Babcock and Marks (2010) reviewed survey data from 2003–2005 and obtained an average of μ = 14 hours per week spent studying by full-time students at four-year colleges in the United States. To determine whether this average has changed in the past 10 years, a researcher selected a sample of n = 64 of today’s college students and obtained an average of M = 12.5 hours. If the standard deviation for the distribution is σ = 4.8 hours per week, does this sample indicate a significant change in the number of hours spent studying?

Answer:

the answer D=1.44

Step-by-step explanation:

if the X= number of red cards observed in 6 trials , since each card observation is independent from the others and the sampling process is done with replacement ( the card is observed, then returned and reshuffled) , X follows an binomial distribution.

X(x)= n!/((n-x)!*x!) *p^x *(1-p)^(n-x)

where n = number of trials = 6 , x= number of red cards observed , p= probability of obtaining a red card in one try

the probability of obtaining the card in one try is

p = number of red cards / total number of cards = 12/ (12+8) = 0.6

since we know that X has a binomial distribution, the variance of this kind of distribution is

variance = σ² = n * p * (1-p)

therefore the variance of X is

variance = σ² = n * p * (1-p) = 6 * 0.6 * (1-0.6) = 1.44

Answer with Step-by-step explanation:

We are given that a set {a,b,c,d}

S={(a,b),(a,c),(c,d),(c,a)}

R={(b,c),(c,b),(a,d),(d,b)]

Composition of relation:Let R and S are two relations on the given set

If ordered pair (a,b) belongs to relation R and (b,c) belongs to S .

Then, SoR={(a,c)}

By using this rule

SoR={(b,d),(b,a)}[/tex]

Because [tex](b,c)\in R[/tex] and [tex](c,d)\in S[/tex].Thus, [tex](b,d)\in SoR[/tex]

[tex](b,c))\in R[/tex] and [tex](c,a)\in S[/tex].Thus, [tex](b,a)\in SoR[/tex]

b.RoS={(a,c),(a,b),(c,b),(c,d)}

Because

[tex](a,b)\in S,(b,c)\in R[/tex] .Therefore, the ordered pair [tex](a,c)\in[/tex] RoS

[tex](a,c)\in S,(c,b)\in R[/tex] .Thus, [tex](a,b)\in RoS[/tex]

[tex](c,d)\in S,(d,b)\in R[/tex].Thus, [tex](c,b)\in RoS[/tex]

[tex](c,a)\in S,(a,d)\in R[/tex].Thus,[tex](c,d)\in RoS[/tex]

c.SoS={(a,d),(a,a),(c,c),(c,b)}

Because

[tex](a,c)\;and\; (c,d)\in S[/tex].Thus, [tex](a,d)\in SoS[/tex]

[tex](c,a),(a,b)\in S[/tex].Thus,[tex](c,b)\in SoS[/tex]

[tex](a,c)\in S[/tex] and [tex](c,a)\in S[/tex].Thus,[tex](a,a)\in SoS[/tex]

[tex](c,a)\in S[/tex] and [tex](a,c)\in S[/tex].Thus ,[tex](c,c)\in SoS[/tex]

Answer:

True

Step-by-step explanation:

The answer is "True" because the data set was picked at random.

Answer:

a) Null hypothesis:[tex]\mu_{s}-\mu_{n}\geq 10[/tex]

Alternative hypothesis:[tex]\mu_{s} - \mu_{n}<10[/tex]

[tex]p_v =P(Z<-1.904)=0.0284[/tex]

Comparing the p value with the significance level [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and the difference between the two groups is significantly lower than 10.

b) The 90% confidence interval would be given by [tex]-21.035 \leq \mu_1 -\mu_2 \leq 7.685[/tex]

c) If we analyze the interval obtained we see that the interval contains the value of -10 so then we agree with the result obtained from the hypothesis, that he alternative hypothesis is true.

Step-by-step explanation:

Data given and notation

[tex]\bar X_{s}=750.2[/tex] represent the mean for the sample standard  process

[tex]\bar X_{n}=756.875[/tex] represent the mean for the sample with the new process

[tex]s_{s}=19.128[/tex] represent the sample standard deviation for the sample standard process

[tex]s_{n}=21.283[/tex] represent the sample standard deviation for the new process

[tex]\sigma_s=\sigma_n=\sigma=20[/tex] represent the population standard deviation for both samples.

[tex]n_{s}=15[/tex] sample size for the group Cincinnati

[tex]n_{n}=8[/tex] sample size for the group Pittsburgh

z would represent the statistic (variable of interest)

Concepts and formulas to use

(a) Formulate and test an appropriate hypothesis using a 0.10. What are your conclusions?

