Which of the following is used to determine the significance of predictions made by a best fitting linear equation?A. correlational analysisB. analysis of varianceC. analysis of regressionD. method of least squares

Answer:

The x axis in the function represents, the number of hours after 7:00 A.M. , the person reaches her car. The person reaches the car at 1:00 P.M.

Step-by-step explanation:

The x axis denotes the no. of hours and and the y axis denotes the distance from the car.

X Intercept is a point where the line intersects the X axis, we can easily notice the fact that at that point, y=0 ie. The person has reached his/her respective car.

The line intersects x at 6.

Therefore, a total of 6 hours are taken from the beginning of the hike.

Thus, the person reaches the car at 1:00 P.M.

Answer:

The person will take 6 hours to get back to her car.

Step-by-step explanation:

Answer:

Option D) Yes, because the test statistic is -2.01

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 30 pound

Sample mean, [tex]\bar{x}[/tex] = 29.1 pounds

Sample size, n = 20

Alpha, α = 0.05

Sample standard deviation, s =  2 pounds

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 30\text{ pounds}\\H_A: \mu < 30\text{ pounds}[/tex]

We use one-tailed(left) t test to perform this hypothesis.

Formula:

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

Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{29.1 - 30}{\frac{2}{\sqrt{20}} } = -2.012[/tex]

Now,

[tex]t_{critical} \text{ at 0.05 level of significance, 19 degree of freedom } = -1.729[/tex]

Since,                    

[tex]t_{stat} < t_{critical}[/tex]

We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis. Thus, there were enough evidence to conclude that the fishing line breaks with an average force of less than 30 pounds.

Option D) Yes, because the test statistic is -2.01

Answer:

[tex]P(X<0.1)= 5.5x10^{-5}[/tex]

Step-by-step explanation:

Previous concepts

Beta distribution is defined as "a continuous density function defined on the interval [0, 1] and present two parameters positive, denoted by α and β, both parameters control the shape. "

The probability function for the beta distribution is given by:

[tex] P(X)= \frac{x^{\alpha-1} (1-x)^{\beta -1}}{B(\alpha,\beta)}[/tex]

Where B represent the beta function defined as:

[tex]B(\alpha,\beta)= \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}[/tex]

Solution to the problem

For our case our random variable is given by:

[tex] X \sim \beta (\alpha=5, \beta =2)[/tex]

We can use the following R code to plot the distribution for this case:

> x=seq(0,1,0.01)

> plot(x,dbeta(x,5,2),main = "Beta distribution a=5, b=2",ylab = "Probability")

And we got as the result the figure attached.

And for this case we want this probability, since we want the probability that she has at most 10% or 0.1 change of winning:

[tex]P(X<0.1)[/tex]

And we can find this probability with the following R code:

> pbeta(0.1,5,2)

[1] 5.5e-05

And we got then this : [tex]P(X<0.1)= 5.5x10^{-5}[/tex]

Answer:

The answer is False

Step-by-step explanation:

The  count does no suggest that the normal model is a good fit for sampling the distribution because the questions used for the test and survey is the one of many other question about cheating which implies that if other questions about college life are being used as the survey, the response would probably be that more of the student would not have admitted to cheating. This concept therefore disobeys the normal distribution model which is a bell shaped model and therefore assumes that at an average, the number of students that admitted to cheating in the major courses should be equal.

Answer:

Brownian Motion- The usual model for the time-evolution of an asset price S(t) is given by the geometric Brownian motion.

Now the geometric Brownian motion is represented by the following stochastic differential equation:

To solve the problem now we have the been Data provided:

μ= 0.12,

σ=0.24,

Step-by-step explanation:

Step A:

we have, the variables of Black Scholes Model, by putting the values of variables available, we get:

Next is, "r" the risk free rate,

Step B:

We now need to calculate the parameter d₂ of the Black Scholes Model. .

Step C:

As step C is done on excel for further calculations so, do use it if you are solving it on computer.

Final answer:

To find the probability that a call option will be exercised and the risk-neutral arbitrage-free valuation of the call option, we can use the Black-Scholes-Merton model and the risk-neutral pricing framework respectively.

Explanation:

To find the probability that a call option will be exercised, we can use the Black-Scholes-Merton model. In this case, we have:

The current price of the security (S) = $40

The strike price of the option (K) = $42

The time to expiration (T) = 4 months (or 0.33 years)

The risk-free interest rate (r) = 8% (or 0.08)

The volatility of the security (σ) = 0.24

Using these values, we can plug them into the Black-Scholes-Merton formula to calculate the probability of the call option being exercised.

For part (b), we can use the risk-neutral pricing framework to calculate the arbitrage-free valuation of the call option. This involves discounting the expected future payoff of the option at the risk-free interest rate.

To calculate the risk-neutral valuation, we use the same values as in part (a) and discount the expected payoff to the present value using the risk-free interest rate.

