So then the correct answer would be:
A region R is revolved about the y-axis. The volume of the resulting solid could (in principle) be found using the disk/washer method and integrating with respect to y or using the shell method and integrating with respect to x.
a chemist examines 17 seawater samples for iron concentration for the sample data is 0.704 cc/cubic meter with a standard deviation of 0.0142. determine the 99% confidence interval for the population mean from concentration. Assume the population is approx. normal.
Step 1. Find the critical value that should be used in constructing the confidence interval ( round your answer to 3 decimal places)
Step 2. Construct the 99% confidence interval (Round answer to 3 decimal places)
Answer:
Critical value: [tex]z= 2.575[/tex]
99% confidence interval: (0.695 cc/cubic meter, 0.713 cc/cubic meter).
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.99}{2} = 0.005[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex] is the critical value
Now, find M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}} = 2.575*\frac{0.0142}{\sqrt{17}} = 0.0089[/tex]
The lower end of the interval is the mean subtracted by M. So it is 0.704 - 0.0089 = 0.695 cc/cubic meter.
The upper end of the interval is the mean added to M. So it is 0.704 + 0.0089 = 0.713 cc/cubic meter.
So
99% confidence interval: (0.695 cc/cubic meter, 0.713 cc/cubic meter).
In a study of annual salaries of employees, random samples were selected from two companies to test if there is a difference in average salaries. For Company "X", the sample was size 65, the sample mean was $47,000 and the population standard deviation is assumed to be $11,000. For Company "Y", the sample size was 55, the sample mean was $44,000 and the population standard deviation is assumed to be $10,000. Test for a difference in average salaries at a 5% level of significance. What is your conclusion?
Answer:
Step-by-step explanation:
The display provided from technology available below results from using data for a smartphone carrier's data speeds at airports to test the claim that they are from a population having a mean less than 4.004.00 Mbps. Conduct the hypothesis test using these results. Use a 0.050.05 significance level. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. LOADING... Click the icon to view the display from technology. What are the null and alternative hypotheses? A. Upper H 0H0: muμequals=4.004.00 Mbps
Answer:
Null hypothesis:[tex]\mu \geq 4.0[/tex]
Alternative hypothesis:[tex]\mu < 4.00[/tex]
[tex]t=\frac{3.48-4.00}{\frac{1.150075}{\sqrt{45}}}=-3.033077[/tex]
[tex]df=n-1=45-1=44[/tex]
Since is a left-sided test the p value would be:
[tex]p_v =P(t_{(44)}<-3.033077)=0.002025[/tex]
If we compare the p value and 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, so we can conclude that the mean is significantly less than 4.00 at 5% of significance.
Step-by-step explanation:
Data given and notation
[tex]\bar X=3.48[/tex] represent the sample mean
[tex]s=1.150075[/tex] represent the sample standard deviation
[tex]n=45[/tex] sample size
[tex]\mu_o =4.00[/tex] represent the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is less than 4.00 :
Null hypothesis:[tex]\mu \geq 4.0[/tex]
Alternative hypothesis:[tex]\mu < 4.00[/tex]
Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{3.48-4.00}{\frac{1.150075}{\sqrt{45}}}=-3.033077[/tex]
P-value
First we need to calculate the degrees of freedom given by:
[tex]df=n-1=45-1=44[/tex]
Since is a left-sided test the p value would be:
[tex]p_v =P(t_{(44)}<-3.033077)=0.002025[/tex]
Conclusion
If we compare the p value and 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, so we can conclude that the mean is significantly less than 4.00 at 5% of significance.
Sandra normally takes 2 hours to drive from her house to her grandparents' house driving her usual speed. However, on one particular trip, after 40% of the drive, she had to reduce her speed by 30 miles per hour, driving at this slower speed for the rest of the trip. This particular trip took her 228 minutes. What is her usual driving speed, in miles per hour?
Answer:
Her usual driving speed is 38 miles per hour.
Step-by-step explanation:
We know that:
[tex]s = \frac{d}{t}[/tex]
In which s is the speed, in miles per hour, d is the distance, in miles, and t is the time, in hours.
We have that:
At speed s, she takes two hours to drive. So
[tex]s = \frac{d}{2}[/tex]
[tex]d = 2s[/tex]
However, on one particular trip, after 40% of the drive, she had to reduce her speed by 30 miles per hour, driving at this slower speed for the rest of the trip. This particular trip took her 228 minutes.
