If $x^5 - x^4 x^3 - px^2 qx 4$ is divisible by $(x 2)(x - 1),$ find the ordered pair $(p,q).$

Answer:

Step-by-step explanation:

Given

mean [tex]\mu =20\ ounces[/tex]

standard deviation [tex]\sigma =1\ ounce[/tex]

The no of sample boxes weigh Every morning is 25

Average weight is 1 % more than average

i.e. [tex]20+20\times 0.1=20.2 ounce[/tex]

The company re-calibrates the machine

[tex]P\left ( \bar{x}> 20.2\right )=P\left ( z-\frac{20.2-20}{\frac{1}{\sqrt{25}}}\right )[/tex]

[tex]=P\left ( z> 1\right )=1-P\left ( z\leq 1\right )[/tex]

[tex]=1-0.8413[/tex]

[tex]=0.1587[/tex]

Therefore the Probability that the average weight of box is more than 20.2 ounce is 0.1587

No of days the machine is expected to re-calibrate is [tex]100\times 0.1587=16\ days[/tex]

Answer: Option 'c' is correct.

Step-by-step explanation:

Since we have given that

Number of colors that the arrow can land on = 8

We need to find the degrees of freedom for a goodness of fit test of the fairness of this spinner.

As we know that degree of freedom is 1 less than the size of sample.

So, Mathematically, it is expressed as below:

So, Degree of freedom would be

[tex]v=n-1\\\\v=8-1\\\\v=7[/tex]

Hence, Option 'c' is correct.

Answer: Standard error of the average annual salary SE = $597.6

Step-by-step explanation:

Given;

Standard deviation = r = $5,000

Number of samples = n = 70

Mean = $50,000

To derive the standard error of mean SE. It is given as

SE = r/√n

SE = $5,000/√70

SE = $5,000/8.3666

SE = $597.6

Answer:

Please read the answers below.

Step-by-step explanation:

Let's recall that in a square all its sides are equal length and all the four internal angles measure 90 °

5. If LN = 46, then we have:

OM = 46 (Same length than LN)

PN = LN/2 = 46/2 = 23

ON = √LN²/2 = √ 46²/2 = √ 2,116/2 = √ 1,058 = 32.53 (Rounding to two decimal places)

MN = ON = 32.53

6. m ∠EFG = 90°

m ∠GDH = ∠GDH/2 = 90/2 = 45°

m ∠FEG = ∠DEF/2 = 90/2 = 45°

m ∠DHG = 180 - (∠GDH + ∠DGH) = 180 - (45 + 45)= 180 - 90 = 90°

7. Solve for x

6x - 21 = ∠PQR/2

6x - 21 = 90/2

6x - 21 = 45

6x = 45 +21

6x = 66

x = 11

Answer:

[tex]\S^2_p =\frac{(18-1)(2.5)^2 +(18 -1)(2.3)^2}{18 +18 -2}=5.77[/tex]

[tex]S_p=2.402[/tex]

[tex]t=\frac{(12 -13)-(0)}{2.402\sqrt{\frac{1}{18}+\frac{1}{18}}}=-1.249[/tex]

[tex]p_v =2*P(t_{34}<-1.249) =0.2201[/tex]

Step-by-step explanation:

When we have two independent samples from two normal distributions with equal variances we are assuming that  

[tex]\sigma^2_1 =\sigma^2_2 =\sigma^2[/tex]

And the statistic is given by this formula:

[tex]t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}[/tex]

Where t follows a t distribution with [tex]n_1+n_2 -2[/tex] degrees of freedom and the pooled variance [tex]S^2_p[/tex] is given by this formula:

[tex]\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}[/tex]

This last one is an unbiased estimator of the common variance [tex]\sigma^2[/tex]

The system of hypothesis on this case are:

Null hypothesis: [tex]\mu_1 = \mu_2[/tex]

Alternative hypothesis: [tex]\mu_1 \neq \mu_2[/tex]

Or equivalently:

Null hypothesis: [tex]\mu_1 - \mu_2 = 0[/tex]

Alternative hypothesis: [tex]\mu_1 -\mu_2 \neq 0[/tex]

Our notation on this case :

[tex]n_1 =18[/tex] represent the sample size for group 1

[tex]n_2 =18[/tex] represent the sample size for group 2

[tex]\bar X_1 =12[/tex] represent the sample mean for the group 1

[tex]\bar X_2 =13[/tex] represent the sample mean for the group 2

[tex]s_1=2.5[/tex] represent the sample standard deviation for group 1

[tex]s_2=2.3[/tex] represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

[tex]\S^2_p =\frac{(18-1)(2.5)^2 +(18 -1)(2.3)^2}{18 +18 -2}=5.77[/tex]

And the deviation would be just the square root of the variance:

[tex]S_p=2.402[/tex]

And now we can calculate the statistic:

[tex]t=\frac{(12 -13)-(0)}{2.402\sqrt{\frac{1}{18}+\frac{1}{18}}}=-1.249[/tex]

Now we can calculate the degrees of freedom given by:

[tex]df=18+18-2=34[/tex]

And now we can calculate the p value using the altenative hypothesis:

[tex]p_v =2*P(t_{34}<-1.249) =0.2201[/tex]

If we compare the p value obtained and using the significance level assumed [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.  

