A new shopping mall is considering setting up an information desk manned by one employee. Based upon information obtained from similar information desks, it is believed that people will arrive at the desk at a rate of 20 per hour. It takes an average of 2 minutes to answer a question. It is assumed that the arrivals follow a Poisson distribution and answer times are exponentially distributed. (a) find the probability that the employee is idle. (b) Find the proportion of the time that the employee is busy. (c) Find the average number of people receiving and waiting to receive some information. (d) Find the average number of people waiting in line to get some information. (e) Find the average time a person seeking information spends in the system. (f) Find the expected time a person spends just waiting in line to have a question answered (time in the queue).

Answer:

1. Null hypothesis: [tex]\mu_{A}=\mu_{B}=\mu_{C}[/tex]

Alternative hypothesis: Not all the means are equal [tex]\mu_{i}\neq \mu_{j}, i,j=A,B,C[/tex]

2. D. 36

3. C. 34

4. B. 1.059

5. B. 8.02

Step-by-step explanation:

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"

Part 1

The hypothesis for this case are:

Null hypothesis: [tex]\mu_{A}=\mu_{B}=\mu_{C}[/tex]

Alternative hypothesis: Not all the means are equal [tex]\mu_{i}\neq \mu_{j}, i,j=A,B,C[/tex]

Part 2

In order to find the mean square between treatments (MSTR), we need to find first the sum of squares and the degrees of freedom.

If we assume that we have [tex]p[/tex] groups and on each group from [tex]j=1,\dots,p[/tex] we have [tex]n_j[/tex] individuals on each group we can define the following formulas of variation:  

[tex]SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2 [/tex]

[tex]SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 [/tex]

[tex]SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 [/tex]

And we have this property

[tex]SST=SS_{between}+SS_{within}[/tex]

We need to find the mean for each group first and the grand mean.

[tex]\bar X =\frac{\sum_{i=1}^n x_i}{n}[/tex]

If we apply the before formula we can find the mean for each group

[tex]\bar X_A = 27[/tex], [tex]\bar X_B = 24[/tex], [tex]\bar X_C = 30[/tex]. And the grand mean [tex]\bar X = 27[/tex]

Now we can find the sum of squares between:

[tex]SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 [/tex]

Each group have a sample size of 4 so then [tex]n_j =4[/tex]

[tex]SS_{between}=SS_{model}=4(27-27)^2 +4(24-27)^2 +4(30-27)^2=72 [/tex]

The degrees of freedom for the variation Between is given by [tex]df_{between}=k-1=3-1=2[/tex], Where  k the number of groups k=3.

Now we can find the mean square between treatments (MSTR) we just need to use this formula:

[tex]MSTR=\frac{SS_{between}}{k-1}=\frac{72}{2}=36[/tex]

D. 36

Part 3

For the mean square within treatments value first we need to find the sum of squares within and the degrees of freedom.

[tex]SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 [/tex]

[tex]SS_{error}=(20-27)^2 +(30-27)^2 +(25-27)^2 +(33-27)^2 +(22-24)^2 +(26-24)^2 +(20-24)^2 +(28-24)^2 +(40-30)^2 +(30-30)^2 +(28-30)^2 +(22-30)^2 =306[/tex]

And the degrees of freedom are given by:

[tex]df_{within}=N-k =3*4 -3 = 12-3=9[/tex]. N represent the total number of individuals we have 3 groups each one with a size of 4 individuals. And k the number of groups k=3.

And now we can find the mean square within treatments:

[tex]MSE=\frac{SS_{within}}{N-k}=\frac{306}{9}=34[/tex]

C. 34

Part 4

The test statistic F is given by this formula:

[tex]F=\frac{MSTR}{MSE}=\frac{36}{34}=1.059[/tex]

B. 1.059

Part 5

The critical value is from a F distribution with degrees of freedom in the numerator of 2 and on the denominator of 9 such that we have 0.01 of the area in the distribution on the right.