We need to conduct a hypothesis in order to check if the difference in mean batch viscosity is 10 or less system of hypothesis would be:

Null hypothesis:[tex]\mu_{s}-\mu_{n}\geq 10[/tex]

Alternative hypothesis:[tex]\mu_{s} - \mu_{n}<10[/tex]

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

[tex]z=\frac{(\bar X_{s}-\bar X_{n})-\Delta}{\sqrt{\frac{\sigma^2_{s}}{n_{s}}+\frac{\sigma^2_{n}}{n_{n}}}}[/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

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

[tex]z=\frac{(750.2-756.875)-10}{\sqrt{\frac{20^2}{15}+\frac{20^2}{8}}}}=-1.904[/tex]  

Statistical decision

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

[tex]p_v =P(Z<-1.904)=0.0284[/tex]

Comparing the p value with the significance level [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and the difference between the two groups is significantly lower than 10.

(b) Find a 90% confidence interval on the difference in mean batch viscosity resulting from the process change.

The confidence interval for the difference of means is given by the following formula:  

[tex](\bar X_s -\bar X_n) \pm z_{\alpha/2}\sqrt{\sigma^2(\frac{1}{n_s}+\frac{1}{n_s})}[/tex] (1)  

The point of estimate for [tex]\mu_1 -\mu_2[/tex] is just given by:

[tex]\bar X_s -\bar X_n =750.2-756.875=-6.675[/tex]

Since the 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]  

The standard error is given by the following formula:

[tex]SE=\sqrt{\sigma^2(\frac{1}{n_s}+\frac{1}{n_n})}[/tex]

And replacing we have:

[tex]SE=\sqrt{20^2(\frac{1}{15}+\frac{1}{8})}=8.756[/tex]

Confidence interval

Now we have everything in order to replace into formula (1):  

[tex]-6.675-1.64\sqrt{20^2(\frac{1}{15}+\frac{1}{8})}=-21.035[/tex]  

[tex]-6.675+1.64\sqrt{20^2(\frac{1}{15}+\frac{1}{8})}=7.685[/tex]  

So on this case the 90% confidence interval would be given by [tex]-21.035 \leq \mu_1 -\mu_2 \leq 7.685[/tex]

(c) Compare the results of parts (a) and (b) and discuss your findings.

If we analyze the interval obtained we see that the interval contains the value of -10 so then we agree with the result obtained from the hypothesis, that he alternative hypothesis is true.

Final answer:

The viscosity analysis involves a hypothesis test to check for significant changes in mean batch viscosity and constructing a confidence interval to estimate the range of this difference following a process change in the manufacturing of a polymer.

Explanation:

The question involves conducting a hypothesis test and constructing a confidence interval to analyze the change in mean batch viscosity of a polymer before and after a process change. A hypothesis test will be used to determine if the mean batch viscosity has changed significantly, and the confidence interval will provide a range in which the true difference in means likely falls.

Hypothesis Test:

Calculate the mean of both sets of viscosity measurements.

Set up the null hypothesis (H0) and the alternative hypothesis (H1): H0: μ1 - μ2 <= 10 (no significant change), H1: μ1 - μ2 > 10 (significant change).

Determine the test statistic using a t-test for two independent samples.

Calculate the p-value associated with the test statistic.

Compare the p-value with the significance level (α = 0.10). If p <= α, reject H0; otherwise, fail to reject H0.

Calculate the standard deviation of both sets of measurements.

Determine the standard error of the difference in means.

Find the appropriate t-distribution critical value based on the given confidence level (90%) and degrees of freedom.

Calculate the 90% confidence interval using the difference in means ± the margin of error.

Compare the results of the hypothesis test and the confidence interval. If the hypothesis test leads to rejection of the null hypothesis while the confidence interval for the difference in means includes values greater than 10, there is evidence suggesting a significant change in mean batch viscosity. However, if the hypothesis test does not lead to rejection of H0 and the confidence interval includes 10, it suggests that the process change did not have a significant effect on mean batch viscosity.

Answer:

0.546 , -4.71

Step-by-step explanation:

Given:

An angle's initial ray points in the 3-o'clock direction and its terminal ray rotates counter -clock wise.

Here, Slope = tan\theta

If θ = 0.5

Then, Slope = tan(θ) = tan(0.5) = 0.546

If θ = 1.78

Then, Slope = tan(θ) = tan(1.78) = - 4.71

The expression (in terms of θ) that represents the varying slope of the terminal ray.

Slope = m = tanθ, where θ is the varying angle

A) The slope of the terminal ray when θ = 0.5 radians is; 0.5463

B) The slope of the terminal ray when θ = 1.78 radians is; -4.71

C) The expression that will represent the varying slope of the terminal ray is;

tan (Δy/Δx)

We are given that θ represents the angle's varying measure (in radians).

Now, in mathematics, slope is simply the tangent of an angle. Thus;

A) At θ = 0.5 radians ,

Slope of terminal ray = tan θ

Slope = tan 0.5

Using radians calculator, tan 0.5 = 0.5463

Thus, slope = 0.5463

B) At θ = 1.78

Slope of terminal ray = tan θ

Slope = tan 1.78

Using radians calculator, tan 1.78 = -4.71

Thus, slope = -4.71

C) The expression that will represent the varying slope of the terminal ray is;

tan (Δy/Δx)

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Answer:

(a) The p-value is 0.9554

(b) We cannot reject the null hypothesis at the significance level of 0.06

Step-by-step explanation:

We are dealing with a lower-tail alternative [tex]H_{1}: p < 0.4[/tex]. Because the test statistic is z = +1.7 which comes from a normal distribution, the p-value is the probability of getting a value as extreme as the already observed.  

