A random sample of 20 students yielded a mean of ¯x = 72 and a variance of s2 = 16 for scores on a college placement test in mathematics. Assuming the scores to be normally distributed, construct a 98% confidence interval for σ2.

On average, when placing a $1 bet on five numbers in roulette, you would expect to lose about 53 cents due to the game's probability structure.

In order to calculate the expected winnings in a roulette game, consider a $1 bet on five numbers. The roulette wheel has 38 possibilities (1 to 36, 0, and 00). So, the chance that your bet will win is 5 out of 38. If it wins, the game pays off 6 to 1, which means you get your $1 bet back and win $6 additional, for a total of $7. The expected value of this bet can be calculated as follows: (probability of winning * amount won if bet is successful) - (probability of losing * amount lost if bet is not successful). In other words, E(X) = [5/38 * $7] - [33/38 * $1]. This equates to an expected value of -$0.53 (rounded to the nearest cent). So, on average, with each $1 bet on five numbers, you would expect to lose about 53 cents.

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In the game of roulette with 38 outcomes, if you place a $1 bet on five numbers, your expected winnings are -$0.08, which means you lose 8 cents per game on average.

To calculate the expected winnings of a bet in this roulette scenario, you need to know the probability of winning and the associated payout, and also the probability of losing. In a roulette wheel with numbers 1-36 plus 0 and 00, there are a total of 38 possible outcomes. When you place a 5-number bet, the probability of winning is 5 out of 38 or approximately 0.1316, and the probability of losing is 33 out of 38 or approximately 0.8684.

Now we add the expected winnings and losses. The expected gain from a win is the probability of winning times the payout, which is $6 in this case. So, 0.1316 * $6 = $0.79, rounded to the nearest cent. The expected loss from a bet is the probability of loss times the amount of the bet which is $1. So, 0.8684 * -$1 = -$0.87, rounded to the nearest cent. Total expected value or earnings from a single $1 bet on five numbers would be the sum of expected gains and losses, or $0.79 - $0.87 = -$0.08.

So, with a $1 bet on five numbers on this roulette wheel, you can expect, on average, to lose 8 cents per game. This negative figure indicates that this is not a good bet if you're expecting to make money on average.

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Answer:

(1.4328, 1.622)

Claim is supported by evidence.

Step-by-step explanation:

Given that a  recent study was conducted to determine the curing efficiency (time to harden) of dental composites (resins for the restoration of damaged teeth) using two different types of lights.

Let X be the halogen and y the Luxomax light

[tex]H_0: \bar x =\bar y\\H_a: \bar x > \bar Y[/tex]

(Two tailed test)

we are given data as:

Group   Group One     Group Two  

Mean   5.3500            3.9000

SD             0.7000     0.8000

SEM       0.2214     0.2530

N                10                 10      

The mean of Group One minus Group Two equals 1.4500

Std error for difference =  0.336

Test statistic t=4.3135

 df = 18

p value = 0.0004

Since p <0.01 at 1% level, reject H0

There is significant difference and hence the claim is valid.

There  is evidence to support this claim at 1% significance level

Margin of error =1.172

99% confidence interval = [tex](1.45-1.172, 1.45+1.172)\\\\=(1.4328, 1.622)[/tex]

To determine if the Halogen light produces a larger cure depth, a two-sample t-test with equal variances is used. The test statistic is calculated and compared with the critical value at α = 0.01 to test the claim. Additionally, a 99% confidence interval for the difference in mean cure depths is constructed.

To assess whether the Halogen light produces a larger cure depth after 40 seconds than LuxOMax lights, we perform a two-sample t-test assuming equal variances. The hypotheses are:

Using the given data, we calculate the test statistic:

μ1 = 5.35, μ2 = 3.90, n1 = n2 = 10, s1 = 0.7, s2 = 0.8.

The pooled standard deviation (Sp) is computed as: Sp = √[((n1-1)s1² + (n2-1)s2²) / (n1+n2-2)] = √[(9×0.7² + 9×0.8²) / 18].