228 minutes is 3.8 hours. So
[tex]0.4s + 0.6(s - 30) = \frac{d}{3.8}[/tex]
So
[tex]3.8(0.4s + 0.6s - 18) = d[/tex]
[tex]3.8s - 68.4 = 2s[/tex]
[tex]1.8s = 68.4[/tex]
[tex]s = 38[/tex]
Her usual driving speed is 38 miles per hour.
A circle of fourths is generated by starting at any note and stepping upward by intervals of a fourth (five half-steps). By what factor is the frequency of a tone increased if it is raised by a fourth? How many fourths are required to complete the entire circle of fourths? How many octaves are covered in a complete circle of fourths?
Answer:
a) Q₀ × 1.335 cps
b) 12 fourths
c) 5 octaves
Step-by-step explanation:
The circle of fourth is generated by starting at any note and stepping upward by intervals of a fourth(five half-steps)
(a) By what factor is the frequency of a tone increased if it is raised by a fourth?
consider the following exponential growth formula :
Q = Q₀ × f °
Q = Q₀ × 1.05946 ⁿ
Substituting n = 5
Q = Q₀ × 1.05946 ⁿ
Q = Q₀ × 1.05946⁵
Q = Q₀ × 1.335 cps
Therefore, note that five half-step higher will be increased by factor 1.335
(b) How many fourths are required to complete the entire circle of fourths?
SOLUTION
In each step, number of half steps = 5
Total number of half steps in one octave = 12 half- steps
Therefore, total number of fourths required to complete the entire circle of fourths = 12 fourths
(c) Total number of half steps in one octave = 12 half-steps
Total half steps in complete circle of fourths
= 12×5
= 60 half-steps
Calculating number of octaves (dividing it by 12)
= 60/12
= 5 octaves
It is given that the circle of fourth is generated by starting at any note and stepping upward by intervals of a fourth(five half-steps)
Consider the exponential growth formula :[tex]Q = Q_0 \times f ^0\\Q = Q_0 \times 1.05946^n[/tex]
Put n = 5 in the above formula, we get
[tex]Q = Q_0 ( 1.05946 ^n)\\Q = Q_0 ( 1.05946 ^5)\\Q = Q_0 (1.335) cps[/tex]
So, by 1.335 factor the frequency of a tone increased if it is raised by a fourth.
In each step, number of half steps = 5Total number of half steps in one octave = 12 half- steps
So, total number of fourths required to complete the entire circle of fourths = 12 fourths
Total number of half steps in one octave = 12 half-stepsTotal half steps in complete circle of fourths will be 12×5 = 60 half-steps
Now, the number of octaves is [tex]\frac{60}{12} =5[/tex] octaves.
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An educator evaluates the effects of small, medium, and large class sizes on academic performance among male and female students. Identify each factor and the levels of each factor in this example. (Select all that apply.)
a. gender (two levels: male, female)
b. academic performance (three levels: above average, average, below average) class size (two levels: small, large)
c. academic performance (two levels: passing, failing)
d. class size (three levels: small, medium, large)
e. gender (three levels: male, female, trans)
Answer:
a. gender (two levels: male, female)
d. class size (three levels: small, medium, large)
Step-by-step explanation:
The study is going to be made to identify how male and female students react to the class size(small, medium, large). So there are two correct options, since the gender and the class size are factors in this research.
The correct options are:
a. gender (two levels: male, female)
d. class size (three levels: small, medium, large)
In this example, the factors and levels are gender, academic performance, and class size.
Explanation:The factors and levels in this example are:
a. Gender (two levels: male, female)
b. Academic performance (three levels: above average, average, below average)
c. Class size (two levels: small, large)
Therefore, the answer is a, b, and c.
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evaluate M2 + MNP if M=3,N= 4, and P=7
M^2 + MNP
(3)^2 + (3)(4)(7)
9 + 12(7)
9 + 84
93
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Suppose f(x) = 1/4 over the range a ≤ x ≤ b, and suppose P(X > 4) = 1/2.
What are the values for a and b?
-0 and 4
-2 and 6
-Can be any range of x values whose length (b − a) equals 4.
-Cannot answer with the information given.
Answer:
a=2
b=6
Step-by-step explanation:
Assuming a uniform distribution and that a ≤ x ≤ b, if f(x) = 1/4, then:
[tex]f(x) =\frac{1}{4}=\frac{1}{b-a}\\b-a = 4[/tex]
If P(X > 4) = 1/2, then:
[tex]\frac{1}{2} = 1-(\frac{x-a}{b-a})=1-(\frac{4-a}{4})\\a-4+4 =\frac{1}{2}*4\\ a=2\\b=4+a\\b=6[/tex]
The values for a and b are, respectively, 2 and 6.