To find the test statistic for comparing the sales of two varieties of truffles, we use the formula for a two-sample t-test with assumed equal population variances. The test statistic is calculated using the sample means, sample standard deviations, and sample sizes for both types of truffles.

The student is asking how to find the value of the test statistic when comparing the means of two independent samples with unknown but equal population variances. Using the provided sample means and standard deviations for the two varieties of milk chocolate truffles—one plain and one filled with mint—we can apply the formula for the test statistic in a two-sample t-test where population variances are assumed to be equal.

The formula is given by:
t = (X₁ - X₂) / S_p * sqrt(1/n₁ + 1/n₂)
where:

First, calculate the pooled standard deviation (S_p) using the formula:
S_p = sqrt(((n₁-1) * S₁² + (n₂-1) * S₂²) / (n₁ + n₂ - 2))
Then, plug the values into the above formula to find the t-statistic.

Answer:

92.79% probability that the mean clock life would be greater than 15.1 years.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex].

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 16, \sigma = 4, n = 42, s = \frac{4}{\sqrt{42}} = 0.6172[/tex]

What is the probability that the mean clock life would be greater than 15.1 years?

This probability is 1 subtracted by the pvalue of Z when [tex]X = 15.1[/tex]. So:

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{15.1 - 16}{0.6172}[/tex]

[tex]Z = -1.46[/tex]

[tex]Z = -1.46[/tex] has a pvalue of 0.0721.

So there is a 1-0.0721 = 0.9279 = 92.79% probability that the mean clock life would be greater than 15.1 years.

Final answer:

Using the Central Limit Theorem and the given standard deviation and mean, the standard error was calculated, and then the z-score was determined. After finding the z-score, the area to the right of it on the standard normal distribution curve gives the probability that the mean clock life is greater than 15.1 years, which is 0.9274.

Explanation:

To find the probability that the mean clock life of a sample of 42 wall clocks is greater than 15.1 years, we can use the Central Limit Theorem which tells us that the sampling distribution of the sample mean will be normally distributed if the sample size is large enough. Since we are dealing with a sample of 42, which is generally considered sufficiently large, the sampling distribution of the sample mean is normally distributed. Given the population mean (μ) is 16 years, the standard deviation (σ) is 4 years, the sample size (n) is 42, and we want to find the probability of the sample mean (μ_x) > 15.1 years, we first need to find the standard error (SE) which is σ/sqrt(n). Then we can find the z-score which is (15.1 - μ) / SE. Lastly, we use the z-score to find the probability from z-tables or a calculator with normal distribution functions.

The standard error (SE) is:
SE = 4 / sqrt(42) ≈ 0.6172
The z-score (z) is:
z = (15.1 - 16) / 0.6172 ≈ -1.4573

To find the probability of the sample mean being greater than 15.1 years, we want the area to the right of the z-score on the standard normal distribution curve. Looking up the z-score of -1.4573 on the z-table or using a normal distribution calculator gives us the probability to the left of the z-score. We subtract this value from 1 to find the probability to the right:
P(X > 15.1) = 1 - P(Z < -1.4573) ≈ 1 - 0.0726 = 0.9274

Therefore, the probability that the mean clock life is greater than 15.1 years is 0.9274 when rounded to four decimal places.

The random variable X represents the sum of numbers on the balls chosen. The probability distribution for X is X=3 with 1/3 probability, X=4 with 1/3 probability and X=5 with 1/3 probability. The probability that the sum is at least 4 is 2/3. The expected value or mean of X is 4.

In this problem, the random variable X represents the sum of the numbers on the two balls we pick from the box. The possible sum values, excluding the order in which we pick the balls, could be (1+2)=3, (2+3)=5, or (1+3)=4.

(a) Thus, the probability distribution for X would be:
X = 3 with probability 1/3 (occurs when we pick balls 1,2)
X = 4 with probability 1/3 (occurs when we pick balls 1,3)
X = 5 with probability 1/3 (occurs when we pick balls 2,3)

(b) A sum of at least 4 can occur either when X=4 or X=5. Because each of these two outcomes is equally likely with a probability of 1/3, we sum these probabilities to find that the probability that the sum is at least 4 is 2/3.

(c) The mean of X is the expected value, which is calculated by multiplying each outcome by its probability and summing these products. In this case, that would be (3*(1/3)) + (4*(1/3)) + (5*(1/3)) = 4

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Answer:

A.

Step-by-step explanation:

Hello!

First a little reminder:

The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).

In the example, the results of a drug test were reported, being the null hypothesis "the drug has no effect" and the alternative hypothesis "the drug decreases vision loss in people with Macular Degeneration"

After conducting the test, the researchers obtained the p-value 0.004

Taking the previous definition of the p-value, the correct answer is A.

To tell whether B. and C. are correct or incorrect there should be specified what signification level was used in the analysis. Remember, the p-value is the probability of the statistic value under the null hypothesis and to use it to make a decision ver the null hypothesis you have to compare it with the signification level. If the p-value is greater than the level of significance, then you don't reject the null hypothesis (Then you can conclude that the drug has no effect) If the p-value is equal or less than the level of significance, then you reject the null hypothesis. (Then you can conclude that the drug decreases the vision loss)

Using a level of signification of 0.01 then the decision is to not reject the null hypothesis but with levels of 0.05 or 0.1 then the decision is to reject it. This is why it is important to know what level of significance was used in the test when interpreting the p-value.