The test statistic t is calculated by: t = (x1 - x2) / (Sp×√(1/n1 + 1/n2)).

Using α = 0.01, we compare the calculated t value to the critical t value from a t-distribution table. If t > t_critical, we reject H0, providing evidence to support the Halogen light's claim.

Next, we construct a 99% confidence interval (CI) for the difference in population mean cure depths:

CI = (x1 - x2) ± (t_critical×Sp×√(1/n1 + 1/n2)).

This interval estimates the range within which the true difference in mean cure depths between the two lights lies with 99% confidence.

Answer:

Which is the output of the formula =AND(12>6;6>3;3>9)?

A.

TRUE

B.

FALSE

C.

12

D.

9

Step-by-step explanation:

Answer: t = 1.9287

Step-by-step explanation:

Let [tex]\mu[/tex] be the average number of pages should be sent by 2-day mail instead.

As per given we have,

[tex]H_0: \mu \leq7\\\\H_a:\mu>7[/tex]

Sample mean : [tex]\overline{x}=8.52[/tex]

Sample standard deviation : s=3.81

sample size : n= 24

Since , the sample size is less than 30 and populations standard deviation is unknown , so we use t-test.

The test statistic for population mean :-

[tex]t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}[/tex]

[tex]t=\dfrac{8.5-7}{\dfrac{3.81}{\sqrt{24}}}\\\\=\dfrac{1.5}{\dfrac{3.81}{4.8990}}\\\\=\dfrac{1.5}{0.777712993333}=1.92873208093\approx1.9287[/tex]

Hence, the test statistics : t = 1.9287

Answer:

B and D

Step-by-step explanation:

sin x cos x = 1/4

Multiply both sides by 2:

2 sin x cos x = 1/2

Use double angle formula:

sin(2x) = 1/2

Solve:

2x = π/6 + 2nπ or 2x = 5π/6 + 2nπ

x = π/12 + nπ or x = 5π/12 + nπ

Answer:

76

Step-by-step explanation:

Final answer:

The worst exam score when compared to the other students is the one with the lowest z-score. In this case, with a z-score of -0.8, the 78 in Math is the worst score.

Explanation:

To determine which exam score was the worst compared to the other students, we need to calculate the number of standard deviations the student's score falls from the class mean. This calculation is known as a z-score. A z-score allows us to understand the position of a score within the context of a distribution.

For the Math exam:
Z = (Your score - Class mean) / Standard deviation
= (78 - 82) / 5
= -0.8.

For the History exam:
Z = (85 - 87) / 4
= -0.5.

For the English exam:
Z = (80 - 76) / 2.5
= 1.6.

The score with the lowest z-score is the worst because it is the farthest below the mean. Comparing the z-scores, the Math exam has the lowest z-score at -0.8, so the 78 in Math is the worst score when compared to the other students in your classes.

Answer:

1 because almost certain event

Step-by-step explanation:

whenever a fair die is rolled the number of getting a 6 is having probability 1/6. Each throw is independent of the other

Hence no of sixes would be binomial with p=1/6

when 48000 dice are rolled, using binomial would be a hectic task.

Hence we approximate to normal distribution

X - no of sixes in 48000 throws would be normal

with mean = np = 8000

Var =npq = 6666.667

Std dev = 81.650

Now it is easier to find out

[tex]P(7500<x<8500)\\=P(7499.5<x<8499.5)[/tex](using continuity correctin)

=[tex]P(|z|<6.1295)\\=1[/tex]

we get this probability almost equal to 1

d) Normal distribution is a good choice because when no of trials increase using binomial and combination formulae would not be easy

Answer:

z=0.77

Step-by-step explanation:

If 22% of the area under the standard normal curve lies to the right of z, 78% lies to the left.