The time that it takes a randomly selected job applicant to perform a certain task has a distribution that can be approximated by a normal distribution with a mean value of 150 sec and a standard deviation of 25 sec. The fastest 10% are to be given advanced training. What task times qualify individuals for such training? (Round the answer to one decimal place.)
________ seconds or less
Final answer:
The cutoff time to be in the fastest 10% of job applicants is approximately 118.0 seconds or less, calculated using a z-score of the 10th percentile in a normal distribution.
Explanation:
The student is asking about finding the cutoff time for the fastest 10% of job applicants performing a certain task, given that the time taken is normally distributed with a mean of 150 seconds and a standard deviation of 25 seconds. To solve this problem, we will use the z-score corresponding to the fastest 10% (the 10th percentile) in a normal distribution.
First, we look up the z-score for the 10th percentile in the z-table, which is approximately -1.28. Then we use the z-score formula:
z = (X - μ) / σ
Plugging in the known values:
-1.28 = (X - 150) / 25
Now, we solve for X:
X = -1.28 × 25 + 150
X = -32 + 150
X = 118 seconds
So, job applicants must complete the task in approximately 118.0 seconds or less to qualify for the advanced training.
find a and b from the picture
Answer:a is 40 degrees
b is 140 degrees
Step-by-step explanation:
The given polygon has 5 irregular sides. This means that it is an irregular pentagon. The sum of the exterior angles of a polygon is 360 degrees
The exterior angles of the given pentagon are a, 75, 65, 60 and 120. Therefore
a + 75 + 65 + 60 + 120 = 360
a + 320 = 360
Subtracting 320 from both sides of the equation, it becomes
a = 360 - 320 = 40 degrees
The sum of angles on a straight line is 180/degrees. Therefore,
a + b = 180
b = 180 - a = 180 - 40 = 140 degrees
A city has 1,091,953 residents. A recent census showed that 364,656 of these residents regularly use the city's public transportation system. A survey is being conducted in which 1,047 of the city's 1,091,953 residents will be randomly selected. This question relates to the number of people in the survey that is made up of people that do use the city's public transport. The number of people in the survey that do use public transport approximately follows a normal distribution. Calculate the standard deviation of this distribution. Give your answer rounded to 2 decimal places.
Answer:
The standard deviation of the distribution is 492.847
Step-by-step explanation:
Consider the provided information.
A city has 1,091,953 residents. A recent census showed that 364,656 of these residents regularly use the city's public transportation system.
Therefore the value of n is 1,091,953
The probability of success is: [tex]\frac{364,656 }{1,091,953}=0.33395[/tex]
Calculate the standard deviation using the formula: [tex]\sigma=\sqrt{np(1-p)}[/tex]
Substitute the respective values.
[tex]\sigma=\sqrt{1091953(0.334)(1-0.334)}[/tex]
[tex]\sigma=\sqrt{242898.3931}[/tex]
[tex]\sigma=492.847[/tex]
Hence, the standard deviation of the distribution is 492.847
Suppose that we don't have a formula for g(x) but we know that g(3) = −5 and g'(x) = x2 + 7 for all x.
(a) Use a linear approximation to estimate g(2.9) and g(3.1). g(2.9) ≈ g(3.1) ≈
(b) Are your estimates in part (a) too large or too small? Explain.
The value of the function g(x) at x = 2.9 and x = 3.1 will be -6.57 and -3.37, respectively. And the estimation is too small in part (a).
What is integration?Integration is a way of finding the total by adding or summing the components. It's a reversal of differentiation, in which we break down functions into pieces. This approach is used to calculate the total on a large scale.
Suppose that we don't have a formula for g(x) but we know that g(3) = −5 and g'(x) = x2 + 7 for all x.
Integrate the function, then we have
∫g'(x) = ∫(x² + 7) dx
g(x) = x³/3 + 7x + c
g(3) = 3³ / 3 + 7 (3) + c
- 5 = 9 + 21 + c
c = - 30 - 5
c = -35
Then the function is given as,
g(x) = x³/3 + 7x - 35
At x = 2.9, we have
g(2.9) = (2.9)³/3 + 7(2.9) - 35
g(2.9) = -6.57
At x = 3.1, we have
g(3.1) = (3.1)³/3 + 7(3.1) - 35
g(3.1) = -3.37
The value of the function g(x) at x = 2.9 and x = 3.1 will be -6.57 and -3.37, respectively. And the estimation is too small in part (a).