I hope it helps!

Answer:

The equation of the line with slope is 3  and point (2,17) is [tex]y=3x+11[/tex]

Step-by-step explanation:

Given slope is 3 ie., m=3 and point (2,17)

To find the equation of the line with slope and a point:

Using slope point formula

[tex]y-y_{1}=m(x-x_{1})[/tex]

Let the point (2,17) be [tex](x_{1},y_{1})[/tex]

Substitute the values of m=3 and point (2,17)

[tex]y-y_{1}=m(x-x_{1})[/tex]

[tex]y-17=3(x-2)[/tex]

[tex]y-17=3x-6[/tex]

[tex]y=3x-6+17[/tex]

[tex]y=3x+11[/tex]

Therefore the equation of the line is [tex]y=3x+11[/tex]

The test statistic for a chi-square test with a sample size of 14, sample standard deviation of 20, and null hypothesis variance of 500 should be 10.40. However, since this number does not match any of the options provided, there may be an error in the question or the options.

The subject in question involves performing a hypothesis test to determine whether the variance of a population is greater than or equal to a specified value. Given a sample size (n = 14), sample standard deviation (s = 20), and the null hypothesis being (σ² ≤ 500), we need to calculate the test statistic for a chi-square distribution with degrees of freedom df = n - 1, which in this case is df = 13. The test statistic is calculated using the formula:

chi-square test statistic = χ² = (n - 1)s² / σ²

Plugging in the values, we get:

chi-square test statistic = (14 - 1) * 20² / 500 = 13 * 400 / 500 = 5200 / 500 = 10.40

However, since this value does not match any of the answer options provided in the initial question, it appears that there may be an error in the question itself or in the provided options. It's important to carefully review the calculation and ensure the values and hypotheses given are as intended.

To test the hypothesis that the average debt load of graduating students with a Bachelor's degree is equal to $17,000, a hypothesis test needs to be conducted using the given sample data. The p-value needs to be calculated and compared to the significance level α to determine the conclusion. Without the p-value, we cannot make a definitive conclusion.

To test the hypothesis that the average debt load of graduating students with a Bachelor's degree is equal to $17,000, the Department of Education would conduct a hypothesis test using the given sample data. The null hypothesis, denoted as H0, states that the average debt load is equal to $17,000, while the alternative hypothesis, denoted as Ha, states that the average debt load is not equal to $17,000.

Based on the given information, the sample mean debt load is $18,200, the population standard deviation is $4,200, and the sample size is 34. Using these values, we can calculate the p-value, which represents the probability of obtaining a sample mean as extreme as $18,200 or more extreme, assuming the null hypothesis is true. The p-value is the probability of observing a sample mean of $18,200 or more extreme, given that the average debt load is $17,000.

To determine the conclusion of the hypothesis test, we compare the p-value to the significance level α. In this case, the given α is 0.05. If the p-value is less than α, we reject the null hypothesis and conclude that the average debt load is not equal to $17,000. If the p-value is greater than or equal to α, we fail to reject the null hypothesis and cannot conclude that the average debt load is not equal to $17,000.

In this scenario, the p-value is not given, so we cannot make a definitive conclusion. We would need to calculate the p-value based on the given information to determine which statement is true.

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The correct option is A.  [tex]\text{Because the p-value is greater than}[/tex] [tex]\alpha,[/tex] [tex]\text{ we fail to reject the null hypothesis and conclude that the average debt load is equal to }[/tex] [tex]\$17,000.[/tex]

To determine the correct statement based on the hypothesis test conducted:

Given data:

Population standard deviation[tex](\( \sigma \)): \$4,200[/tex]

Sample size [tex](\( n \)): 34[/tex]

Sample mean [tex](\( \bar{x} \)): \$18,200[/tex]

Hypothesized population mean [tex](\( \mu_0 \))[/tex]: [tex]\$17,000.[/tex]

Significance level [tex](\alpha): 0.05[/tex]

Hypotheses

We are testing whether the average debt load [tex](\( \mu \))[/tex] of graduating students with a Bachelor's degree is equal to [tex]\$17,000.[/tex]

Null hypothesis [tex](\( H_0 \)): \( \mu = 17,000 \)[/tex]

Alternative hypothesis [tex](\( H_1 \)): \( \mu \neq 17,000 \)[/tex]

This is a two-tailed test because [tex]\( H_1 \)[/tex] states that [tex]\( \mu \)[/tex] is not equal to [tex]\$17,000.[/tex]

Test Statistic

Since the population standard deviation [tex](\( \sigma \))[/tex] is known and the sample size [tex](\( n \))[/tex] is greater than [tex]30[/tex], we use a z-test.