According to a one-sided z-score table:

If z-score = 0.77 the area under the curve is 77.94%

If z-score = 0.78 the area under the curve is 78.23%

Since only two decimal plates are required, there is no need to actually interpolate the values, just verify which one is closer to 78%

Since 77.94% is closer, a z-score of 0.77 represents a 22% area under the standard normal curve to the right of z.

To find z such that 22% of the area under the standard normal curve lies to the right of z, the value of z is approximately 0.81.

To find the value of z such that 22% of the area under the standard normal curve lies to the right of z, we can use a z-table or a calculator. First, we need to find the area under the curve to the left of z, which is 1 - 0.22 = 0.78. We can then look up the z-score that corresponds to this area in the table. The z-score is approximately 0.81. Therefore, the value of z is approximately 0.81.

Answer:

$302,500

Step-by-step explanation:

If cost (C) = $7, then Sales (S) = 37,500 units

If cost (C) = $15, then Sales (S) = 17,500 units

The slope of the linear relationship between units sold and cost is:

[tex]m=\frac{37,500-17,500}{7-15}\\m= -2,500[/tex]

The linear equation that describes this relationship is:

[tex]s-s_0 =m(c-c_0)\\s-17500 =-2500(c-15)\\s(c)=-2500c + 55,000[/tex]

The revenue function is given by:

[tex]R(c) = c*s(c)\\R(c)=-2500c^2 + 55,000c[/tex]

The cost at which the derivative of the revenue equals zero is the cost that yields the maximum revenue.

[tex]\frac{d(R(c))}{dc}=0 =-5000c + 55,000\\c=\$11[/tex]

The optimal cost is $11, therefore, the maximum revenue is:

[tex]R(11)=-2500*11^2 + 55,000*11\\R(11)=\$ 302,500[/tex]

Answer: The lower bound of confidence interval would be 0.116.

Step-by-step explanation:

Since we have given that

p = 13.2%= 0.132

n = 1105

At 90%  confidence,

z = 1.645

So, Margin of error would be

[tex]z\sqrt{\dfrac{p(1-p)}{n}}\\\\=1.645\times \sqrt{\dfrac{0.132\times 0.868}{1152}}}\\\\=0.0164[/tex]

So, the lower bound of the confidence interval would be

[tex]p-\text{margin of error}\\\\=0.132-0.0164\\\\=0.116[/tex]

Hence, the lower bound of confidence interval would be 0.116.

The confidence interval for percentage of U.S. adults who were victims of crime at the 90% confidence level is (0.114, 0.15). The lower bound of the confidence interval is 0.114.

To estimate the percentage of U.S. adults who were victims of crime at the 90% confidence level, we can use the formula for a confidence interval:

[tex]CI = p +/- Z * \sqrt{((p * (1-p)) / n)[/tex]

where CI is the confidence interval, p is the sample proportion, Z is the Z-score corresponding to the desired confidence level, and n is the sample size. In this case, p = 0.132, Z = 1.645 (corresponding to a 90% confidence level), and n = 1105. Plugging in these values, we can calculate the confidence interval as:

[tex]CI = 0.132 +/- 1.645 * \sqrt((0.132 * (1-0.132)) / 1105)[/tex]

Simplifying the expression gives us:

CI = 0.132 ± 0.018

Therefore, the confidence interval for the percentage of U.S. adults who were victims of crime at the 90% confidence level is (0.114, 0.15). The lower bound of the confidence interval is 0.114.

Answer:

a) Around 2,052.583 Km.

b) Around 22 days.

c) Around 3.887 Km/h

Step-by-step explanation:

a)

In order to find and approximation of the distance traveled, we have to make some assumptions:

If these assumptions hold, then the distance d in Km can be calculated by using the haversine formula:

[tex]\large d=2R*arcsin\left(\sqrt{sin^2((\varphi_2-\varphi_1)/2)+cos(\varphi_1)cos(\varphi_2)sin^2((\lambda_2-\lambda_1)/2)}\right)[/tex]

where  

[tex]\large (\varphi_1, \lambda_1)[/tex] are the latitude and longitude of the starting point in radians.