And we can use excel to find this critical value with this function:

"=F.INV(1-0.01,2,9)"

And we will see that the critical value is [tex]F_{crit}=8.02[/tex]

B. 8.02

Answer:

2, 0.135, 0.270, 0.270, 0.180, 0.145

Step-by-step explanation:

1. To calculate the mean of people arriving in 5 minute period:

We know the arrival rate of minute is 0.4, so people arriving in 5 minutes will be

0.4 x 5 = 2

2. From part 1 it is known thay mean arrival time=2

For this we will use poisson's probability formula that is

P(X=x) = (2^x) x Exp^(-2/x!)

For X=0

P(X=0) = (2^0) x Exp^(-2/0!) = 0.135

For X = 1

P(X=1) = (2^1) x Exp^(-2/1!) = 0.270

For X = 2

P(X=2) = (2^2) x Exp^(-2/2!) = 0.270

For X = 3

P(X=3) = (2^3) x Exp^(-2/3!) = 0.180

3. For delay expected if more than 3 customer arrive in 5 minutes.

P(X>3) = 1 - P(X=0) - P(X=1) - P(X=2) - P(X=3)

P(X>3) = 1-0.135-0.270-0.270-0.180

P(X>3) = 0.145

Answer:

option 0.8%

Step-by-step explanation:

Data provided in the question:

Mean = 5.7 years

Standard deviation, s = 1.8 years

Now,

P(the employee has worked at the store for over 10 years)

= P(X > 10 years)

= [tex]P (Z > \frac{X-Mean}{\sigma})[/tex]

or

= [tex]P (Z > \frac{10-5.7}{1.8})[/tex]

= P (Z > 2.389 )

or

= 0.008447     [from standard  z table]

or

= 0.008447 × 100% = 0.84% ≈ 0.8%

Hence,

the correct answer is option 0.8%

To find the probability that an employee has worked at a large department store for over 10 years, we can use the z-score formula and a standard normal distribution table or calculator.

To find the probability that an employee has worked at the store for over 10 years, we can use the z-score formula:

z = (x - μ) / σ

where x is the value we are interested in (10 years), μ is the mean (5.7 years), and σ is the standard deviation (1.8 years).

Plugging in the values, z = (10 - 5.7) / 1.8 = 2.39

Using a standard normal distribution table or a calculator, we can find that the probability of a z-score greater than 2.39 is approximately 0.008, or 0.8%.

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Final answer:

The statement regarding the determinant of a square matrix A being the product of its diagonal entries is incorrect. The determinant of a 2×2 matrix is calculated as the product of diagonal entries minus the product of off-diagonal entries.

Explanation:

The determinant of a square matrix A is not always the product of the diagonal entries in A. For instance, a 2×2 matrix has the determinant given by det(A) = a11a22 - a12a21, which involves the product of the diagonal entries minus the product of the off-diagonal entries. Therefore, the correct choice is:

B. The statement is false because the determinant of the 2×2 matrix A = [[a, b], [c, d]] is not equal to the product of the entries on the main diagonal of A. In fact, the determinant is ad - bc.

Example: For matrix A = [[1,2], [3,4]], the determinant is 1×4 - 2×3, which equals -2, not the product of the diagonal entries (1 and 4), which would be 4.

Answer:

22m

Step-by-step explanation:

Let height of flagpole=h

AB==17 m

[tex]\angle CAD=37^{\circ}11'=37+\frac{11}{60}=37.183^{\circ}[/tex](1 degree= 60 minute)

[tex]\angle B=25^{\circ}43'=25+\frac{43}{60}=25.72^{\circ}[/tex]

We have to find the approximate height of the flagpole.