The test statistic [tex]\( z \)[/tex] is calculated as:

[tex]\[ z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}} \][/tex]

Calculate [tex]\( z \)[/tex]

[tex]\[ z = \frac{18,200 - 17,000}{\frac{4,200}{\sqrt{34}}} \][/tex]

[tex]\[ z = \frac{1,200}{\frac{4,200}{5.83}} \][/tex]

[tex]\[ z = \frac{1,200 \times 5.83}{4,200} \][/tex]

[tex]\[ z = 1.66 \][/tex]

P-Value Calculation

The p-value is the probability of observing a sample mean at least as extreme as [tex]18,200[/tex] if the null hypothesis [tex](\( \mu = 17,000 \))[/tex] is true. Since this is a two-tailed test, we consider both tails of the normal distribution.

From the standard normal distribution table:

The area to the right of [tex]\( z = 1.66 \)[/tex] (since [tex]\( z \)[/tex] is positive) is approximately [tex]0.0485.[/tex]

Therefore, the p-value for the two-tailed test is approximately [tex]\( 2 \times 0.0485 = 0.097 \).[/tex]

Conclusion

Compare the p-value [tex](0.097)[/tex] with the significance level [tex]\alpha = 0.05)\( \text{p-value} (0.097) > \alpha (0.05) \)[/tex]

Since the p-value is greater than the significance level ([tex]\alpha[/tex]), we fail to reject the null hypothesis [tex]\( H_0 \)[/tex]. This means we do not have enough evidence to conclude that the average debt load is different from [tex]\$17,000.[/tex]

Answer:

True

Step-by-step explanation:

In Mathematics, A radical symbol is defined as √

It is an expression that is uses a root, such as the square root (√ ) or cube root ( ∛ )

To fully show a radical expression:

[tex]\sqrt{A}[/tex]

A is the radicand

√ is the radical symbol

Here, the degree is 2.

Quotient Rule for Radicals

For non- negative real numbers, x and y:

[tex]\frac{\sqrt[n]{x}}{\sqrt[n]{y} }[/tex] = [tex]\sqrt[n]{\frac{x}{y}}[/tex]

Example :

[tex]\sqrt{\frac{36}{9}}[/tex]= [tex]\frac{\sqrt{36}}{\sqrt{9} }[/tex] = [tex]\frac{6}{3}[/tex] = 2

So, in simplifying radicals, quotient rule is used as demonstrated above. So the condition is true

Answer:

We conclude that the actual average cost per workbook is higher than $27.50.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = $27.50

Sample mean, [tex]\bar{x}[/tex] = $28.90

Sample size, n = 44

Alpha, α = 0.05

Population standard deviation, σ = $5.00

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 27.50\text{ dollars}\\H_A: \mu > 27.50\text{ dollars}[/tex]

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{28.90 - 27.50}{\frac{5.00}{\sqrt{44}} } = 1.8573[/tex]

Now, [tex]z_{critical} \text{ at 0.05 level of significance } = 1.64[/tex]

Since,  

[tex]z_{stat} > z_{critical}[/tex]

We reject the null hypothesis and accept the alternate hypothesis.

Thus, we conclude that the actual average cost per workbook is higher than $27.50.

Answer:

Slope: 3

Y intercept: y = 75

Step-by-step explanation:

The slope of a linear function is the coefficient in front of x. In this case, 3 is being multiplied by x, indicating the slope is 3.

The y-intercept is the y value where the graph of the function crosses the y-axis. In other words, it's the y value when x = 0. To find the y-intercept, simply substitute 0 for x:

[tex]y = 3(0) + 75 \\ y = 75[/tex]

We find that the y-intercept is y = 75.

In the linear function y=3x+75, the slope 3 represents the rate of cost per mile to tow a car ($3/mile), and the y-intercept 75 is the basic fee for the towing service. So, the total cost is a fixed fee of $75 plus $3 per mile towed.

In the linear function y = 3x + 75, the slope, represented by '3', is the cost in dollars per mile to tow a car. This means for each mile the car is towed, the cost increases by $3.

The y-intercept, represented by '75', is the basic fee for the service, or the cost in dollars that one must pay even if the car isn't towed any miles.

So, to interpret the equation function: the total cost (y) of having a car towed is composed of a fixed fee of $75 plus an additional $3 for each mile (x) the car is towed.

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At x = -2, there is a blue dot, which is a local minimum, because it is below the x axis.

The dot is located at y = -4

The answer is Yes there is a local minimum at x = -2

The local minimum is y = -4

Answer:

The answer is Yes there is a local minimum at x = -2

The local minimum is y = -4

Step-by-step explanation:

Found answer online so no work!!!

Answer:

On demand

Step-by-step explanation:

On demand (fund Value Streams, not projects).

Lean Budgets is a collection of methods that significantly reduce overhead by financing and encouraging Value Streams instead of projects while preserving financial and user-fit governance. This is done by comprehensive work program assessment, active management of Epic finances, and complex budget changes.

Hence, the recommended frequency for updating lean budget distribution on demand that is fund value streams.

Lean budget distribution should be updated at least once per quarter, or as often as required by the specific circumstances of your organization. The goal of a lean budget is to optimize resource use and reduce waste to deliver more value to customers.

The recommended frequency for updating lean budget distribution is not fixed and largely depends on the specific needs, context, and circumstances of your organization. However, a common practice is to review and adjust the lean budget allocation at least once per quarter. This allows you to take into account any changes in strategy, project progress, and business environment. It's important to highlight that these revisions should be strategic and informed by data and not just random shifts in resource allocation. Remember, the main goal of a lean budget is to minimize waste and optimize resource utilization to bring more value to customers.