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4. A food company is concerned that its 16-ounce can of sliced pears is being overfilled. The quality-control department took a sample of 35 cans and found that the sample mean weight was 16.04 ounces, with a sample standard deviation of 0.08 ounces. Test at a 5% level of significance to see if the mean weight is greater than 16 ounces. What is your conclusion?
Answer:
The mean weight is not greater than 16 ounces
Step-by-step explanation:
Null hypothesis: Mean weight is greater than 16 ounces
Alternate hypothesis: Mean weight is not greater than 16 ounces.
Sample mean= 16.04 ounces
Assumed mean= 16 ounces
Significance level = 0.05
Sample standard deviation = 0.08 ounces
Number of samples= 35
To test the claim about the mean, Z = (sample mean - assumed mean) ÷ (standard deviation÷ √number of samples)
Z = (16.04 - 16) ÷ (0.08 ÷ √35)
Z = 0.04 ÷ (0.08 ÷ 5.92)
Z= 0.04 ÷ 0.014 = 2.86
Z = 2.86
It is a one-tailed test, the critical region is an area of 0.05 (the significance level). The critical value of the critical region is 1.64
Since the computed Z 2.86 is greater than the critical value 1.64, the null hypothesis is rejected.
Conclusion
The mean weight is not greater than 16 ounces
This is a GCSE maths question which i don’t understand.
The petrol consumption for David's journey, considering bounds, is between 6.5439 km/l and 6.8182 km/l.
To determine the petrol consumption (c) for David's journey, we use the formula c = d/p, where d is the distance (187 km) and p is the amount of petrol used (28 litres).
Firstly, we calculate the maximum and minimum possible values for c by considering the upper and lower bounds of d and p:
Maximum value of c:
Upper bound of distance (d) = 187 + 0.5 (half of the smallest decimal place in 187 km) = 187.5 km
Lower bound of petrol used (p) = 28 - 0.5 (half of the smallest decimal place in 28 litres) = 27.5 litres
Maximum c = (187.5 km) / (27.5 litres) = 6.8182 km/l
Minimum value of c:
Lower bound of distance (d) = 187 - 0.5 = 186.5 km
Upper bound of petrol used (p) = 28 + 0.5 = 28.5 litres
Minimum c = (186.5 km) / (28.5 litres) = 6.5439 km/l
Therefore, considering bounds, the petrol consumption (c) for David's journey is between 6.5439 km/l and 6.8182 km/l.
A manufacturer is interested in determining whether it can claim that the boxes of detergent it sells contain, on average, more than 500 grams of detergent. The firm selects a random sample of 100 boxes and records the amount of detergent (in grams) in each box. The data are provided in the file P09_02.xlsx.
a.Identify the null and alternative hypotheses for this situation.
b.Is there statistical support for the manufacturers claim
Answer:
a) H0=Weight of detergent=500gr (null hypotheses )
H1=Weight of detergent>500 gr. (alternative hypotheses )
b) X test.
Step-by-step explanation:
a)
The statement from which we are going to establish the assumptions is that supplied by the manufacturer, which says that the average weight of the detergent is 500 g. So:
H0=Weight of detergent=500gr (null hypotheses )
H1=Weight of detergent>500 gr. (alternative hypotheses )
b)
You are not providing the source you mention where the sample of detergent boxes taken is located. So I will explain in general terms how the problem should be posed.
For this case, we can use Z test because the sample we are taking is larger than 30.
The first thing we have to do is calculate Z, which is given by:
[tex]Z=\frac{(X-Xav)}{SD}[/tex]
Where:
Xav=Average of x.
SD=Standard deviation.
Then, we look for the table and if the value we have just calculated is greater than that of the table, then we must reject the null hypothesis.
A performer expects to sell 5,000 tickets for an upcoming concert. They want to make a total of $311, 000 in sales from these tickets. What is the price for one ticket?
The price of one ticket is $ 62.2
Solution:Given that a performer expects to sell 5000 tickets for an upcoming event
They want to make a total of $ 311, 000 in sales from these tickets
To find: price of one ticket
Let us assume that all tickets have the same price
Let "a" be the price of one ticket
So the total sales price of $ 311, 000 is obtained from product of 5000 tickets and price of one ticket
[tex]\text {total sales price }=5000 \times \text { price of one ticket }[/tex]
[tex]311000 = 5000 \times a\\\\a = \frac{311000}{5000}\\\\a = 62.2[/tex]
Thus the price of one ticket is $ 62.2
I really need help with this!
Answer:
Step-by-step explanation:
1) in a rhombus, the diagonals bisect each other at the midpoint, forming 4 right angles.