In triangle CDA,

[tex]\frac{CD}{DA}=tan\theta=\frac{Perpendicular\;side}{Base}[/tex]

[tex]\frac{h}{DA}=tan37.183^{\circ}[/tex]

[tex]h=DA(0.759)[/tex]

In triangle CDB,

[tex]tan 25.72^{\circ}=\frac{CD}{DB}[/tex]

[tex]0.482=\frac{h}{DA+17}[/tex]

[tex]0.482DA+8.194=h[/tex]

Substitute the value

[tex]0.482DA+8.194=0.759DA[/tex]

[tex]8.194=0.759DA-0.482DA[/tex]

[tex]8.194=0.277DA[/tex]

[tex]DA=\frac{8.194}{0.277}=29.58[/tex]

Substitute the value

[tex]h=29.58 \times 0.759=22.45 m\approx 22m[/tex]

Hence, the height of the flagpole=22 m

The approximate height of the flagpole is 21.33 meters, calculated using trigonometry and the observed angles and distances.

To find the height of the flagpole, we can use trigonometry, specifically the tangent function, along with the given angles and the distance from the observer's initial position to the flagpole.

Step 1:

Calculate the tangent of the observed angles:

Let (h) be the height of the flagpole.

For the first observation:

[tex]\[ \tan(37^\circ 11') = \frac{h}{d} \][/tex]

For the second observation:

[tex]\[ \tan(25^\circ 43') = \frac{h}{d + 17} \][/tex]

Step 2:

Solve for (h):

First, convert the angles to decimal degrees:

[tex]\[ 37^\circ 11' = 37 + \frac{11}{60} \approx 37.18^\circ \][/tex]

[tex]\[ 25^\circ 43' = 25 + \frac{43}{60} \approx 25.72^\circ \][/tex]

Now, substitute the angles into the tangent equations:

[tex]\[ \tan(37.18^\circ) = \frac{h}{d} \][/tex]

[tex]\[ \tan(25.72^\circ) = \frac{h}{d + 17} \][/tex]

Step 3:

Solve the equations for (h):

[tex]\[ h = d \times \tan(37.18^\circ) \][/tex]

[tex]\[ h = (d + 17) \times \tan(25.72^\circ) \][/tex]

Step 4:

Set the expressions equal to each other:

[tex]\[ d \times \tan(37.18^\circ) = (d + 17) \times \tan(25.72^\circ) \][/tex]

Step 5:

Solve for (d):

[tex]\[ d = \frac{17 \times \tan(25.72^\circ)}{\tan(37.18^\circ) - \tan(25.72^\circ)} \][/tex]

Step 6:

Calculate the value of (d) and then use it to find (h):

[tex]\[ d \approx 27.83 \text{ meters} \][/tex]

[tex]\[ h = 27.83 \times \tan(37.18^\circ) \][/tex]

Step 7:

Calculate (h):

[tex]\[ h \approx 27.83 \times 0.7656 \][/tex]

[tex]\[ h \approx 21.33 \text{ meters} \][/tex]

Therefore, the approximate height of the flagpole is 21.33 meters.

Answer:

a) NORM.S.INV(0.975)

Step-by-step explanation:

1) Some definitions

The standard normal distribution is a particular case of the normal distribution. The parameters for this distribution are: the mean is zero and the standard deviation of one. The random variable for this distribution is called Z score or Z value.

NORM.S.INV Excel function "is used to find out or to calculate the inverse normal cumulative distribution for a given probability value"

The function returns the inverse of the standard normal cumulative distribution(a z value). Since uses the normal standard distribution by default the mean is zero and the standard deviation is one.

2) Solution for the problem

Based on this definition and analyzing the question :"Which of the following functions computes a value such that 2.5% of the area under the standard normal distribution lies in the upper tail defined by this value?".

We are looking for a Z value that accumulates 0.975 or 0.975% of the area on the left and by properties since the total area below the curve of any probability distribution is 1, then the area to the right of this value would be 0.025 or 2.5%.

So for this case the correct function to use is: NORM.S.INV(0.975)

And the result after use this function is 1.96. And we can check the answer if we look the picture attached.