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Answer:

[tex]z=\frac{0.46-0.38}{\sqrt{0.42(1-0.42)(\frac{1}{50}+\frac{1}{50})}}=0.8104[/tex]  

[tex]p_v =P(Z>0.8104)=0.209[/tex]  

If we compare the p value and using any significance level for example [tex]\alpha=0.1[/tex] always [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the proportion of women favored the increase is not significantly higher than the proportion of men favored the increase at 10% of significance.  

Step-by-step explanation:

1) Data given and notation  

[tex]X_{1}=23[/tex] represent the number of women favored the increase

[tex]X_{2}=19[/tex] represent the number of men favored the increase

[tex]n_{1}=50[/tex] sample 1 selected

[tex]n_{2}=50[/tex] sample 2 selected

[tex]p_{1}=\frac{23}{50}=0.46[/tex] represent the proportion of women favored the increase

[tex]p_{2}=\frac{19}{50}=0.38[/tex] represent the proportion of men favored the increase

z would represent the statistic (variable of interest)  

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

[tex]\alpha=0.1[/tex] represent the significance level

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to check if that a larger proportion of women favor the increase than men, the system of hypothesis would be:  

Null hypothesis:[tex]p_{1} \leq p_{2}[/tex]  

Alternative hypothesis:[tex]p_{1} > p_{2}[/tex]  

We need to apply a z test to compare proportions, and the statistic is given by:  

[tex]z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}[/tex]   (1)

Where [tex]\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{23+19}{50+50}=0.42[/tex]

3) Calculate the statistic

Replacing in formula (1) the values obtained we got this:  

[tex]z=\frac{0.46-0.38}{\sqrt{0.42(1-0.42)(\frac{1}{50}+\frac{1}{50})}}=0.8104[/tex]  

4) Statistical decision

We can calculate the p value for this test.  

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

[tex]p_v =P(Z>0.8104)=0.209[/tex]  

If we compare the p value and using any significance level for example [tex]\alpha=0.1[/tex] always [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the proportion of women favored the increase is not significantly higher than the proportion of men favored the increase at 10% of significance.  

Answer:

a) Null hypothesis:[tex]\mu \geq 40[/tex]  

Alternative hypothesis:[tex]\mu < 40[/tex]  

b) The rejection zone for this test would be:

[tex] (-\infty, -2.36)[/tex]

c) If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is significantly less than 40 years at 1% of signficance.  

Step-by-step explanation:

1) Data given and notation  

[tex]\bar X=39.2[/tex] represent the sample mean  

[tex]s=8[/tex] represent the sample standard deviation

[tex]n=100[/tex] sample size  

[tex]\mu_o =40[/tex] represent the value that we want to test

[tex]\alpha=0.01[/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)  

Part a

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is less than 40, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \geq 40[/tex]  

Alternative hypothesis:[tex]\mu < 40[/tex]  

If we analyze the size for the sample is > 30 but we don't know the population deviation so 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{39.2-40}{\frac{8}{\sqrt{100}}}=-1[/tex]    

Part b: Rejection zone

On this case we are conducting a left tailed test so we need to find first the degrees of freedom for the distribution given by:

[tex]df=n-1=100-1=99[/tex]

Now we need to find a critical value on th t distribution with 99 degrees of freedom that accumulates 0.01 of the area on the right tail and this value is given by [tex]t_{crit}=-2.36[/tex]

And we can find it using the following excel code: "=T.INV(0.01,99)"

So then the rejection zone for this test would be:

[tex] (-\infty, -2.36)[/tex]

Part c: P-value and conclusion

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

[tex]p_v =P(t_{(99)}<-1)=0.160[/tex]  

Conclusion  

If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is significantly less than 40 years at 1% of signficance.  

Answer:

[tex]\frac{dV}{dt}=525 cm^{3}/min[/tex]

Step-by-step explanation:

The volume of a sphere is:

[tex]V=\frac{4}{3}\pi r^{3}[/tex] (1)

We know that:

If we take the derivative of (1) with respect of time t, we ca n find the rate of increase of the volume.

[tex]\frac{dV}{dt}=\frac{4}{3}\pi 3r^{2}\frac{dr}{dt}=4\pi r^{2}\frac{dr}{dt}[/tex] (2)

We also know that the volume is 40 cm³, then using the (1) we can get the radius at this value.

Solving (1) for r, we have:

[tex]r=\left(\frac{3\cdot V}{4\pi}\right)^{1/3}=\left(\frac{3\cdot 400}{4\pi}\right)^{1/3}=4.57 cm[/tex]

Finally dV/dt will be:

[tex]\frac{dV}{dt}=4\pi (4.57)^{2}\cdot 2=525 cm^{3}/min[/tex]

I hope it helps you!

Answer:

The correct option is B) 0.0228

Step-by-step explanation:

Consider the provided information.

The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in the sample to check out was 3.1 minutes. The population standard deviation is known to be 0.5 minute.

Thus, the value of n = 100, [tex]\bar x=3.1[/tex] and σ = 0.5

We want to test to determine whether or not the mean waiting time of all customers is significantly more than 3 minutes.