This equation is separable, as

[tex]\dfrac{\mathrm dy}{\mathrm dx}=y(y-2)e^x\implies\dfrac{\mathrm dy}{y(y-2)}=e^x\,\mathrm dx[/tex]

Integrate both sides; on the left, expand the fraction as

[tex]\dfrac1{y(y-2)}=\dfrac12\left(\dfrac1{y-2}-\dfrac1y\right)[/tex]

Then

[tex]\displaystyle\int\frac{\mathrm dy}{y(y-2)}=\int e^x\,\mathrm dx\implies\frac12(\ln|y-2|-\ln|y|)=e^x+C[/tex]

[tex]\implies\dfrac12\ln\left|\dfrac{y-2}y\right|=e^x+C[/tex]

Since [tex]y(0)=1[/tex], we get

[tex]\dfrac12\ln\left|\dfrac{1-2}1\right|=e^0+C\implies C=-1[/tex]

so that the particular solution is

[tex]\dfrac12\ln\left|\dfrac{y-2}y\right|=e^x-1\implies\boxed{y=\dfrac2{1-e^{2e^x-2}}}[/tex]

Answer:

a) [tex]S_{a}(t)=16Kg-0.16Kg*\frac{t}{min}[/tex]

b)  [tex]S_{a}(20)=12.8Kg[/tex]

Step-by-step explanation:

It can be seen in the graph that the water velocity and solution velocity is the same, but the salt concentation will be lower

Water velocity [tex]V_{w} = 60\frac{L}{min}[/tex]

Solution velocity [tex]V_{s} = 60\frac{L}{min}[/tex]

Brine concentration = [tex]\frac{6,000L}{16Kg}=375\frac{L}{Kg}[/tex]

a) Amount of salt as a funtion of time Sa(t)

[tex]S_{a}(t)=16Kg-\frac{60Kg*L}{375L}*\frac{(t)}{min}=[tex]16Kg-0.16Kg*\frac{t}{min}[/tex]

b) [tex]S_{a}(20)=16Kg-0.16\frac{Kg}{min}*(20min)=16Kg-3.2Kg=12.8Kg[/tex]

This value was to be expected since as the time passes the concentration will be lower due to the entrance to the pure water tank

To calculate the amount of salt in the tank after a certain amount of time, we need to consider the rate at which salt enters and leaves the tank. The rate of salt entering the tank is given as 60 L/min and the total volume of the tank is 6000 L. Using these values, we can find the rate of salt entering the tank in kg/min and then calculate the amount of salt in the tank after a specific time.

To calculate the amount of salt in the tank after a certain amount of time, we need to consider the rate at which salt enters and leaves the tank. The rate of salt entering the tank is given as 60 L/min and the total volume of the tank is 6000 L. Using these values, we can find the rate of salt entering the tank in kg/min:

Rate of salt entering = (60 L/min) * (16 kg/6000 L) = 0.16 kg/min

Therefore, the amount of salt in the tank after t minutes is given by:

y = 0.16 kg/min * t min = 0.16t kg

Answer:

a) [tex]df=n-1=16-1=15[/tex]

The statistic calculated is given by t=3.3  

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

[tex]p_v =P(t_{15}>3.3)=0.0024[/tex]  

So since the p value is lower than the significance level [tex]pv<\alpha[/tex] we reject the null hypothesis.

b) [tex]df=n-1=8-1=7[/tex]

The statistic calculated is given by t=1.8  

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

[tex]p_v =P(t_{7}>1.8)=0.057[/tex]  

So since the p value is higher than the significance level [tex]pv>\alpha[/tex] we FAIL to reject the null hypothesis.

c) [tex]df=n-1=26-1=25[/tex]

The statistic calculated is given by t=-0.6  

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

[tex]p_v =P(t_{25}>-0.6)=0.723[/tex]  

So since the p value is higher than the significance level [tex]pv>\alpha[/tex] we FAIL 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 standard deviation for the sample

[tex]n[/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)  

State the null and alternative hypotheses.      

We need to conduct a hypothesis in order to determine if the true mean is higher than 20, the system of hypothesis would be:      

Null hypothesis:[tex]\mu \leq 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{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".  

(a) n = 16, t = 3.3, a = 0.05, P-value =

First we need to calculate the degrees of freedom given by:  

[tex]df=n-1=16-1=15[/tex]

The statistic calculated is given by t=3.3  

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

[tex]p_v =P(t_{15}>3.3)=0.0024[/tex]  

So since the p value is lower than the significance level [tex]pv<\alpha[/tex] we reject the null hypothesis.

(b) n = 8, t = 1.8, a = 0.01, P-value =

First we need to calculate the degrees of freedom given by:  

[tex]df=n-1=8-1=7[/tex]

The statistic calculated is given by t=1.8  

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

[tex]p_v =P(t_{7}>1.8)=0.057[/tex]  

So since the p value is higher than the significance level [tex]pv>\alpha[/tex] we FAIL to reject the null hypothesis.

(c) n = 26,t = -0.6, P-value =

 First we need to calculate the degrees of freedom given by:  

[tex]df=n-1=26-1=25[/tex]

The statistic calculated is given by t=-0.6  

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

[tex]p_v =P(t_{25}>-0.6)=0.723[/tex]  

So since the p value is higher than the significance level [tex]pv>\alpha[/tex] we FAIL to reject the null hypothesis.

The P-value helps decide whether to reject the null hypothesis in a t-test. If the P-value is less than the significance level α, the null hypothesis is rejected. For each given scenario, the P-value is found from the t-distribution considering the provided t-statistic and degrees of freedom (n-1).

The problem is about conducting a t-test to check if the reflectometer reading (μ) for a new type of road paint is greater than a specified value (20). The P-value of the t-test will tell us if we should reject the null hypothesis H0: μ = 20 in favor of the alternative hypothesis Ha: μ > 20. The P-value is the probability of observing a t-score as extreme as the one calculated from the sample data (or more extreme), given that the null hypothesis is true.

(a) For n = 16, t = 3.3, and α = 0.05, we can use the t-distribution table or a statistical software to find the P-value. Since t is positive and we are dealing with a one-tailed test (because Ha: μ > 20), the P-value is the area to the right of the t-score (3.3) under the t-distribution. If the calculated P-value is less than α (0.05), we reject the null hypothesis.
(b) The same process applies for n = 8, t = 1.8, and α = 0.01. However, due to the smaller α level, we would need a larger t statistic (and thus a smaller P-value) to reject the null hypothesis.
(c) For n = 26, t = -0.6, if the calculated P-value is greater than the chosen α value, we do not reject the null hypothesis believing that the true mean reflectometer reading is 20.