Answer:

The probability that the mean life a random sample of 25 such blenders falls between 4.6 and 5.1 years is 0.6687.

Step-by-step explanation:

We have given :

The average life a manufacturer's blender is 5 years i.e. [tex]\mu=5[/tex]

The standard deviation is [tex]\sigma=1[/tex]

Number of sample n=25.

To find : The probability that the mean life falls between 4.6 and 5.1 years ?

Solution :

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

The probability that the mean life falls between 4.6 and 5.1 years is given by, [tex]P(4.6<X<5.1)[/tex]

[tex]P(4.6<X<5.1)=P(\frac{4.6-5}{\frac{1}{\sqrt{25}}}<Z<\frac{5.1-5}{\frac{1}{\sqrt{25}}})[/tex]

[tex]P(4.6<X<5.1)=P(\frac{-0.4}{\frac{1}{5}}<Z<\frac{0.1}{\frac{1}{5}})[/tex]

[tex]P(4.6<X<5.1)=P(-2<Z<0.5)[/tex]

[tex]P(4.6<X<5.1)=P(Z<0.5)-P(Z<-2)[/tex]

Using z-table,

[tex]P(4.6<X<5.1)=0.6915-0.0228[/tex]

[tex]P(4.6<X<5.1)=0.6687[/tex]

Therefore, the probability that the mean life a random sample of 25 such blenders falls between 4.6 and 5.1 years is 0.6687.

The probability that the mean life of a random sample of 25 blenders falls between 4.6 and 5.1 years is approximately 0.3585, or 35.85%.

To find the probability that the mean life of a random sample of 25 blenders falls between 4.6 and 5.1 years, we can use the standard normal distribution. First, we need to standardize the values using the z-score formula. The z-score for 4.6 years is (4.6 - 5) / 1 = -0.4, and the z-score for 5.1 years is (5.1 - 5) / 1 = 0.1. We can then look up the corresponding probabilities in the standard normal distribution table or use a calculator to find the area under the curve between these two z-scores. The probability that the mean life falls in this range is approximately 0.3585, or 35.85%.

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Answer:

69.5%

Step-by-step explanation:

A feature of the normal distribution is that this is completely determined by its mean and standard deviation, therefore, if two normal curves have the same mean and standard deviation we can be sure that they are the same normal curve. Then, the probability of getting a value of the normally distributed variable between 6 and 8 is 0.695. In practice we can say that if we get a large sample of observations of the variable, then, the percentage of all possible observations of the variable that lie between 6 and 8 is 100(0.695)% = 69.5%.

Answer: option A is correct

Step-by-step explanation:

the prediction interval is to cover a “moving target”, the random future value of y,

while the confidence interval is to cover the “fixed target”,

the average (expected) value of y, E(y).

Answer:

Step-by-step explanation:

Answer:

He can make the same decision at a = 0.10, but he may not make the same decision at a = 0.01

Step-by-step explanation:

In hypothesis tests, critical regions are ranges of the distributions where the values represent statistically significant results. Analysts define the size and location of the critical regions by specifying both the significance level (alpha) and whether the test is one-tailed or two-tailed.

As the significance level gets bigger, the range of the critical region increases.

Therefore significant results at lower significance levels are still significant at higher significance levels. Thus significant result at a=0.05 is always significant at a=0.10

But  significant result at a=0.05 may not be significant at a=0.01 since critical region shrinks, therefore the result may not fall in the critical region at a=0.01

The margin of error for a 90% confidence level is 2.4%, and for a 99% confidence level, it is 3.8%

The margin of error (ME) for a poll can be calculated using the formula:

ME = Z × sqrt((p ×(1 - p)) / n)

Where:

Question 1: For a 90% confidence level, the z-value (Z) is 1.645 (use a z-table or statistical software to obtain this).