The null and alternative hypotheses are,

[tex]H_0:\mu\leq3[/tex] and [tex]H_a:\mu>3[/tex]

Compute the test statistic [tex]z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]z=\frac{3.1-3}{\frac{0.5}{\sqrt{100}}}[/tex]

[tex]z=\frac{0.1}{0.05}[/tex]

[tex]z=2[/tex]

By using the table.

P value = P(Z>2) = 0.0228

Hence, the correct option is B) 0.0228

Answer:

0% probability of a guest arriving by taxi and requesting parking for a car.

Step-by-step explanation:

Two events are said to be mutually exclusive if one event precludes the other. That means that if A and B are mutually exclusive, the probability of A and B happening at the same time is 0%.

In this problem, we have that a guest arriving via taxi and requesting parking at the hotel are mutually exclusive.

So there is a 0% probability of a guest arriving by taxi and requesting parking for a car.

Answer:

a) Null hypothesis:[tex]\mu \geq 10[/tex]  

Alternative hypothesis:[tex]\mu < 10[/tex]  

b) [tex]p_v =P(t_{17}<-2.3)=0.017[/tex]

And on this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis.

c) [tex]p_v =P(t_{17}<-1.8)=0.0448[/tex]

And on this case since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis.

d) [tex]p_v =P(t_{17}<-3.6)=0.0011[/tex]

And on this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation  

[tex]\bar X[/tex] represent the sample mean

[tex]s[/tex] represent the sample deviation

[tex]n=18[/tex] sample size  

[tex]\mu_o =10[/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)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is at least 10 hours, the system of hypothesis would be:  

Part a

Null hypothesis:[tex]\mu \geq 10[/tex]  

Alternative hypothesis:[tex]\mu < 10[/tex]  

If we analyze the size for the sample is < 30 and we don't know the population deviation so 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)  

Part b

For this case we have t=-2.3 , [tex]\alpha=0.05[/tex]

First we need to find the degrees of freedom [tex]df=n-1=18-1=17[/tex]

Now since we are conducting a left tailed test the p value is given by:

[tex]p_v =P(t_{17}<-2.3)=0.017[/tex]

And on this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis.

Part c

For this case we have t=-1.8 , [tex]\alpha=0.01[/tex]

First we need to find the degrees of freedom [tex]df=n-1=18-1=17[/tex]

Now since we are conducting a left tailed test the p value is given by:

[tex]p_v =P(t_{17}<-1.8)=0.0448[/tex]

And on this case since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis.

Part d

For this case we have t=-3.6 , [tex]\alpha=0.05[/tex]

First we need to find the degrees of freedom [tex]df=n-1=18-1=17[/tex]

Now since we are conducting a left tailed test the p value is given by:

[tex]p_v =P(t_{17}<-3.6)=0.0011[/tex]

And on this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis.

The student is learning about conducting a one-sample t-test in a hypothesis testing problem related to the lifetime of a pen. The null and alternative hypothesis are defined, and conclusions are drawn based on various test statistics and significance levels.

This situation relates to hypothesis testing in statistics, in particular the usage of a one-sample t-test.

The null hypothesis (H0) and alternative hypothesis (Ha) to be tested would be:

The t value represents the test statistic for conducting the hypothesis test, and the 'α' and 'u' are the significance levels.

With t = -23 and α = 0.05,  this falls within the rejection region, as the t value is less than the critical value for a one-tailed test at significance level α = 0.05. Hence, we reject the null hypothesis. There is evidence to suggest that the pen's lifetime is less than 10 hours.

Similarly, with t = -1.8 and u = 0.01, the null hypothesis would be retained, as this value is within the acceptance region.The conclusion would be that there is not sufficient evidence to suggest that the pen's lifetime is less than 10 hours.

Finally,  with t = -3.6, the null hypothesis is rejected, suggesting that the pen's lifetime is less than 10 hours. However, this is only applicable if the significance level is above the calculated p-value for t = -3.6.

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Answer:

1a) Plot on the graph.

1b) Its proportional if y (weight) varies directly with variation in x(age) and fits the form y=k*x

Step-by-step explanation:

1a)

draw graph with x axis in increments of 1 month

draw y-axis in increments of 1 pound.

put 1 dot at (1, 4)

put 1 dot at (2,7)

connect the 2 dots, and continue the line to hit one of the axes.

It will not hit the origin.

1b) show that y=kx

4=k*1 , so k=4

7=k*2 , so k=3.5

Since we have 2 different values for k (4 & 3.5), it's not proportional.

Answer:

We conclude that at a 5% significance level that the mean delivery time is longer than what the pizza shop claims.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 30 minutes

Sample mean, [tex]\bar{x}[/tex] = 31.5 minutes

Sample size, n = 41

Alpha, α = 0.05

Sample standard deviation, s = 3.5 minutes

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 30\text{ minutes}\\H_A: \mu > 30\text{ minutes}[/tex]

We use one-tailed z test to perform this hypothesis.

Formula:

[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]z_{stat} = \displaystyle\frac{31.5 - 30}{\frac{3.5}{\sqrt{41}} } = 2.744[/tex]

Now, [tex]z_{critical} \text{ at 0.05 level of significance } = 1.64[/tex]

Since,  

[tex]z_{stat} > z_{critical}[/tex]

We reject the null hypothesis and accept the alternate hypothesis.