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Answer:

c. Interval or ratio

Step-by-step explanation:

There exists the following measurement scales:

Nominal: A variable is linked to a number. For example, Buffalo Bills players, 27 is Tre'Davious White, 49 Tremanine Edmunds, and then on...

Ordinal: Ranks the intensity of something. For example, grading some pain on a 1 to 10 scale.

Interval or ratio: Represents quantity and has an equality of units. One example is the rates of unemployment.

Descriptive: Tries to attribute qualities to quantitative data. For example, rates of unemployment being classified as very low, low, medium, and then on...

So the correct answer is:

c. Interval or ratio

The unemployment rate is measured using an interval or ratio scale, represented as a percentage. Various methods and measures are considered for accurate calculation, though it has limitations in fully capturing unemployment's societal impact.

The reported unemployment rate of 5.5% of the population uses an interval or ratio scale for measurement. This type of scale is used because the unemployment rate is a percentage that represents a proportion of the population. The measurement of unemployment involves a ratio of two quantities: the number of unemployed individuals and the total labor force. Different methods such as Labor Force Sample Surveys, Official Estimates, Social Insurance Statistics, and Employment Office Statistics are used to calculate this figure. Moreover, the U.S. Bureau of Labor Statistics employs six different measures (U1 - U6) to capture various aspects of unemployment. While the rate is informative, there are shortcomings in how it represents the real impact on society, as it does not account for underemployment or those who have stopped looking for work. Understanding these statistics is crucial as unemployment has significant economic and social consequences, such as increasing inequality and potentially leading to civil unrest.

Answer:

Step-by-step explanation:

Answer: The number of scores between 21 and 25 is 2872

Step-by-step explanation:

Since the test scores are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - u)/s

Where

x = test scores

u = mean test score

s = standard deviation

From the information given,

u = 23

s = 4

We want to find the probability test scores between 21 points and 25. It is expressed as

P(21 lesser than or equal to x lesser than or equal to 25)

For x = 21,

z = (21 - 23)/4 = - 0.5

Looking at the normal distribution table, the probability corresponding to the z score is 0.30854

For x = 25,

z = (25 - 23)/4 = 0.5

Looking at the normal distribution table, the probability corresponding to the z score is 0.69146

P(21 lesser than or equal to x lesser than or equal to 25)

= 0.69146 - 0.30854 = 0.38292

The number of scores between 21 and 25 would be

0.38292 × 7500 = 2872

The requirements necessary for a normal probability distribution to be a standard normal probability​ distribution is that The mean and standard deviation have the values of mu equals 1and sigma equals 1. Hence the correct answer is A.

A probability distribution is one whose mean data points are symmetric. That is, most values from such a distribution cluster around the mean.

Another name for Normal Probability Distribution is Gaussian Distribution.

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Final answer:

A standard normal probability distribution requires a mean of 0 and a standard deviation of 1. The correct answer is B, which reflects these necessary conditions for standardization, allowing for the comparison of z-scores across different distributions.

Explanation:

To transform a normal probability distribution into a standard normal probability distribution, certain requirements must be met. Specifically, the distribution must have a mean (mu) of 0 and a standard deviation (sigma) of 1. Among the given options, the correct answer is B, where the mean and standard deviation have the values of mu equals 0 and sigma equals 1.

This standardization process allows any normal distribution to be compared on a common scale, and it is fundamental for calculating z-scores, which indicate how many standard deviations an element is from the mean.For example, if we have a normally distributed variable 'x' from a distribution with any mean µ and standard deviation o, the standardized value or z-score is calculated as follows:z = (x - µ) / o

Answer:

Step-by-step explanation:

Given

span of bridge [tex]L=1270\ ft[/tex]

height of span [tex]h=157\ ft[/tex]

Equation of Parabola

[tex]y=0.00039x^2[/tex]

[tex]|x|<635[/tex] i.e.

[tex]-635<x<635[/tex]

[tex]\frac{dy}{dx}=2\times 0.00039[/tex]

length of Arc[tex]=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}[/tex]

[tex]=\int_{-635}^{635}\sqrt{1+(\frac{dy}{dx})^2}[/tex]

[tex]=\int_{-635}^{635}\sqrt{1+(0.00078x)^2}[/tex]

[tex]=2\times \int_{0}^{635}\sqrt{1+(0.00078x)^2}[/tex]

[tex]=2\times (660.08)[/tex]

[tex]=1320.16\ ft[/tex]

The approximate length of the cables is approximately 4534.24 feet.

To approximate the length of the cables that stretch between the tops of the two towers of the suspension bridge, we can use the integral calculus to find the length of the curve defined by the equation [tex]\(y = 0.00039x^2\).[/tex]

The formula for finding the length of a curve between two points [tex]\([a, b]\)[/tex] is given by the integral:

[tex]\[L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx\][/tex]

Where:

- L is the length of the curve.

- a and b are the x-coordinates of the two points between which we want to find the length.

- f(x) is the function representing the curve.

- f'(x) is the derivative of the function.

In this case, we want to find the length of the cables between the two towers, which corresponds to the x-values from -635 to 635 since the width of the bridge is 1270 feet. The curve is defined by [tex]\(y = 0.00039x^2\)[/tex], so:

- a = -635

- b = 635

- [tex]\(f(x) = 0.00039x^2\)[/tex]

- [tex]\(f'(x) = 2 \cdot 0.00039x\)[/tex]

Now, let's calculate the length:

[tex]\[L = \int_{-635}^{635} \sqrt{1 + (2 \cdot 0.00039x)^2} \, dx\][/tex]

[tex]\[L = \int_{-635}^{635} \sqrt{1 + 0.0001536x^2} \, dx\][/tex]

Now, we can evaluate this integral:

[tex]\[L = \int_{-635}^{635} \sqrt{1 + 0.0001536x^2} \, dx \approx 4534.24\][/tex]

So, the approximate length of the cables that stretch between the tops of the two towers of the suspension bridge is approximately 4534.24 feet.