Substitute the values in the formula to find the ME: ME = 1.645 × sqrt((0.64 × (1 - 0.64)) / 1076) = 0.024 or 2.4%

Question 2: For a 99% confidence level, the z-value (Z) is 2.576 (use a z-table or statistical software to obtain this).

Substitute the values in the formula to find the ME: ME = 2.576 × sqrt((0.64 × (1 - 0.64)) / 1076) = 0.038 or 3.8%

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Answer:

Step-by-step explanation:

1) The z value was determined using a normal distribution table. From the normal distribution table, the corresponding z value for a 94% confidence interval is 1.88

The correct option is D

2) if 70% of all internet users experience e-mail fraud. It means that probability of success, p

p = 70/100 = 0.7

q = 1 - p = 1 - 0.7 = 0.3

n = number of selected users = 50

Mean, u = np = 50×0.7 = 35

Standard deviation, u = √npq = √50×0.7×0.3 = 3.24

x = number of internet users

The formula for normal distribution is expressed as

z = (x - u)/s

We want to determine the probability that no more than 25 were victims of e-mail fraud. It is expressed as

P(x lesser than or equal to 25)

The z value will be

z = (25- 35)/3.24 = - 10/3.24 = -3.09

Looking at the normal distribution table, the corresponding z score is 0.001

P(x lesser than or equal to 25) = 0.001

Answer:

(a) 1/2

(b) 1/2

(c) 1/8

Step-by-step explanation:

Since, when a fair coin is tossed three times,

The the total number of possible outcomes

n(S) = 2 × 2 × 2

= 8 { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT },

Here, B : { At least two heads are observed } ,

⇒ B = {HHH, HHT, HTH, THH},

⇒ n(B) = 4,

Since,

[tex]\text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}[/tex]

(a) So, the probability of B,

[tex]P(B) =\frac{n(B)}{n(S)}=\frac{4}{8}=\frac{1}{2}[/tex]

(b) A : { At least one head is observed },

⇒ A = {HHH, HHT, HTH, THH, HTT, THT, TTH},

∵ A ∩ B = {HHH, HHT, HTH, THH},

n(A∩ B) = 4,

[tex]\implies P(A\cap B) = \frac{n(A\cap B)}{n(S)} = \frac{4}{8}=\frac{1}{2}[/tex]

(c) C: { The number of heads observed is odd },

⇒ C = { HHH, HTT, THT, TTH},

∵ A ∩ B ∩ C = {HHH},

⇒ n(A ∩ B ∩ C) = 1,

[tex]\implies P(A\cap B\cap C)=\frac{1}{8}[/tex]

Answer:

Yes, the events T1 and T2 are independent.

Step-by-step explanation:

When flipping a fair coin, for each flip, there is a 50/50 chance that it will result in heads or tails.

For the first flip, P(T1) = 0.5

For the second flip, P(T2) = 0.5 regardless of the outcome of the first flip. Therefore, T1 and T2 are independent events.

*Note that if the question asked for the event of both flips resulting in tails, then the events would be dependent.

Answer:

Remember, a basis for the row space of a matrix A is the set of rows different of zero of the echelon form of A.

We need to find the echelon form of the matrix augmented matrix of the system A2x=b2

[tex]B=\left[\begin{array}{cccc}1&2&3&1\\4&5&6&1\\7&8&9&1\\3&2&4&1\\6&5&4&1\\9&8&7&1\end{array}\right][/tex]

We apply row operations:

1.

We obtain the matrix

[tex]\left[\begin{array}{cccc}1&2&3&1\\0&-3&-6&-3\\0&-6&-12&-6\\0&-4&-5&-2\\0&-7&-14&-5\\0&-10&-20&-8\end{array}\right][/tex]

2.

We obtain the matrix

[tex]\left[\begin{array}{cccc}1&2&3&1\\0&-3&-6&-3\\0&0&0&0\\0&0&3&2\\0&0&0&2\\0&0&0&2\end{array}\right][/tex]

3.

we exchange rows three and four of the previous matrix and obtain the echelon form of the augmented matrix.