[tex]\large (\varphi_2, \lambda_2)[/tex] are the latitude and longitude of the end point in radians.

R = radius of  Earth in kilometers.

Be careful to convert the angles into radians before computing the trigonometric functions. This can be done by cross-multiplication knowing that 180° ≅ 3.141592654 radians.

In the problem we have

[tex]\large \varphi_1[/tex] = 48° = 0.837758041 radians

[tex]\large \lambda_1[/tex] = 161° = 2.809980096 radians

[tex]\large \varphi_2[/tex] = 54° = 0.942477796 radians

[tex]\large \lambda_2[/tex] = 133° = 2.321287905 radians

so

[tex]\large (\varphi_2-\varphi_1)/2[/tex] = 3° = 0.052359878 radians

[tex]\large (\lambda_2-\lambda_1)/2[/tex] = -14° = -0.2443461 radians

Replacing in the formula for the distance:

[tex]\large d=2R*arcsin\left(\sqrt{sin^2(0.052359878)+cos(0.837758041)cos(0.942477796)sin^2(-0.2443461)}\right)=\\\\=0.32237835*R[/tex]

According to NASA, the radius of  Earth at the poles is around 6,356 Km and at the equator is 6,378 Km.

Since they traveled around the middle point between the equator and the North pole, a better estimate of the radius in this case would be the average (6,378+6,356)/2 = 6,367 Km

We have that an approximation to the distance traveled would be  

0.32237835*6,367 = 2,052.583 Km

b)

Assuming that the shoes where left and found the same year, there are 22 days from April 30th  to May 22nd , so they traveled for around 22 days (this may vary slightly depending on the exact time the shoes were left and found)

c)

They traveled 2,052.583 Km in 22 days.

22 days equals 22*24 = 528 hours.

By cross-multiplication

528 h _______  2,052.583 Km

1 h __________ x Km

x =  2,052.583/528 ≅ 3.887 Km

So they traveled at a rate of 3.887 Km/h

Answer:

(B) ​If the sampling procedure were repeated many times, 95% of the resulting confidence intervals would contain the population mean systolic blood pressure.

Step-by-step explanation:

A confidence interval of 95% means that there is a 95% certainty that for a given sample, the population mean will be within the confidence interval estimated.

This is the same as saying that if he sampling procedure were repeated many times, 95% of the time the population mean would be contained in the resulting confidence interval.

Therefore, the answer is B)

The correct statement is (B) If the sampling procedure were repeated many times, 95% of the resulting confidence intervals would contain the population mean systolic blood pressure.

A 95% confidence interval is a range of values that is likely to contain the population mean. This does not mean that there is a 95% probability that the population mean falls within the interval calculated from a single sample. Instead, it means that if we were to take many samples and calculate a confidence interval from each sample, approximately 95% of those intervals would contain the true population mean.

Let's evaluate the other options:

(A) 95% of the sample of employees has a systolic blood pressure between 123 and 139. - This statement is incorrect because the confidence interval refers to the population mean, not the distribution of individual sample values. It does not imply that 95% of the sample has values within the interval.

(C) If the sampling procedure were repeated many times, 95% of the sample means would be between 123 and 139. - This statement is incorrect because it misinterprets the confidence interval. The confidence interval gives us a range where we expect the population mean to lie, not the sample means. While the sample means will vary, the confidence interval tells us about the variability of the estimate of the population mean, not the variability of the sample means themselves.

(D) 95% of the population of employees has a systolic blood pressure between 123 and 139. - This statement is incorrect because it generalizes the confidence interval to the entire population. The confidence interval does not tell us what proportion of the population falls within the interval; it only provides a range that is likely to contain the population mean.

Therefore, the correct interpretation of the 95% confidence interval is that if we were to take many samples from the population and calculate a 95% confidence interval for each, approximately 95% of those intervals would contain the true population mean systolic blood pressure. This is why option (B) is the valid statement.