MK = NK + NM
Since MK = 24 and NK = NM, then,
NK = 24/2 = 12
JL = NJ + NL
Since JL = 20 and NJ = NMlL, then,
NL = 20/2 = 10
Looking at triangle LMN, it is a right angle triangle. Applying Pythagoras theorem,
ML^2 = 10^2 + 12^2 = 100 + 144 = 244
ML = √244 = 15.62
Since all the sides of the rhombus are equal, then
MJ = ML = 15.62
Angle KNL = angle JML = 90 degrees. This is so because the diagonals are perpendicular)
Angle KJL = 90 - angle MJL
Angle KJL = 90 - 50 = 40 degrees.
Since the opposite angles of the rhombus are equal,
Angle MLK = angle angle MJK = 50 + 40 = 90 degrees
2) since all the sides of a rhombus are equal,
5x + 16 = 9x - 32
9x - 5x = 32 + 16
4x = 48
x = 48/4 = 12
Since PQ = NR = 5x + 16, then
PQ = 5×12 + 16 = 60 + 16
PQ = 76
3) Triangle XYZ is an isosceles triangle. This means that its base angles, XZY and ZXY are equal. The sum of angles in a triangle is 180 degrees. Therefore,
Angle XZY + angle ZXY = 180 - 136
Angle XZY + angle ZXY = 44
Angle XZY = 44/2 = 22
Therefore,
10x - 8 = 22
10x = 22 + 8 = 30
x = 30/10 = 3
In 1997, 46% of Americans said they did not trust the media when it comes to reporting the news and fairly. In 2007 poll of 1010 adults nationwide, 525 stated they did not trust media. At the α=0.05 level of significance, is there evidence to support the claim that the percentage of Americans that do not trust the media to report fully and accurately has increased since 1997
Answer:
We can conclude that the percentage of Americans that do not trust the media to report fully and accurately has increased since 1997 (P-value=0.00014).
Step-by-step explanation:
We have to perform an hypothesis test on a proportion.
The null and alternative hypothesis are:
[tex]H_0: \pi=0.46\\\\H_1: \pi>0.46[/tex]
The significance level is α=0.05.
The standard deviation is estimated as:
[tex]\sigma=\sqrt{\frac{\pi(1-\pi)}{N} } =\sqrt{\frac{0.46(1-0.46)}{1010} }=0.0157[/tex]
The z value for this sample is
[tex]z=\frac{p-\pi-0.5/N}{\sigma} =\frac{525/1010-0.46-0.5/1010}{0.0157} =\frac{0.52-0.46-0.00}{0.0157}=\frac{0.06}{0.0157} =3.822[/tex]
The P-value for z=3.822 is P=0.00014.
The P-value is smaller than the significance level, so the effect is significant. The null hypothesis is rejected.
We can conclude that the percentage of Americans that do not trust the media to report fully and accurately has increased since 1997.
A researcher wants to estimate the mean weekly family expenditure on clothing purchases and maintenance. First, she needs to determine the number of families that must be sampled, in order to estimate the mean weekly expenditure on clothing to within $15 at alpha = 0.01. From prior similar studies she is confident that the standard deviation of weekly family expenditures on clothing purchases and maintenance is $125. The required sample size to be taken is___________.
A. 21
B. 22
C. 461
D. 460
E. 460.46
Answer: C. 461
Step-by-step explanation:
We know that the formula to find the sample size is given by :-
[tex]n=(\dfrac{z^*\cdot \sigma}{E})^2[/tex]
, where [tex]\sigma[/tex] = Population standard deviation from prior study.
E = margin of error.
z* = Critical value.
As per given , we have
[tex]\sigma=125[/tex]
E= 15
Significance level : [tex]\alpha=0.01[/tex]
Critical value (Two tailed)=[tex]z^*=z_{\alpha/2}=z_{0.005}=2.576[/tex]
Now , Required sample size = [tex]n=(\dfrac{(2.576)\cdot 125}{15})^2[/tex]
[tex]n=(21.4666666667)^2\\\\ n=460.817777779\approx461[/tex] [Round to next integer]
Hence, the required sample size to be taken is 461.
Correct answer = C. 461
Put the equation in slope intercept form y-10=-2(x-3)
Answer:
y= -2x + 16
Step-by-step explanation:
multiple the -2 with everything in parentheses then get -2x and 6 add 10 to both sides then then bring everything down to get y= -2x + 16
A hemispherical bowl of radius r contains water to a depth h. Give a formula that you can use to measure the volume of the water in the bowl.