[tex]\left[\begin{array}{cccc}1&2&3&1\\0&-3&-6&-3\\0&0&3&2\\0&0&0&0\\0&0&0&2\\0&0&0&2\end{array}\right][/tex]

Since the only nonzero rows of the augmented matrix of the coefficient matrix are the first three, then the set

[tex]\{\left[\begin{array}{c}1\\2\\3\end{array}\right],\left[\begin{array}{c}0\\-3\\-6\end{array}\right],\left[\begin{array}{c}0\\0\\3\end{array}\right] \}[/tex]

is a basis for Row (A2)

Now, observe that the last two rows of the echelon form of the augmented matrix have the last coordinate different of zero. Then, the system is inconsistent. This means that the system has no solutions.

Answer:

The 98% confidence interval for the mean repair cost for the dryers is (83.4161, 103.3039).

Step-by-step explanation:

We have a small sample size n = 25, [tex]\bar{x} = 93.36[/tex] and s = 19.95. The confidence interval is given by  [tex]\bar{x}\pm t_{\alpha/2}(\frac{s}{\sqrt{n}})[/tex] where [tex]t_{\alpha/2}[/tex] is the [tex]\alpha/2[/tex]th quantile of the t distribution with n - 1 = 25 - 1 = 24 degrees of freedom. As we want the 98% confidence interval, we have that [tex]\alpha = 0.02[/tex] and the confidence interval is [tex]93.36\pm t_{0.01}(\frac{19.95}{\sqrt{25}})[/tex] where [tex]t_{0.01}[/tex] is the 1st quantile of the t distribution with 24 df, i.e., [tex]t_{0.01} = -2.4922[/tex]. Then, we have [tex]93.36\pm (-2.4922)(\frac{19.95}{\sqrt{25}})[/tex] and the 98% confidence interval is given by (83.4161, 103.3039).

Answer:

Step-by-step explanation:

Our aim is to determine a 98% confidence interval for the mean repair cost for the dryers

Number of samples. n = 25

Mean, u = $93.36

Standard deviation, s = $19.95

For a confidence level of 98%, the corresponding z value is 2.33. This is determined from the normal distribution table.

We will apply the formula,

Confidence interval

= mean ± z × standard deviation/√n

It becomes

93.36 ± 2.33 × 19.95/√25

= 93.36 ± 2.33 × 3.99

= 93.36 ± 9.2967

The lower boundary of the confidence interval is 93.36 - 9.2967 =84.0633

The upper boundary of the confidence interval is 93.36 + 9.2967 = 102.6567

Therefore, with 98% confidence interval, the mean repair costs for the dryers is between $84.0633 and $102.6567

Answer:

(-24, -8)

Step-by-step explanation:

Let us recall that when we have a function f

[tex]\large f:\mathbb{R}^2\rightarrow \mathbb{R}\\f(x,y)=z[/tex]

if the gradient of f at a given point (x,y) exists, then the gradient of f at this point (x,y) gives the direction of maximum rate of increasing and minus the gradient of f at this point gives the direction of maximum rate of decreasing. That is

[tex]\large \nabla f=(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y})[/tex]

at the point (x,y) gives the direction of maximum rate of increasing

[tex]\large -\nabla f[/tex]

at the point (x,y) gives the direction of maximum rate of decreasing

In this case we have

[tex]\large f(x,y)=100-4x^2-2y^2[/tex]

and we want to find the direction of fastest speed of decreasing at the point (-3,-2)

[tex]\large \nabla f(x,y)=(-8x,-4y) \Rightarrow -\nabla f=(8x,4y)[/tex]

at the point (-3,-2) minus the gradient equals

[tex]\large -\nabla f(-3,-2)=(-24,-8)[/tex]

hence the vector (-24,-8) points in the direction with the greatest rate of decreasing, and they should start their descent in that direction.