Thus, we conclude that at a 5% significance level that the mean delivery time is longer than what the pizza shop claims.

Answer:

[tex]z=\frac{0.540-0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=2.53[/tex]  

[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=0.01[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the true proportion is not significantly higher than 0.5.  

Step-by-step explanation:

1) Data given and notation

n=1000 represent the random sample taken

X=540 represent the people indicated that they would be willing to give up some personal time in order to make more money

[tex]\hat p=\frac{540}{1000}=0.54[/tex] estimated proportion of people indicated that they would be willing to give up some personal time in order to make more money

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

[tex]\alpha=0.01[/tex] represent the significance level

Confidence=99% or 0.99

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 the majority of women age 22 to 35 who work full-time would be willing to give up some personal time for more money :  

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].

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

[tex]np_o =100*0.5=500>10[/tex]

[tex]n(1-p_o)=1000*(1-0.5)=500>10[/tex]

3) Calculate the statistic  

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

[tex]z=\frac{0.540-0.5}{\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=0.01[/tex]. The next step would be calculate the p value for this test.  

Since is a 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=0.01[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the true proportion is not significantly higher than 0.5.  

The sample data provides convincing evidence that the majority of women age 22 to 35 who work full-time would be willing to give up some personal time for more money.

To test whether the majority of women age 22 to 35 who work full-time would be willing to give up some personal time for more money, we need to conduct a hypothesis test. The null hypothesis is that the proportion of women willing to give up personal time is 0.50, and the alternative hypothesis is that it is greater than 0.50. We can use a one-sample proportion test.

In this case, the sample proportion is 540/1000 = 0.54. The sample size is large enough for conducting a test as the expected count for both categories is greater than 10. The test statistic can be calculated as (sample proportion - hypothesized proportion) / sqrt((hypothesized proportion * (1 - hypothesized proportion)) / sample size). The Z-score can then be compared to the critical Z-value at a significance level of 0.01 to determine if we reject or fail to reject the null hypothesis. The p-value can also be calculated using the Z-score.

In this specific case, the test statistic is (0.54 - 0.50) / sqrt((0.50 * (1 - 0.50)) / 1000) = 7.21. The corresponding p-value is less than 0.0001, which is smaller than the significance level of 0.01. Therefore, we reject the null hypothesis and conclude that the majority of women age 22 to 35 who work full-time would be willing to give up some personal time for more money.

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Answer:

H0: [tex]\sigma \geq 4.1[/tex]

H1: [tex]\sigma <4.1[/tex]

[tex]\Chi^2_{crit} =10.851[/tex]

[tex] t=(21-1) [\frac{2.5}{4.1}]^2 =7.436[/tex]

[tex]p_v = P(\Chi^2_{20}<7.436)=0.0050[/tex]

True, since our p value is lower than the significance level we have enough evidence to reject the null hypothesis at 5% of significance. And we can conclude that the population standard dviation is less than 4.1 (Or the variance is less than 16.81)

Step-by-step explanation:

Previous concepts and notation

The chi-square test is used to check if the standard deviation of a population is equal to a specified value. We can conduct the test "two-sided test or a one-sided test".

[tex]\bar X [/tex] represent the sample mean

n = 21 sample size

[tex]s^2 =6.25[/tex] represent the sample variance

[tex]s=2.5[/tex] represent the sample deviation

[tex]\sigma^2_0 =16.81[/tex] represent the target variance

[tex]\sigma_o =4.1[/tex] the value that we want to test

[tex]p_v [/tex] represent the p value for the test

t represent the statistic

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

State the null and alternative hypothesis

On this case we want to check if the population variance is lower (or if the deviation is lower than 4.1) than 16.81, so the system of hypothesis are:

H0: [tex]\sigma \geq 4.1[/tex]

H1: [tex]\sigma <4.1[/tex]

In order to check the hypothesis we need to calculate th statistic given by the following formula:

[tex] t=(n-1) [\frac{s}{\sigma_o}]^2 [/tex]

This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

What is the critical value for the test statistic at an α = 0.05 significance level?

Since is a left tailed test the critical zone it's on the left tail of the distribution. On this case we need a quantile on the chi square distribution with 20 degrees of freedom that accumulates 0.05 of the area on the left tail and 0.95 on the right tail.  

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.05,20)". And our critical value would be [tex]\Chi^2_{crit} =10.851[/tex]

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

[tex] t=(21-1) [\frac{2.5}{4.1}]^2 =7.436[/tex]

What is the approximate p-value of the test?

For this case since we have a left tailed test the p value is given by:

[tex]p_v = P(\Chi^2_{20}<7.436)=0.0050[/tex]

Based on your answer for the p value, answer True or False as to whether the null hypothesis will be rejected.