Answer:
V = π (rh² − ⅓h³)
Step-by-step explanation:
Draw a cross section of the bowl. Cut a thin, horizontal slice of the water. This slice is a circular disc of radius x and thickness dy. It is position a distance of y from the bottom of the bowl. The volume of this slice is:
dV = πx² dy
By drawing a right triangle, we can define x in terms of y:
x² + (r−y)² = r²
x² + r² − 2ry + y² = r²
x² = 2ry − y²
Substitute:
dV = π (2ry − y²) dy
The total volume of the water is the sum of all the slices from y=0 to y=h.
V = ∫₀ʰ π (2ry − y²) dy
V = π ∫₀ʰ (2ry − y²) dy
V = π (ry² − ⅓y³) |₀ʰ
V = π (rh² − ⅓h³)
Final answer:
The formula to measure the volume of water in a hemispherical bowl is V = (2/3)πr²h.
Explanation:
The formula to measure the volume of water in a hemispherical bowl is V = (2/3)πr2h, where V is the volume, π is a constant approximately equal to 3.14159, r is the radius of the bowl, and h is the depth of the water.
1. The volume of a half of the globe is given by V = 2/3 * π * r³. However, since we are only interested in measuring the water's volume—which represents a portion of the hemisphere—we will need to modify the formula accordingly.
2. Within the larger hemisphere, a smaller one is created by the water's depth, h. To ascertain the volume of this more modest half of the globe, we can take away it from the volume of the bigger side of the equator.
3. The span of the more modest half of the globe is likewise r, since it has a similar ebb and flow as the bigger side of the equator.
4. The level of the more modest side of the equator, h, is equivalent to the profundity of the water.
5. To find the volume of the more modest side of the equator, we utilize the recipe V = 2/3 * π * r³ and substitute the sweep r and level h.
6. The volume of the water is around 50% of the volume of the more modest half of the globe since it just fills one side of the bowl.
7. In this way, the volume of the water in the hemispherical bowl is given by 1/2 * (2/3 * π * r³) = 1/3 * π * r³.
8. To represent the profundity of the water, we increase the volume by the proportion of the level h to the sweep r, giving us 1/3 * π * r² * h.
9. We can further simplify by rewriting the formula as 1/2 * * h2 * (3r - h).
Allen runs at an average rate of 9 mi/hr and walks at a average rate of 3 mi/hr. Write an equation in standard form to relate the times he can spend walking and running if he travels 30 miles. If he walks for 4 hours, for how long will he run?
Answer:
The equation would be [tex]30=12+9x[/tex].
Allen will Run for 2 hours.
Step-by-step explanation:
Given:
Total Distance traveled = 30 miles
Average rate of walking = 3 mi/hr
Average rate of running = 9 mi/hr
Time for walking = 4 hours
W need to find the time for running.
Solution:
Let Number of hrs required for running be 'x'.
Total Distance traveled is equal Distance Traveled in walking plus Distance traveled in Running
But Distance is equal Rate multiplied by Time.
Framing in equation form we get;
Hence Total Distance Traveled = Rate of walking × Hours of walking + Rate of Running × Hours of running.
Substituting the values we get;
[tex]30 = 3\times4 +9x\\\\30=12+9x[/tex]
Hence The equation would be [tex]30=12+9x[/tex]
Solving above equation we get;
[tex]9x=30-12\\\\9x=18\\\\x=\frac{18}{9}=2\ hrs[/tex]
Hence Allen will Run for 2 hours.
Suppose the weights of seventh‑graders at a certain school vary according to a Normal distribution, with a mean of 100 pounds and a standard deviation of 7.5 pounds. A researcher believes the average weight has decreased since the implementation of a new breakfast and lunch program at the school. She finds, in a random sample of 35 students, an average weight of 98 pounds. What is the P ‑value for an appropriate hypothesis test of the researcher’s claim?
Answer:
We conclude that there is no change in the average weight since the implementation of a new breakfast and lunch program at the school.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 100 pounds
Sample mean, [tex]\bar{x}[/tex] = 98 pounds
Sample size, n = 35
Alpha, α = 0.05
Population standard deviation, σ = 7.5 pounds
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 100\text{ pounds}\\H_A: \mu < 100\text{ pounds}[/tex]
We use one-tailed(left) z test to perform this hypothesis.
Formula:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]z_{stat} = \displaystyle\frac{98 - 100}{\frac{7.5}{\sqrt{35}} } = -1.577[/tex]
Now, [tex]z_{critical} \text{ at 0.05 level of significance } = -1.64[/tex]
We calculate the p-value with the help of standard normal table.