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Answer: The function is constant from x = -2 to x=1

Final answer:

The function is decreasing from x = 1 to x = 8, as the segment in that interval shows a decrease in y values as x increases.

Explanation:

To determine where the function is decreasing, we need to look at the segments provided. A function is decreasing if, as x increases, the y value of the function decreases. From the description of the segments, we can analyze behavior in each interval:

Therefore, the function is decreasing from x = 1 to x = 8.

Answer:

The firm's sample size must be of at least 171 adults.

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1-0.95}{2} = 0.025[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].

So it is z with a pvalue of [tex]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]

Now, find the margin of error M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the length of the sample.

In this problem, we have that:

[tex]M = 60, \sigma = 400[/tex]. So

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

[tex]60 = 1.96*\frac{400}{\sqrt{n}}[/tex]

[tex]60\sqrt{n} = 784[/tex]

[tex]\sqrt{n} = 13.07[/tex]

[tex]n = 170.7[/tex]

The firm's sample size must be of at least 171 adults.

Answer:

[tex]E[R][/tex] = 99 Ω

[tex]\sigma_R[/tex] = 2.3094 Ω

P(98<R<102) = 0.5696

Step-by-step explanation:

The mean resistance is the average of edge values of interval.

Hence,

The mean resistance, [tex]E[R] = \frac{a+b}{2}  = \frac{95+103}{2} = \frac{198}{2}[/tex] = 99 Ω

To find the standard deviation of resistance, we need to find variance first.

[tex]V(R) = \frac{(b-a)^2}{12} =\frac{(103-95)^2}{12} = 5.333[/tex]

Hence,

The standard deviation of resistance, [tex]\sigma_R = \sqrt{V(R)} = \sqrt5.333[/tex] = 2.3094 Ω

To calculate the probability that resistance is between 98 Ω and 102 Ω, we need to find Normal Distributions.

[tex]z_1 = \frac{102-99}{2.3094} = 1.299[/tex]

[tex]z_2 = \frac{98-99}{2.3094} = -0.433[/tex]

From the Z-table, P(98<R<102) = 0.9032 - 0.3336 = 0.5696

Answer:

The probability that can be applied to the given situation is Poisson distribution

Step-by-step explanation:

The probability that can be applied to the given situation is Poisson distribution because this distribution applied when the duration of occurrence of an event is known. In the given question laws have occurred every 100 feet therefore we have the number of events that have occurred. Moreover, Poisson distribution predicts the probability of events in fixed time interval

The probability distribution that has the greatest chance of applying to the number of blemishes or flaws occurring in each 100 feet of material in the textile industry is the Poisson distribution.

The probability distribution that has the greatest chance of applying to this situation is the Poisson distribution.

The Poisson distribution is commonly used to model the number of events that occur in a fixed interval of time or space. In this case, the manufacturer is interested in the number of blemishes or flaws occurring in each 100 feet of material, which can be seen as events occurring in a fixed space.

The Poisson distribution is appropriate when the events occur randomly and independently, and the average rate of occurrence is known. It allows for both discrete and continuous distributions.

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Answer:

E(A)= E[E(A|B)]= 4*0.6 +5*0.4 =4.4

Var(A)= E[Var(A|B)] +Var[E(X|Y)]]=4.4+19.6=24

Step-by-step explanation:

Previous concepts

The Poisson process is useful when we want to analyze the probability of ocurrence of an event in a time specified. The probability distribution for a random variable X following the Poisson distribution is given by:

[tex]P(X=x) =\lambda^x \frac{e^{-\lambda}}{x!}[/tex]

For this distribution the expected value is the same parameter [tex]\lambda[/tex]

[tex]E(X)=\mu =\lambda[/tex]

Other equations useful:

Let X and Y random variables

E(X) =E[E(X|Y)] (conditional expectation)

Var(X)=E[Var(X|Y)]+Var[E(X|Y)] (Total variance)

Solution to the problem

Let A the random variable that represent the number of winter storms next year

B a binary variable, B=1 if the next year is a good year and B=0 in the other case. Then we have this:

E(A|B=1) = 4  and E(A|B=0)=5

We can use the propoerties of conditional expectation like this:

E(A)= E[E(A|B)]= E(A|B=1)P(B=1) +E(A|B=0)P(B=0)

E(A)= E[E(A|B)]= 4*0.6 +5*0.4 =4.4

And we can use also the properties for conditional variance we have the following values:

Var(A|B=1)=4 Var(A|B=0)=5, by the propertis of the Poisson distribution

And then the conditional variance is givne by:

[tex]Var[E(A|B)]= E(A|B=1)^2 P(B=1) +E(A|B=0)^2 P(B=0)[/tex]

And if we replace we got:

[tex]Var[E(A|B)]= 4^2 *0.6 +5^2 *0.4 =19.6[/tex]

And we have also that the expected value for the conditional variance is given by:

E[Var(A|B)]= 4*0.6 +5*0.4 =4.4

And then finally the variance for the random variable A is given by:

Var(A)= E[Var(A|B)] +Var[E(X|Y)]]=4.4+19.6=24

The expected value and variance in the number of storms are 4.4 and 4.4 respectively.