Answer:

Step-by-step explanation:

To find median and mode for

a) In a uniform distribution median would be

(a+b)/2 and mode = any value

b) X is N

we know that in a normal bell shaped curve, mean = median = mode

Hence mode = median = [tex]\mu[/tex]

c) Exponential with parameter lambda

Median = [tex]\frac{ln2}{\lambda }[/tex]

Mode =0

The median of a distribution is the middle value while the mode is the highest occuring value

The median (M) of a uniform distribution is:

[tex]M = \frac{a +b}2[/tex]

A uniform distribution has no mode

For a normal distribution with the given parameters, we have:

Median = Mean = Mode = μ

Hence, the median and the mode are μ

For an exponential distribution with the given parameter, we have:

[tex]Median = \frac{\ln 2}{\lambda}[/tex]

The mode of an exponential distribution is 0

Read more about median and mode at:

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Answer:

x=-3+4t\\

y =5-7t

Step-by-step explanation:

we are given that when two point (x1,y1) and (x2,y2) are joined the parametric equation representing the line segment would be

[tex]x = x_1 + (x_2 - x_1)t y = y_1 + (y_2 - y_1)t, 0 \leq ≤t \leq ≤ 1[/tex]

Our points given are

[tex](-3, 5) to (1, -2)[/tex]

So substitute the values to get

[tex]x=-3+(1+3)t =-3+4t\\y = 5+(-2-3)t =5-7t[/tex]

Hence parametric equations for the line segment joining the points

(-3,5) and (1,-2) is

[tex]x=-3+4t\\y =5-7t[/tex]

Final answer:

The parametric equations for the line segment from (-3, 5) to (1, -2) are x = -3 + 4t and y = 5 - 7t for 0 ≤ t ≤ 1.

Explanation:

The question asks for parametric equations to describe the line segment joining the points P1(-3, 5) and P2(1, -2). According to the formula provided, we have:

x = x1 + (x2 – x1)t

y = y1 + (y2 – y1)t

Substituting the values of P1 and P2 into the equations:

x = -3 + (1 – (-3))t = -3 + 4t

y = 5 + (-2 – 5)t = 5 - 7t

This means for 0 ≤ t ≤ 1, the parametric equations representing the line segment from (-3, 5) to (1, -2) are:

x = -3 + 4t

y = 5 - 7t

The probability a student gets a Bachelor of Arts and Science degree is 0.18, the probability a student gets a Bachelor of Arts degree or a Bachelor of Science degree is 0.9, and the probability a student gets only a Bachelor of Arts degree is 0.5.

The probability a student gets a Bachelor of Arts and Science degree can be found by multiplying the probabilities of getting a Bachelor of Arts degree and a Bachelor of Science degree.

So, P(Arts and Science) = P(Arts) * P(Science) = 0.6 * 0.3 = 0.18. Therefore, the probability a student gets a Bachelor of Arts and Science degree is 0.18, which is not listed as an option in the given choices.

The probability a student gets a Bachelor of Arts degree or a Bachelor of Science degree can be found by adding the probabilities of getting each degree. So, P(Arts or Science) = P(Arts) + P(Science) = 0.6 + 0.3 = 0.9.

Therefore, the probability a student gets a Bachelor of Arts degree or a Bachelor of Science degree is 0.9, which is listed as option a) 0.9.

The probability a student gets only a Bachelor of Arts degree can be found by subtracting the probability of getting a Bachelor of Science degree and the probability of getting no degree from 1.

So, P(Only Arts) = 1 - P(Science) - P(no) = 1 - 0.3 - 0.2 = 0.5. Therefore, the probability a student gets only a Bachelor of Arts degree is 0.5, which is listed as option b) 0.5.

Answer:

The answer is $64.25 or you could say its B.

Hope this helps :)

Answer:

a sample of 29 network will have a voltage of 232V

Step-by-step explanation:

N = 66

variate = 232 V.

mean voltage = 231.7 V

standard deviation = 2.19 V.