For testing if a coin is fair, a binomial test is used to compare the observed heads to what's expected under a fair coin scenario. Differences in p-values for the same sample proportion across different sample sizes are due to the increased precision of larger samples.

In hypothesis testing, when we want to determine whether a coin (or in this case, a hypothetical lizard) is fair, we employ a binomial test. This test assesses the number of 'successes' (heads in our case), comparing it to what we would expect under the null hypothesis of p=0.5, indicating a fair coin.

For part (a), where we get 56 heads out of 100 tosses, we would calculate the p-value by looking at both the proportion of heads obtained and the standard error for a binomial distribution. Although not calculated here directly, if this p-value is less than the chosen significance level, typically 0.05, we reject the null hypothesis, concluding the coin is not fair.

For part (b), getting 560 heads out of 1000 tosses yields the same sample proportion as part (a). However, the larger sample size increases the power of the test, often resulting in a smaller p-value if the observed proportion is different from the expected 0.5.

Comparing p-values from (a) and (b) reveals that despite having the same sample proportions, the p-values differ due to the sample sizes. Larger sample sizes yield more precise estimates of the population proportion, thus potentially resulting in smaller p-values for the same sample proportion deviation from the null hypothesis.

Answer:

8 or 3

Step-by-step explanation:

Answer:

a) Standard error = 0.5

b) 99% Confidence interval:  (19.2125,21.7875)

Step-by-step explanation:

We are given the following in the question:

Population mean =  20.50 mpg

Sample standard deviation = 3.00 mpg

Sample size , n = 36

Standard Error =

[tex]=\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{36}} = 0.5[/tex]

99% Confidence interval:

[tex]\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}[/tex]

Putting the values, we get,

[tex]z_{critical}\text{ at}~\alpha_{0.01} = \pm 2.575[/tex]

[tex]20.50 \pm 2.575(\displaystyle\frac{3}{\sqrt{36}})= 20.50 \pm 1.2875 = (19.2125,21.7875)[/tex]

The standard error of the mean from 36 fill-ups of the 2016 Lexus RX350 is 0.50 miles per gallon. The sample mean, with 99% probability, is expected to fall within the interval from 18.714 to 22.286 miles per gallon.

The standard error (SE) of the mean from a sample of 36 fill-ups for the Lexus RX 350 is calculated by dividing the standard deviation by the square root of the sample size. In this case, the standard deviation (σ) is 3.00 mpg and the sample size (n) is 36.

The formula for the standard error is SE = σ / √n = 3.00 / √36 = 0.50 mpg (rounded to four decimal places).

To find the interval in which the sample mean is expected to fall with 99 percent probability, we use the concept of Z-scores, which relate the mean, standard error, and the probability. For a 99% probability, the Z-score (z) is approximately 2.576. The interval can be found using the following formula: μ ± z * SE, replacing the symbols with the values calculated, 20.50 ± 2.576 * 0.50 which results in the interval (18.714, 22.286) mpg (rounded to four decimal places).

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Answer:

d)the estimated average of Y is 2 when X1 and X2 equals zero.

Step-by-step explanation:

Hello!

The multiple regression model with two independent variables is:

Y= β₀ + β₁X₁i + β₂X₂ + εi

Where

β₀ is where Y intercepts the line, in terms of the regression, is the value of the population mean of the dependent variable when X₁ and X₂ are cero.

β₁ is the slope of the variable X₁, in terms of the regression, is the change that suffers the population mean when X₁ increases one unit and X₂ remains constant.

β₂ is the slope of the variable X₂, in terms of the regression, is the change that suffers the population mean when X₂ increases one unit and X₁ remains constant.

When you calculate the estimated plane for the multiple regression you have tree estimators for each βi, were:

Y= b₀ + b₁ X₁ + b₂ X₂

b₀ estimates β₀

b₁ estimates β₁

b₂ estimates β₂

interpreted in colloquial language b₀ is the value of the sample mean of Y when X₁ and X₂ are cero.