P-value = 0.057398
Since the p-value is greater than the significance level, we fail to reject the null hypothesis and accept is.
We conclude that there is no change in the average weight since the implementation of a new breakfast and lunch program at the school.
To calculate the p-value for the hypothesis test, we calculate the t-score using the given values and the formula. Then, we use the t-distribution table or a calculator to find the p-value associated with the t-score and degrees of freedom. The calculated p-value is approximately 0.1093.
Explanation:In this problem, we are given the mean and standard deviation of a normal distribution for the weights of seventh-graders. The researcher believes that the new breakfast and lunch program has decreased the average weight. We are asked to calculate the p-value for an appropriate hypothesis test.
To calculate the p-value, we can use the t-distribution since the population standard deviation is unknown. We can calculate the t-score using the formula:
t = (sample mean - population mean) / (sample standard deviation / √sample size)
Plugging in the given values, we get t = (98 - 100) / (7.5 / √35) = -1.654. The degrees of freedom for this test is (sample size - 1) = (35 - 1) = 34.
Using the t-distribution table or a calculator, we can find the p-value associated with a t-score of -1.654 and 34 degrees of freedom. The p-value is approximately 0.1093.
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The amount of water in a bottle is approximately normally distributed with a mean of 2.55 liters with a standard deviation of 0.035 liter. Complete parts (a) through (d) below. a. What is the probability that an individual bottle contains less than 2.52 liters? Round to three decimal places as needed.) b. If a sample of 4 bottles is selected, what is the probability that the sample mean amount contained is less than 2.52 liters? (Round to three decimal places as noeded) c. If a sample of 25 bottles is selected, what is the probability that the sample mean amount contained is less than 2.52 liters? (Round to three decimal places as needed.) d. Explain the difference in the results of (a) and (c)
Answer:
a) [tex]P(X<2.52)=P(Z<\frac{2.52-2.55}{0.035})=P(Z<-0.857)=0.196[/tex]
b) [tex]P(\bar X <2.52) = P(Z<-1.714)=0.043[/tex]
c) [tex]P(\bar X <2.52) = P(Z<-4.286)=0.000[/tex]
d) For part a we are just finding the probability that an individual bottle would have a value of 2.52 liters or less. So we can't compare the result of part a with the results for parts b and c.
If we see part b and c are similar but the difference it's on the sample size for part b we just have a sample size 4 and for part c we have a sample size of 25. The differences are because we have a higher standard error for part b compared to part c.
Step-by-step explanation:
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean". The letter [tex]\phi(b)[/tex] is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: [tex]\phi(b)=P(z<b)[/tex]
Let X the random variable that represent the amount of water in a bottle of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(2.55,0.035)[/tex]
a. What is the probability that an individual bottle contains less than 2.52 liters?
We are interested on this probability
[tex]P(X<2.52)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X<2.52)=P(\frac{X-\mu}{\sigma}<\frac{2.52-\mu}{\sigma})[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(X<2.52)=P(Z<\frac{2.52-2.55}{0.035})=P(Z<-0.857)=0.196[/tex]
b. If a sample of 4 bottles is selected, what is the probability that the sample mean amount contained is less than 2.52 liters? (Round to three decimal places as noeded)
And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:
[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]
On this case [tex]\bar X \sim N(2.55,\frac{0.035}{\sqrt{4}})[/tex]
The z score on this case is given by this formula:
[tex]z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And if we replace the values that we have we got:
[tex]z=\frac{2.52-2.55}{\frac{0.035}{\sqrt{4}}}=-1.714[/tex]
For this case we can use a table or excel to find the probability required:
[tex]P(\bar X <2.52) = P(Z<-1.714)=0.043[/tex]
c. If a sample of 25 bottles is selected, what is the probability that the sample mean amount contained is less than 2.52 liters? (Round to three decimal places as needed.)
The z score on this case is given by this formula:
[tex]z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And if we replace the values that we have we got:
[tex]z=\frac{2.52-2.55}{\frac{0.035}{\sqrt{25}}}=-4.286[/tex]
For this case we can use a table or excel to find the probability required:
[tex]P(\bar X <2.52) = P(Z<-4.286)=0.0000091[/tex]
d. Explain the difference in the results of (a) and (c)
For part a we are just finding the probability that an individual bottle would have a value of 2.52 liters or less. So we can't compare the result of part a with the results for parts b and c.