True, since our p value is lower than the significance level we have enough evidence to reject the null hypothesis at 5% of significance. And we can conclude that the population standard dviation is less than 4.1 (Or the variance is less than 16.81)

Answer:

Step-by-step explanation:

1) 1) 97 – 39. It becomes

100 - 40 = 60

2)2) 812,344 – 187,675. It becomes

800000 - 200000 = 600000

3)321 + 79. It becomes

300 + 80 = 410

4) 315 x 821. It becomes

300 + 800 = 1100

5) 562 x 791. It becomes

600× 800 = 480000

6) 82 x 156. It becomes

80×200 = 16000

7) 711 x 884. It becomes

700×900 = 630000

8) 126 x 952. It becomes

100×1000 = 100000

9) 824 x 541. It becomes

800× 500 = 400000

10) 4027 x 78. It becomes

4000×80 = 320000

11) 796 x 123. It becomes

800×100 = 80000

12) 817 19. It becomes

800/20 = 40

13) 3615 72. It becomes

4000/70 = 57.1 = 60

14) 232 64. It becomes

200/60 = 33.3 = 30

15) 559 81. It becomes

600/80 = 7.5 = 8

16) 2986 222. It becomes

3000/200 = 15

17) 10275 232. It becomes

10000/200 = 50

18) 7428 286. It becomes

7000/300 = 23.3 = 20

19) 7143 369. It becomes

7000/400 = 17.5 = 18

20) 628,597 1525, it becomes

600000 - 2000 = 300

Answer:

1) 100 - 40 = 60

2) 800000 - 200000 = 600000

3) 300 + 80 = 410

4) 3000 + 800 = 1100

5) 600× 800 = 480000

6) 80×200 = 16000

7) 700×900 = 630000

8) 100×1000 = 100000

9) 800× 500 = 400000

10) 4000×80 = 320000

11) 800×100 = 80000

12) 800/20 = 40

13) 4000/70 = 57.1 = 60

14) 200/60 = 33.3 = 30

15) 600/80 = 7.5 = 8

16) 3000/200 = 15

17) 10000/200 = 50

18) 7000/300 = 23.3 = 20

19) 7000/400 = 17.5 = 18

20) 600000 / 2000 = 300

Step-by-step explanation:

1) 1) 97 – 39. It becomes

100 - 40 = 60

2)2) 812,344 – 187,675. It becomes

800000 - 200000 = 600000

3)321 + 79. It becomes

300 + 80 = 410

4) 315 x 821. It becomes

300 + 800 = 1100

5) 562 x 791. It becomes

600× 800 = 480000

6) 82 x 156. It becomes

80×200 = 16000

7) 711 x 884. It becomes

700×900 = 630000

8) 126 x 952. It becomes

100×1000 = 100000

9) 824 x 541. It becomes

800× 500 = 400000

10) 4027 x 78. It becomes

4000×80 = 320000

11) 796 x 123. It becomes

800×100 = 80000

12) 817 19. It becomes

800/20 = 40

13) 3615 72. It becomes

4000/70 = 57.1 = 60

14) 232 64. It becomes

200/60 = 33.3 = 30

15) 559 81. It becomes

600/80 = 7.5 = 8

16) 2986 222. It becomes

3000/200 = 15

17) 10275 232. It becomes

10000/200 = 50

18) 7428 286. It becomes

7000/300 = 23.3 = 20

19) 7143 369. It becomes

7000/400 = 17.5 = 18

20) 628,597 1525, it becomes

600000 - 2000 = 300

Answer:

a) [tex]T(x)=300x+18x=318x[/tex]

b) [tex]D(x)=318x-(0.1)(318)+10x\\=296.2x[/tex]

c) Yes.

Step-by-step explanation:

We have that:

[tex]Tires=$300\\Tax=0.006\\Discount=0.1\\[/tex]

And we have $10 free of taxes.

Making x= number of tires to buy, then we have that the total cost of tires is:

[tex]Total_{Tires}=300x[/tex]

So, what we pay for taxes is given by:

[tex]Taxes=(300x)(0.06)=18x[/tex]

a) Then, according to the above, we can write down the total cost before the discount as:

[tex]T(x)=300x+18x=318x[/tex]

b) And the total cost after discounts, is then given by:

[tex]D(x)=318x-(0.1)(318)+10x\\=296.2x[/tex]

c) If the discount is added first, then less tax will be paid because the amount on which it is paid is lower. If the discount is added later, then the taxes will have been taxed on a higher amount, so it does matter whether they are added first or later.

Answer:

B. The true population percentage is likely to be within 5% of the sample percentage.

Step-by-step explanation:

These samples have a confidence level of x%, which leads us to a confidence interval with a margin of error is M%.

This means that we are x% sure that the true population percentage is within M% of the sample percentage.

In this problem, the true population percentage is likely to be within 5% of the sample percentage.

So the correct answer is:

B. The true population percentage is likely to be within 5% of the sample percentage.

Answer:

[tex]n=(\frac{1.64(14.7)}{3})^2 =64.57 \approx 65[/tex]

So the answer for this case would be n=65 rounded up to the nearest integer

And the sample size would decrease by 160-65=95 subjects

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

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

[tex]\mu[/tex] population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

ME represent the margin of error

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

2) Solution to the problem

Since the new Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that [tex]z_{\alpha/2}=1.64[/tex]

Assuming that the deviation is known we can express the margin of error is given by this formula:

[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]    (a)

And on this case we have that ME =\pm 3 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex]   (b)

Replacing into formula (b) we got:

[tex]n=(\frac{1.64(14.7)}{3})^2 =64.57 \approx 65[/tex]

So the answer for this case would be n=65 rounded up to the nearest integer

And the sample size would decrease by 160-65=95 subjects