The z-value given by = (variate -mean)/ standard deviation

= (232-231.7)/2.19 =

z = 0.3/2.19 = 0.137

From of normal distribution table, the z of 0.137 found between the area of z=0 and z=0.5 is given as 0.0557

Thus the area to the right of the z=0.0557 ordinate is 0.5000−0.0557=0.4443 = 44.43%

thus, this is the probability of Network voltage = 44.43%.

for sample of 66 network, it is likely that 66×0.4443 = 29.32, i.e.

a sample of 29 network will have a voltage of 232V

Answer:

Step-by-step explanation:

She uses 1/4 ounces of green paint to draw to draw a green line on her painting. This means if she draws 2 green lines on her painting, she will require ( 1/4 + 1/4 ) ounces of green paint. This means that if she draws a total of 7 green lines, the correct options would be the options that apply and they are just two viz:

1/4+1/4+1/4+1/4+1/4+1/4+1/4 or

7/4 because they are both the same. Adding each term in the longer oftion gives 7/4

Answer:

c and d

Step-by-step explanation:

Answer with Step-by-step explanation:

We are given that a demand function

[tex]x=D(p)=3100-20p[/tex]

Where x=The quantity of sodas sold

p=Per can price (in cents)

We  have to find the price p for which  the demand inelastic.

Differentiate the demand function w.r.t p

[tex]D'(P)=-20[/tex]

Price elasticity of demand=[tex]E(p)=\frac{-pD'(p)}{D(p)}[/tex]

Price elasticity of demand=[tex]E(p)=\frac{-p(-20)}{3100-20p}[/tex]

When demand  is inelastic then

E(p)<1

[tex]\frac{20p}{3100-20p}[/tex] <1

Multiply by (3100-20p) on both sides

[tex]20p<3100-20p[/tex]

Adding 20p on both side of inequality

[tex]40p<3100[/tex]

Divide by 40 on both sides

[tex]p<77.5[/tex]

When the value of price is less than 77.5 then the demand of elasticity is inelastic.

The demand function of a beverage company is inelastic when the percentage change in the quantity of demanded soda is less than 1% for a 1% change in price. If prices between points A and B change by 1%, and the quantity demanded changes by only 0.45% (less than 1%), the demand is inelastic.

In the given demand function, x = D(p) = 3100 - 20p, the elasticity of demand is considered inelastic if the absolute value of elasticity is less than 1. Elasticity of demand reflects how sensitive the demand for a good is to changes in its price. With a demand curve, we'd consider whether a 1% change in price changes the quantity of the good demanded.

Should the quantity demanded change by less than 1% for a 1% price change (for example, a 0.45% change), we have inelastic demand. This means, if prices between points A and B alter by 1%, the change in the quantity demanded will be only 0.45%. This percentage indicates an inelastic demand because the price change will result in a comparatively smaller percentage change in quantity demanded. Price elasticities of demand are usually negative, showcasing the negative relationship between price and demanded quantity (law of demand). However, while interpreting, the focus lays on the absolute value.

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Answer:

Null hypothesis:[tex]\mu \geq 20[/tex]        

Alternative hypothesis:[tex]\mu < 20[/tex]

[tex]P(t_{68}<-1.06)=0.146[/tex]

If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL reject the null hypothesis, and the the actual sample average lateral recumbency is not significantly lower than 20.        

Step-by-step explanation:

1) Data given and notation        

[tex]\bar X=18.94[/tex] represent the mean for the sample    

[tex]s=8.3[/tex] represent the standard deviation for the sample        

[tex]n=69[/tex] sample size        

[tex]\mu_o =20[/tex] represent the value that we want to test      

[tex]\alpha[/tex] represent the significance level for the hypothesis test.      

t would represent the statistic (variable of interest)        

[tex]p_v[/tex] represent the p value for the test (variable of interest)    

2) State the null and alternative hypotheses.        