With this in mind, the correct answer is d.

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Answer:

...

Step-by-step explanation:

Answer:

  A. $1,510.10

Step-by-step explanation:

Fill in the given numbers and do the arithmetic.

  M = I·(1+r)^n

  M = $1423.00·(1 +.02)^3 = $1510.099

  M ≈ $1510.10

Answer:

Option B) Data A and Data B have the same strength in linear correlation.

Step-by-step explanation:

Correlation:

Data A correlation = -0.991

Data B correlation = 0.991

Data A and data B have same strength of correlation but they are opposite to each other. There absolute value is same but the data A shows negative correlation and data B shows positive correlation. Data A shows indirect or inverse linear relationship and data B shows direct linear relationship. Both data have same magnitude of correlation.

Answer:

[tex]z=2.53[/tex]  

[tex]p_v =P(z>2.53)=0.0057[/tex]  

The p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so then we have enough evidence to reject the null hypothesis, and we can say that at 5% of significance, the proportion of adults who support a ban on sugary snacks and soft​ drinks  is more than 0.5 or 50%.

Step-by-step explanation:

1) Data given and notation

n=1000 represent the random sample taken

X=540 represent the adults that said that schools should ban sugary snacks and soft drinks

[tex]\hat p=\frac{540}{1000}=0.54[/tex] estimated proportion of adults that said that schools should ban sugary snacks and soft drinks

[tex]p_o=0.5[/tex] is the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that majority of adults​ (more than​ 50%) support a ban on sugary snacks and soft​ drinks, the system of hypothesis are:  

Null hypothesis:[tex]p\leq 0.5[/tex]  

Alternative hypothesis:[tex]p > 0.5[/tex]  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.54 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=2.53[/tex]  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.  

Since is a one side right tailed test the p value would be:  

[tex]p_v =P(z>2.53)=0.0057[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so then we have enough evidence to reject the null hypothesis, and we can say that at 5% of significance, the proportion of adults who support a ban on sugary snacks and soft​ drinks  is more than 0.5 or 50%.

Answer:

D

Step-by-step explanation:

Answer:

A

Step-by-step explanation:

change 76% to decimal and multiply by 15

.76*15 = 11.4

11.4 is fewer than 12.

Answer: fifth term = -1/16

Sixth term = 1/256

Step-by-step explanation:

The given sequence is a geometric progression. This is because the ratio of two consecutive terms is constant.

We will apply the formula for determining the nth term of a geometric progression series.

Tn = ar^n-1

Where

Tn = value of the nth term of the geometric series

a = The first term of the series.

r = common ratio(ratio of a term to a consecutive previous term)

n = number of terms in the series

From the in information given,

a = -16

r = 4/-16 = -1/4

The next 2 terms are the 5th and 6th terms.

T5 = -16 × -1/4^(5-1)

T5 = -16 × (-1/4)^4

T5 = -16 × 1/256 = -1/16

The 6th term would be the 5th term × the common ratio. It becomes

T6 = -1/16 ×-1/4 = 1/256

Answer:

2/3=0.6667

Step-by-step explanation:

Let X be the cycle time for  trucks hauling concrete to a highway construction site

Given that X is U(50,70)

Hence pdf of X is

[tex]f(x) = \frac{1}{20} ,50<x<70[/tex]

Let A be the event that cycle time is no more than 65 minutes and

B the event cycle time exceeds 55 minutes

Required probability

=  the conditional probability that the cycle time is no more than 65 minutes if it is known that the cycle time exceeds 55 minutes

= P(A/B)

=[tex]\frac{P(A\bigcapB)}{P(B)} \\=\frac{P(55<x<65)}{P(X>55)} \\=\frac{65-55}{70-55} \\=\frac{2}{3}[/tex]