To calculate the expected number of winter storms, we can use the probability of a good or bad year along with their respective average storms. The expected value is then given by:

E(X) = (0.6 * 4) + (0.4 * 5) = 2.4 + 2 = 4.4 storms.

For the variance, since the variance of a Poisson distribution is equal to its mean, we have:

Variance in a good year = 4

Variance in a bad year = 5

The overall variance is a weighted average:

V(X) = (0.6 * 4) + (0.4 * 5) = 2.4 + 2 = 4.4

Answer:the variable expression to represent the sum of their circumferences would be

(8xy)cm + (5xy)cm

Step-by-step explanation:

The circumference of the first circle is 8xy cm

The circumference of the second circle is 5xy cm

the variable expression to represent the sum of their circumferences would be

(8xy)cm + (5xy)cm

This variable expression can also be simplified further because both terms contain xy.

Answer:

Student needs pens= n = 27.04

Rounding off with upper floor function ⇒ n =28

Rounding off with lower floor function ⇒ n =27

Step-by-step explanation:

Given that lifetime of each pen is a exponential random variable with mean 1 week.

Let [tex]S_{n}[/tex] be total sum of lifetime of n pens.

So mean of [tex]S_{n}[/tex] = μ = n.1

Standard deviation of [tex]S_{n}[/tex] =[tex]\sigma=\sqrt{n}[/tex]

Probability that he doesnot run out of pens= 0.99

Considering Sn be sum of n lifetimes, using central limit theorem

[tex]\frac{S_{n}-n}{\sqrt{n}}\approx N(0,1)\\\\P(S_{n}>15)=[P(\frac{S_{n}-n}{\sqrt{n}})>\frac{15-n}{\sqrt{n}}]\\ 1-\phi(\frac{15-n}{\sqrt{n}})=\phi(-(\frac{15-n}{\sqrt{n}}))=0.99\\[/tex]

From table of standard normal distribution

[tex]\frac{15-n}{\sqrt{n}}=-2.3263\\15-n=-2.3263\sqrt{n}\\n-2.3263-15\sqrt{n}[/tex]

Solving the quadratic Equation in variable x we get

n=27.04

Answer:  c)[50,60]

Step-by-step explanation:

The Empirical rule says that , About 68% of the population lies with the one standard deviation from the mean (For normally distribution).

We are given that , The heights of students in a class are normally distributed with mean 55 inches and standard deviation 5 inches.

Then by Empirical rule, about 68% of the heights of students lies between one standard deviation from mean.

i.e. about 68% of the heights of students lies between [tex]\text{Mean}\pm\text{Standard deviation}[/tex]

i.e. about 68% of the heights of students lies between [tex]55\pm5[/tex]

Here, [tex]55\pm5=(55-5, 55+5)=(50,60)[/tex]

i.e.  The required interval that contains the middle 68% of the heights. = [50,60]

Hence, the correct answer is c) (50,60)

Final answer

The interval containing the middle 68 of a typically distributed class height is one standard divagation from the mean, which is( 50, 60) elevation for the given mean of 55 elevation and standard divagation of 5 elevation.

Explanation

The Empirical Rule countries that for a typically distributed set of data, roughly 68 of data values will fall within one standard divagation of the mean, 95 within two standard diversions, and99.7 within three standard diversions. In this case, the mean height is 55 elevation and the standard divagation is 5 elevation. thus, to find the interval that contains the middle 68 of the heights, we add and abate one standard divagation from the mean.

55 elevation 5 elevation = 60 elevation( Mean height plus one standard divagation)

55 elevation- 5 elevation = 50 elevation( Mean height minus one standard divagation)

This means the interval that contains the middle 68 of the heights is( 50, 60) elevation. Hence, the correct answer is option( c).

Answer:

C.

Step-by-step explanation:

Hypothesis testing procedure:

Hypothesis:

The null hypothesis will be the variances of sales of two musical stores are equal and the alternative hypothesis will be the variances of sales of two musical stores are not equal

Level of significance: alpha=0.05

Test statistic: F=variance A/variance B=(30)^2/(20)^2=900/400=2.25

P-value: p=0.107

As the alternative hypothesis mentioned that the variances are not equal this leads to two tailed test. so the p-value is calculated using excel function 2*F.DIST.RT(2.25,24,15).

Conclusion:

The p-value seems to exceed the alpha=0.05 and this depicts that the null hypothesis should not be rejected.

At 95% confidence, the null hypothesis should not be rejected. The correct answer is option c. should not be rejected.

Step 1

To test whether the variances of the sales at music stores A and B are equal, we perform an F-test. The null hypothesis [tex](\(H_0\))[/tex] states that the variances are equal, while the alternative hypothesis [tex](\(H_a\))[/tex] states that they are not equal.

The F-statistic is calculated as the ratio of the larger sample variance to the smaller sample variance:

[tex]\[ F = \frac{s_1^2}{s_2^2} \][/tex]

Where [tex]\( s_1^2 \)[/tex] is the variance of store A and [tex]\( s_2^2 \)[/tex] is the variance of store B.

Step 2

In this case, the F-statistic is [tex]\( \frac{30^2}{20^2} = \frac{900}{400} = 2.25 \)[/tex].

Using a significance level of 0.05 and degrees of freedom (df) as [tex]\( n_1 - 1 \)[/tex] and [tex]\( n_2 - 1 \)[/tex], we compare this value to the critical F-value from the F-distribution table.

Since the calculated F-statistic of 2.25 is less than the critical F-value, we fail to reject the null hypothesis.

Therefore, at 95% confidence, the correct answer is c. should not be rejected.

Answer:

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

Or in the alternative way would be:

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

Step-by-step explanation:

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".  

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".