We need to conduct a hypothesis in order to determine if the true average lateral recumbency time under these conditions is less than 20 min:      

Null hypothesis:[tex]\mu \geq 20[/tex]        

Alternative hypothesis:[tex]\mu < 20[/tex]        

We don't know the population deviation, so for this case 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{\sigma}{\sqrt{n}}}[/tex] (1)        

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

3) Calculate the statistic        

We can replace in formula (1) the info given like this:        

[tex]t=\frac{18.94-20}{\frac{8.3}{\sqrt{69}}}=-1.06[/tex]        

4) Calculate the P-value        

First we need to calculate the degrees of freedom

[tex]df=n-1=69-1=68[/tex]

The critical value for this case would be :

[tex]P(t_{68}<-1.06)=0.146[/tex]

5) Conclusion      

If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL reject the null hypothesis, and the the actual sample average lateral recumbency is not significantly lower than 20.        

Answer:

We use z-test for this hypothesis.

[tex]z_{stat} = 2.23[/tex]

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 4.88

Sample mean, [tex]\bar{x}[/tex] = 5.91

Sample size, n = 48

Alpha, α = 0.05

Population standard deviation, σ = 3.2

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 4.88\\H_A: \mu > 4.88[/tex]

The null hypothesis states that the mean score of successful managers on a psychological test is 4.88 and the alternate hypothesis says that the mean score of successful managers on a psychological test is greater than 4.88.

We use One-tailed 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{5.91 - 4.88}{\frac{3.2}{\sqrt{48}} } = 2.23[/tex]

Answer:

5

Step-by-step explanation:

27-22= 22-17 = 17-12 =12-7=d

d= 5

The common difference of the arithmetic sequence 7, 12, 17, 22, 27,... is 5, which is the constant difference between consecutive terms of the sequence.

The sequence given is an arithmetic sequence, which is a sequence of numbers where the difference between consecutive terms is constant. To find the common difference in the arithmetic sequence 7, 12, 17, 22, 27,..., we subtract any term from the preceding term in the sequence. Let's perform this calculation using the first two terms:

Thus, the common difference is 5. We can confirm this by checking the difference between other consecutive terms:

All differences are the same, confirming that the common difference of the sequence is indeed 5.

Answer:

A"fishbone" diagram

Step-by-step explanation:

A"fishbone" diagram, may help to determine potential causes of an issue and organise concepts into valuable groups of brain storming. A depiction of the fishbone is a graphical way of looking at causes and effects.

It's a more organised solution than other brain storming tools that cause problems. The issue or effect is shown on the fish's head or mouth.

Answer:

[tex]P(\bar X>308)=1-0.0721=0.928[/tex]

Step-by-step explanation:

1) Previous concepts

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".  

Let X the random variable that represent the GRE score at the University of Pennsylvania for the incoming class of 2016-2017, and for this case we know the distribution for X is given by:

[tex]X \sim N(\mu=311,\sigma=13)[/tex]  

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(311,\frac{13}{\sqrt{40}})[/tex]

2) Calculate the probability

We want this probability:

[tex]P(\bar X>308)=1-P(\bar X<308)[/tex]

The best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

If we apply this formula to our probability we got this:

[tex]P(\bar X >308)=1-P(Z<\frac{308-311}{\frac{13}{\sqrt{40}}})=1-P(Z<-1.46)[/tex]

[tex]P(\bar X>308)=1-0.0721=0.9279[/tex]  and rounded would be 0.928

This answer includes instructions for approximating a function using a third degree Taylor polynomial at a specific point, using Taylor's Inequality to estimate the accuracy of the approximation, and checking the results by graphing the function |Rn(x)|.

For part (a), to approximate the function f(x) = ln(1 + 2x) using a Taylor polynomial of degree 3 at a = 4, we first need to find the first 4 derivatives of the function evaluated at a=4. Substituting the values will give the Taylor polynomial - T3(x).