If we see part b and c are similar but the difference it's on the sample size for part b we just have a sample size 4 and for part c we have a sample size of 25. The differences are because we have a higher standard error for part b compared to part c.
The problem involves using z-scores, normal distribution, central limit theorem and law of large numbers for probability calculations related to the amount of water in sampled bottles. The increase in sample size from 1 to 25 should show a higher probability for the sample's mean to be close to the population mean.
Explanation:In this problem, we will apply the concept of the normal distribution and the central limit theorem to calculate probabilities and compare results. First, we will calculate the z-scores for (a), (b), and (c). The z-score formula is Z = (X - μ) / σ. In (a), X is 2.52 liters, μ (mean) is 2.55 liters, and σ (standard deviation) is 0.035 liters.
In (b) and (c), the σ of a sample mean is σ/√n, where n is the sample size (4 in (b) and 25 in (c)).
Next, we look up the z-score in the z-score table (or use a normal distribution calculator) to get the probabilities.
To answer part (d), probabilistically, as sample size increases, the sample mean tends to get closer to the population mean according to the law of large numbers. Hence, the probabilities in (c) should be higher than (a).
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A study was conducted to examine whether the proportion of females was the same for five groups (Groups A, B, C, D, and E). How many degrees of freedom would the χ2 test statistic have when testing the hypothesis that the proportions in each group are all equal?a. 4 b. 1 c. 0.20 d. 5
Answer: a. 4
Step-by-step explanation:
The number of degrees of freedom represents the number of data-values in the evaluation of a test-statistic that are independent to vary.
The degree of freedom for chi-square test = n-1 , n= Sample size.
Given : A study was conducted to examine whether the proportion of females was the same for five groups (Groups A, B, C, D, and E).
i.e. n= 5
Then, the number of degrees of freedom the χ2 test statistic have when testing the hypothesis that the proportions in each group are all equal = 5-1=4
Hence, the correct answer is OPTION a. 4 .
What's the formula for the standard error of the difference between the estimates of the population proportions, used in a confidence interval for the difference between two proportions?
Answer:
[tex]s_{p_1-p_2}=\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2} }[/tex]
Step-by-step explanation:
The formula for the standard error of the difference between the estimates od the population proportions is:
[tex]s_{p_1-p_2}=\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2} }[/tex]
This is expected, as the variance of a sum (or a substraction) of two random variables is equal to the sum of the variance of the two variables.
Then, the standard error (or standard deviation) is the square root of this variance.
You want to determine whether the amount of coffee that "coffee drinkers" consume on a weekly basis differs depending on whether the coffee drinker also smokes cigarettes. You seek 20 volunteers from your college campus to participate. Ten college students reported that they consume coffee but do not smoke cigarettes; these comprise the Coffee-Only group, or Group 1. Another ten college students reported that they consumed coffee and also smoked cigarettes; these comprise the Coffee + Cigarettes group, or Group 2. Then you ask them to monitor the frequency of 8 ounce cups of coffee they consumed over a seven day period. Which of the following are the correct statements of the null and alternate hypotheses, H0 and HA?
H0: µ1 > µ2 and HA: µ1 ≤ µ2
H0: µ1 ≤ µ2 and HA: µ1 > µ2
H0: µ1 < µ2 and HA: µ1 ≥ µ2
H0: µ1 ≥ µ2 and HA: µ1 < µ2
H0: µ1 = µ2 and HA: µ1 ≠ µ2
H0: µ1 ≠ µ2 and HA: µ1 = µ2
Answer:
The correct option is e) [tex]H_0: \mu_1=\mu_2\ and\ H_a:\mu_1\neq \mu_2[/tex]
Step-by-step explanation:
Consider the provided information.
Ten college students reported that they consume coffee but do not smoke cigarettes; these comprise the Coffee-Only group, or Group 1. Another ten college students reported that they consumed coffee and also smoked cigarettes; these comprise the Coffee + Cigarettes group, or Group 2.
The null hypothesis tells the population parameter is equal to the claimed value.
If there is no statistical significance in the test then it is know as the null which is denoted by [tex]H_0[/tex], otherwise it is known as alternative hypothesis which denoted by [tex]H_a[/tex].
The amount of coffee that "coffee drinkers" consume on a weekly basis differs depending on whether the coffee drinker also smokes cigarettes.
Thus, the required hypothesis are: [tex]H_0: \mu_1=\mu_2\ and\ H_a:\mu_1\neq \mu_2[/tex]
Therefore, the correct option is e) [tex]H_0: \mu_1=\mu_2\ and\ H_a:\mu_1\neq \mu_2[/tex]