On this case the claim that they want to test is: "The means for the two groups (American Idol and 60 Minutes) is the same". So we want to check if we have significant differences between the two means, so this needs to be on the alternative hypothesis and on the null hypothesis we need to have the complement of the alternative hypothesis.

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

This null hypothesis can be expressed like this:

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

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

Or in the alternative way would be:

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

Final answer:

Lucy Baker's alternate hypothesis for her demographic analysis of American Idol and 60 Minutes would suggest a difference in the mean ages of the two audiences, represented as either (Ha: µ₁ ≠ µ₂), (Ha: µ₁ < µ₂), or (Ha: µ₁ > µ₂), depending on the direction of the difference she anticipates.

Explanation:

Lucy Baker's hypothesis test within her demographic analysis involves comparing the mean ages of audiences of two television programs: American Idol and 60 Minutes. This is a hypothesis test for two independent sample means, assuming that population standard deviations are unknown and that the samples are random.

Given that the null hypothesis (
H) posits no difference in the mean ages of the two audiences (
µ₁ = µ₂), the alternate hypothesis (
Ha) Lucy should consider would suggest that there is a difference. Her alternate hypothesis could be that the mean age of one audience is either higher or lower than the other, which can be denoted as either (
Ha: µ₁ ≠ µ₂). However, if she has a specific direction in mind (e.g., assuming one program's audience is younger than the other), she might opt for (
Ha: µ₁ < µ₂) or (
Ha: µ₁ > µ₂), accordingly.

It's crucial to select an appropriate alternate hypothesis as it signifies the anticipated outcome that stands opposed to the assumption of the null hypothesis. In this case, given that the previous study indicates no difference, Lucy's alternative hypothesis should represent the possibility that a difference indeed exists.

Answer:

a)[tex]\chi^2 = \frac{(40-35)^2}{35}+\frac{(35-40)^2}{40}+\frac{(25-25)^2}{25}+\frac{(25-24.5)^2}{24.5}+\frac{(35-28)^2}{28}+\frac{(25-17.5)^2}{17.5}+\frac{(5-10.5)^2}{10.5}+\frac{(10-12)^2}{12}+\frac{(15-7.5)^2}{7.5} =17.03[/tex]

[tex]p_v = P(\chi^2_{4} >17.03)=0.0019[/tex]

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(17.03,4,TRUE)"

Since the p value is lower than the significance level we can reject the null hypothesis at 5% of significance, and we can conclude that we have association or dependence between the two variables.

b)

P(E|Ex)= P(EΛEx )/ P(Ex) = (40/215)/ (70/215)= 40/70=0.5714

P(E|Gx)= P(EΛGx )/ P(Gx) = (35/215)/ (80/215)= 35/80=0.4375

P(E|Fx)= P(EΛFx )/ P(Fx) = (25/215)/ (50/215)= 25/50=0.5

P(G|Ex)= P(GΛEx )/ P(Ex) = (25/215)/ (70/215)= 25/70=0.357

P(G|Gx)= P(GΛGx )/ P(Gx) = (35/215)/ (80/215)= 35/80=0.4375

P(G|Fx)= P(GΛFx )/ P(Fx) = (10/215)/ (50/215)= 10/50=0.2

P(F|Ex)= P(FΛEx )/ P(Ex) = (5/215)/ (70/215)= 5/70=0.0714

P(F|Gx)= P(FΛGx )/ P(Gx) = (10/215)/ (80/215)= 10/80=0.125

P(F|Fx)= P(FΛFx )/ P(Fx) = (15/215)/ (50/215)= 15/50=0.3

And that's what we see here almost all the conditional probabilities are higher than 0.2 so then the conclusion of dependence between the two variables makes sense.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

Quality management        Excellent      Good     Fair    Total

Excellent                                40                35         25       100

Good                                      25                35         10         70

Fair                                         5                   10          15        30

Total                                       70                 80         50       200

Part a

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is independence between the two categorical variables

H1: There is association between the two categorical variables

The level of significance assumed for this case is [tex]\alpha=0.05[/tex]

The statistic to check the hypothesis is given by:

[tex]\chi^2 = \sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]

The table given represent the observed values, we just need to calculate the expected values with the following formula [tex]E_i = \frac{total col * total row}{grand total}[/tex]

And the calculations are given by:

[tex]E_{1} =\frac{70*100}{200}=35[/tex]

[tex]E_{2} =\frac{80*100}{200}=40[/tex]

[tex]E_{3} =\frac{50*100}{200}=25[/tex]

[tex]E_{4} =\frac{70*70}{200}=24.5[/tex]

[tex]E_{5} =\frac{80*70}{200}=28[/tex]

[tex]E_{6} =\frac{50*70}{200}=17.5[/tex]

[tex]E_{7} =\frac{70*30}{200}=10.5[/tex]

[tex]E_{8} =\frac{80*30}{200}=12[/tex]

[tex]E_{9} =\frac{50*30}{200}=7.5[/tex]

And the expected values are given by:

Quality management        Excellent      Good     Fair       Total

Excellent                                35              40          25         100

Good                                      24.5           28          17.5        85

Fair                                         10.5            12           7.5         30

Total                                       70                 80         65        215

And now we can calculate the statistic:

[tex]\chi^2 = \frac{(40-35)^2}{35}+\frac{(35-40)^2}{40}+\frac{(25-25)^2}{25}+\frac{(25-24.5)^2}{24.5}+\frac{(35-28)^2}{28}+\frac{(25-17.5)^2}{17.5}+\frac{(5-10.5)^2}{10.5}+\frac{(10-12)^2}{12}+\frac{(15-7.5)^2}{7.5} =17.03[/tex]

Now we can calculate the degrees of freedom for the statistic given by:

[tex]df=(rows-1)(cols-1)=(3-1)(3-1)=4[/tex]

And we can calculate the p value given by:

[tex]p_v = P(\chi^2_{4} >17.03)=0.0019[/tex]

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(17.03,4,TRUE)"

Since the p value is lower than the significance level we can reject the null hypothesis at 5% of significance, and we can conclude that we have association or dependence between the two variables.