For part (b), Taylor's Inequality states that the remainder term Rn(x) is less than or equal to M|X-a|^(n+1) / (n+1)! where M is the max value of the |f^(n+1)(t)| where t lies in the interval between a and x. By substituting the values we can find |R3(x)|.

For the last part (c), by using graphing software, you can graph the function |Rn(x)| in the given interval to visually confirm your earlier result. These graphs will show the magnitude of the error.|

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Answer:

We failed to accept null hypothesis

Step-by-step explanation:

x= 5.74

Standard deviation = [tex]0.26[/tex]

[tex]H_0:\mu = 5.8\\H_a:\mu < 5.8[/tex]

n = 65

We will use z test

Formula : [tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

Substitute the values :

[tex]z=\frac{5.74-5.8}{\frac{0.26}{\sqrt{65}}}[/tex]

[tex]z=-1.86[/tex]

Refer the z table for p value

p value = 0.0314

α = 0.10

p value < α

So, we failed to accept null hypothesis

A function is a relation that has one output for a given input.

For the first one, there is no one x value with two or more y values so it is a function.

The second example is also a function because a certain value can only have one cube root.

For problem number 3 input "-3" for every instance of x in h(x).

So, h(-3)=2(3^2)-1= 17

To calculate the test statistic for a hypothesis test on population proportion, you need the size of the sample, sample proportion, level of significance, and proposed population proportion.

In order to calculate the test statistic for a hypothesis test on the proportion of people with a characteristic of interest, the following pieces of information are needed:

These pieces of information are necessary to calculate the test statistic, which is used to determine whether the sample data provides enough evidence to support or reject the hypothesis about the population proportion.

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The information needed to calculate the test statistic for a hypothesis test of a population proportion includes the size of the sample selected, the sample proportion calculated, and the proposed population proportion.

To calculate the test statistic for a hypothesis test about a population proportion, several pieces of information are required:

Note that while C. the level of significance used and F. the actual population proportion are important for different hypothesis testing steps such as defining the null and alternative hypotheses, calculating the p-value, or making the final decision for the hypothesis test, they do not directly factor into the calculation of the test statistic. Also, E. the characteristic of interest is not required to calculate the test statistic itself, but it is necessary to determine which is the appropriate test to apply and to structure the hypotheses correctly.

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The value of the test statistic is 1.583.

To find the value of the test statistic, we need to calculate the z-score using the formula:

z = (x - μ) / (σ / √n)

where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.

In this case, the sample mean is 6.9 ppm, the population mean is 6.5 ppm, the population variance is 1.21, and the sample size is 27. Calculate the test statistic:

z = (6.9 - 6.5) / (√(1.21 / 27)) = 0.4 / 0.252951 = 1.583

Rounding to three decimal places, the value of the test statistic is 1.583.

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The setup for the surface integral is achieved by using the parameterized surface and the given function along with the norm of the cross product of partial derivatives of the parameterized surface. The integral is evaluated to represent the net flux through the surface that denotes the volume enclosed by the parameterized surface.

The surface integral of a given function can be calculated using the parameterization of the surface and the given function. In this case, our surface is parameterized by r(u,v)=⟨ucos(3v),usin(3v),v⟩ and our function is f(x,y,z)=x²+y²+z². We first need to calculate the partial derivatives ∂r/∂u and ∂r/∂v, and then cross them to find the norm of the cross product, which gives us the differential area element for the parameterized surface. With these, we can set up the surface integral as follows: ∫σ∫f(x(u,v),y(u,v),z(u,v))∥∥∥∂r∂u×∂r∂v∥∥∥dA = ∫0 to 2π/3 ∫0 to 7 [u²cos²(3v)+u²sin²(3v)+v²]∥∂r/∂u×∂r/∂v∥ du dv.

Please note that the evaluated result of the integral provides the net flux through the rectangular surface which represents the volume enclosed by the parameterized surface.

The above set up integral needs to be solved to find the exact value which is not required in this scenario. However, the integral can be solved using suitable integral calculus methods if needed.

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