Part b

We can find the probabilities that Quality of Management and the Reputation of the Company would be the same like this:

Let's define some notation first.

E= Quality Management excellent     Ex=Reputation of company excellent

G= Quality Management good     Gx=Reputation of company good

F= Quality Management fait     Ex=Reputation of company fair

P(EΛ Ex) =40/215=0.186

P(GΛ Gx) =35/215=0.163

P(FΛ Fx) =15/215=0.0697

If we have dependence then the conditional probabilities would be higher values.

P(E|Ex)= P(EΛEx )/ P(Ex) = (40/215)/ (70/215)= 40/70=0.5714

P(E|Gx)= P(EΛGx )/ P(Gx) = (35/215)/ (80/215)= 35/80=0.4375

P(E|Fx)= P(EΛFx )/ P(Fx) = (25/215)/ (50/215)= 25/50=0.5

P(G|Ex)= P(GΛEx )/ P(Ex) = (25/215)/ (70/215)= 25/70=0.357

P(G|Gx)= P(GΛGx )/ P(Gx) = (35/215)/ (80/215)= 35/80=0.4375

P(G|Fx)= P(GΛFx )/ P(Fx) = (10/215)/ (50/215)= 10/50=0.2

P(F|Ex)= P(FΛEx )/ P(Ex) = (5/215)/ (70/215)= 5/70=0.0714

P(F|Gx)= P(FΛGx )/ P(Gx) = (10/215)/ (80/215)= 10/80=0.125

P(F|Fx)= P(FΛFx )/ P(Fx) = (15/215)/ (50/215)= 15/50=0.3

And that's what we see here almost all the conditional probabilities are higher than 0.2 so then the conclusion of dependence between the two variables makes sense.

Answer:

[tex]z=\frac{(23-18)-0}{\sqrt{\frac{4^2}{4}+\frac{5^2}{4}}}}=1.5617[/tex]  

[tex]p_v =P(Z>1.5617)=0.059[/tex]

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the difference between the true mean of group A and B is not significantly higher than 0 at 5% of significance.

Step-by-step explanation:

Data given and notation

[tex]\bar X_{A}=23[/tex] represent the mean for the sample A

[tex]\bar X_{B}=18[/tex] represent the mean for the sample B

[tex]\sigma_{A}=4[/tex] represent the population standard deviation for the sample A

[tex]\sigma_{B}=5[/tex] represent the population standard deviation for the sample B

[tex]n_{A}=4[/tex] sample size selected A

[tex]n_{B}=4[/tex] sample size selected B

[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

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

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean for A is higher than the mean for B, the system of hypothesis would be:

Null hypothesis:[tex]\mu_{A}-\mu_{B}\leq 0[/tex]

Alternative hypothesis:[tex]\mu_{A}-\mu_{B}>0[/tex]

We know the population deviations, so for this case is better apply a z test to compare means, and the statistic is given by:

[tex]z=\frac{(\bar X_{A}-\bar X_{B})-0}{\sqrt{\frac{\sigma^2_{A}}{n_{A}}+\frac{\sigma^2_{B}}{n_{B}}}}[/tex] (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

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

[tex]z=\frac{(23-18)-0}{\sqrt{\frac{4^2}{4}+\frac{5^2}{4}}}}=1.5617[/tex]  

P-value

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

[tex]p_v =P(Z>1.5617)=0.059[/tex]

Conclusion

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the difference between the true mean of group A and B is not significantly higher than 0 at 5% of significance.

To determine the appropriate conclusion at the 5% significance level, conduct a hypothesis test for the difference in means between the two fertilizer distributors.

To determine the appropriate conclusion at the 5% significance level, we need to conduct a hypothesis test for the difference in means between the two fertilizer distributors. The test statistic calculated in Excel is 1.5617. We compare this test statistic to the critical value of the t-distribution at the desired significance level of 5% with 6 degrees of freedom (8 samples - 2). If the test statistic is greater than the critical value, we reject the null hypothesis that the means are equal and conclude that there is evidence to suggest that distributor A's fertilizer contains more nitrogen than distributor B's.

https://brainly.com/question/34171008

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Answer:

[tex]work \ done= 45937.5[/tex]

Step-by-step explanation:

Work done is given by

[tex]work \ done=\int_a^b w(d-x) \ dx[/tex] , where d = length of cable and w = weight of cable.

Here, d = 175 ft and w = 3 lb/ft

Now, [tex]work \ done=\int_0^{175} 3(175-x) \ dx[/tex]

[tex]work \ done= 3\left [175x-\frac{x^2}{2}  \right ]_0^{175}[/tex]

[tex]work \ done= 3\left [175^2-\frac{175^2}{2}  \right ][/tex]

[tex]work \ done= 3\cdot \frac{175^2}{2}[/tex]

[tex]work \ done= 45937.5[/tex]

Answer:

0.02% probability that the average amount billed on the sample bills is greater than $500.

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 = 465, \sigma = 300, n = 900, s = \frac{300}{\sqrt{900}} = 10[/tex].

What is the probability that the average amount billed on the sample bills is greater than $500?

This probability is 1 subtracted by the pvalue of Z when [tex]X = 500[/tex]. So

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

[tex]Z = \frac{500 - 465}{10}[/tex]

[tex]Z = 3.5[/tex]

[tex]Z = 3.5[/tex] has a pvalue of 0.9998.

So there is a 1-0.9998 = 0.0002 = 0.02% probability that the average amount billed on the sample bills is greater than $500.

The probability that the average amount billed in a sample of 900 bills is greater than $500 is approximately 0.0002.

Given the mean amount billed is $465

standard deviation of $300,

sample size of 900,

the probability that the average amount billed exceeds $500.

First, we need to calculate the standard error of the mean (SEM):

[tex]SEM = \(\frac{\sigma}{\sqrt{n}}\)[/tex]

Substituting the given values:

[tex]SEM = \frac{300}{\sqrt{900}} = \frac{300}{30} = 10[/tex]

Now, we can use the Z-score formula to find the probability.

The Z-score is calculated as follows:

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

(X = 500),

mu = 465

SEM=10

[tex]Z = \frac{500 - 465}{10} = 3.5[/tex]

Using Z-tables or a standard normal distribution calculator, we can find the probability corresponding to a Z-score of 3.5:

P(Z > 3.5) ≈ 0.0002

Therefore, the probability that the average amount billed in a sample of 900 bills is greater than $500 is approximately 0.0002 (rounded to four decimal places).

Answer:The margin of error is 0.01323 and 95% confidence interval is (0.674,0.73).

Step-by-step explanation:

Since we have given that

p = 0.70

n = 1200

We need to find the margin of error;

Margin of error would be

[tex]\sqrt{\dfrac{p(1-p)}{n}}\\\\=\sqrt{\dfrac{0.7\times 0.3}{1200}}\\\\=0.01323[/tex]

At 95% confidence level, α = 1.96

so, 95% confidence interval would be

[tex]p\pm z\times 0.01323\\\\=0.7\pm 1.96\times 0.01323\\\\=0.7\pm 0.02593\\\\=(0.7-0.02593,0.7+0.02593)\\\\=(0.674,0.73)[/tex]

Hence, the margin of error is 0.01323 and 95% confidence interval is (0.674,0.73).

Answer:

The key difference is that the dependent sample test uses usually the same individuals to obtain the info for two different moments. And the independent sample test uses two different groups in order to compare a parameter on specific, usually the mean.

Step-by-step explanation:

A paired t-test is used to compare two population means when we have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations in order to see an improvement or not for a method we can use it ( Without treatment and With specific treatment).  

An independent sample test is used when we want to compare "two sample means to determine whether the population means are significantly different". For example if we want to compare the scores for male and female in a test.  

The key difference is that the dependent sample test uses usually the same individuals to obtain the info for two different moments. And the independent sample test uses two different groups in order to compare a parameter on specific, usually the mean.

When conducting hypothesis testing, the choice between using a related sample or an independent sample depends on the nature of the data being analyzed. A related sample is used when the two samples are dependent or paired, while an independent sample is used when the two samples are not related and can be considered as separate groups.

When conducting hypothesis testing, the choice between using a related sample or an independent sample depends on the nature of the data being analyzed. A related sample is used when the two samples are dependent or paired, meaning that there is a one-to-one correspondence between the data points in the two samples. An independent sample is used when the two samples are not related and can be considered as separate groups.

For example, in a related sample scenario, you might compare the blood pressure of the same group of individuals before and after a treatment. The paired samples would be the pre-treatment and post-treatment measurements of each individual. In an independent sample scenario, you might compare the test scores of two different groups of students who were taught using different teaching methods.

Answer:

The relation is antisymmetric and transitive

Step-by-step explanation:

Let a,b,c be elements of the set of all people.

1) Let a be a person who is 20 years old. aRa means that this person is younger than themselves, which it's false because 20<20 is false. Then R is not reflexive.

2) Let a be a person who is 20 years old and b a person who is 30 years old. Then a is younger than b, that is, aRb.

However, it is not true that b is younger than a, as 30<20 is false, therefore bRa is false and R is not symmetric.

3) Suppose that aRb, so that a is younger than b. Then, b is not younger than a. If n denotes the age of a and m denotes the age of b, we have that n<m which implies that m<n is false. Then bRa is false, thus R is antisymmetric.

4) Suppose that aRb and bRc. Let n,m,p denote the ages of a,b,c respectively. Then n<m and m<p (a is younger than b and b is younger than c), and by transitivity of the ordering of numbers, n<p, that is, a is younger than c. Thus aRc, and R is transitive.

Answer:

There is sufficient evidence to conclude that child booster seats meet the specific requirement.

Step-by-step explanation:

Sample: 697, 759, 1266, 621, 569, 432

Formula:

[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]  

where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.  

[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]

[tex]Mean =\displaystyle\frac{4344}{6} = 724[/tex]

Sum of squares of differences = 415616

[tex]S.D = \sqrt{\frac{415616}{5}} = 288.31[/tex]

We are given the following in the question:  

Population mean, μ = 1000 hic

Sample mean, [tex]\bar{x}[/tex] = 724

Sample size, n = 16

Alpha, α = 0.05

Sample standard deviation, s = 288.31

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 1000\text{ hic}\\H_A: \mu < 1000\text{ hic}[/tex]

We use one-tailed(left) t test to perform this hypothesis.

Formula:

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

Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{724 - 1000}{\frac{288.31}{\sqrt{6}} } = -2.344[/tex]

Now, [tex]t_{critical} \text{ at 0.05 level of significance, 5 degree of freedom } = -2.015[/tex]

Calculation the p-value from table,

P-value = 0.033

Since,                  

Since, the p value is lower than the significance level, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

We conclude that the measurement is less than 1000 hic.

Thus, there is sufficient evidence to conclude that child booster seats meet the specific requirement.