(a) P(Z < +1.7) = 0.9554

(b) The p-value is very large, and 0.9554 > 0.06, so, we cannot reject the null hypothesis at the significance level of 0.06

Answer:

There is no sufficient evidence to reject the company's claim at the significance level of 0.05

Step-by-step explanation:

Let [tex]\mu[/tex] be the true mean weight per apple the company ship.  We want to test the next hypothesis

[tex]H_{0}:\mu=120[/tex] vs [tex]H_{1}:\mu\neq 120[/tex] (two-tailed test).

Because we have a large sample of size n = 49 apples randomly selected from a shipment, the test statistic is given by

[tex]Z=\frac{\bar{X}-120}{\sigma/\sqrt{n}}[/tex] which is normally distributed. The observed value is  

[tex]z_{0}=\frac{122.5-120}{12/\sqrt{49}}=1.4583[/tex]. The rejection region for [tex]\alpha = 0.05[/tex] is given by RR = {z| z < -1.96 or z > 1.96} where the area below -1.96 and under the standard normal density is 0.025; and the area above 1.96 and under the standard normal density is 0.025 as well. Because the observed value 1.4583 does not fall inside the rejection region RR, we fail to reject the null hypothesis.

To determine whether there is sufficient evidence to reject the company's claim, we can perform a hypothesis test.

To determine whether there is sufficient evidence to reject the company's claim, we can perform a hypothesis test. The null hypothesis, H0, is that the mean weight per apple is 120 grams, and the alternative hypothesis, Ha, is that the mean weight per apple is not 120 grams. With a sample of 49 apples, we can calculate the test statistic using the formula:

t = (sample mean - population mean) / (standard deviation / sqrt(sample size))

Once we have the test statistic, we can compare it to the critical value from the t-distribution to determine if we reject or fail to reject the null hypothesis. As per calculation, yhe area below -1.96 and under the standard normal density is 0.025; and the area above 1.96 and under the standard normal density is 0.025 as well.

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Final answer:

Null Hypothesis (H0): P >= 0.80, Alternative Hypothesis (Ha): P < 0.80, Test Statistic (z): -1.1112, P-value: 0.1335, Conclusion: Fail to reject H0.

Explanation:

Null Hypothesis (H0): P ≥ 0.80

Alternative Hypothesis (Ha): P < 0.80

Test Statistic (z): -1.1112

P-value: 0.1335

Conclusion: Fail to reject H0. There is not sufficient evidence to support the claim that polygraph test results are correct less than 80% of the time.

Answer:

As x approaches 1 from the left, the value of [tex]arcsin(x)[/tex] is [tex]\frac{\pi }{2}[/tex]

Step-by-step explanation:

To find the value of [tex]arcsin(x)[/tex] when x approaches 1 from the left, you can use the graph of the function [tex]arcsin(x)[/tex]

Examine what happens as x approaches from the left.

As x approaches 1 from the left, the function seems to be approaching [tex]\frac{\pi }{2}[/tex]

Therefore, as x approaches 1 from the left, the value of [tex]arcsin(x)[/tex] is [tex]\frac{\pi }{2}[/tex]

As x approaches 1 from the left, the value of arcsin(x) approaches π/2 radians.

The expression x → 1− refers to the limit as x approaches 1 from the left. For the function presented, arcsin(x), we need to consider the value that this function approaches as x gets infinitely close to 1, but is still less than 1. This is asking what angle in radians has a sine of 1, or almost 1 from the left side. The sine of an angle is 1 when that angle is π/2 (or 90° in degree measure), hence as x → 1−, the value of arcsin(x) approaches π/2.

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Answer:

d=0.0167mille

Step-by-step explanation:

we must determine its position (X-Y).

ship 1

[tex]v_{1}=8\frac{m}{h}\\t_{1}=12.5 h\\t_{2}=14h\\t_{t2}=t_{2}-t_{1}=14-12.5=1.5h\\d_{s1}=v_{1}*t_{t1}=8\frac{m}{h}*1.5h=12m[/tex]

ship 2

[tex]v_{2}=10\frac{m}{h}\\t_{1}=12.5 h\\t_{2}=14h\\t_{t2}=t_{2}-t_{1}=14-13.5=0.5h\\d_{s2}=v_{1}*t_{t2}=10\frac{m}{h}*0.5h=5m[/tex]

component  x-y

[tex]sin55=\frac{y_{1}}{h};y_{1}=12*sin55=9.82\\   cos55=\frac{x_{1} }{h};x_{1}=5*cos55=6,88[/tex]

[tex]sin150=\frac{y_{2}}{h};y_{2}=5*sin150=2.5\\cos150=\frac{x_{2} }{h};x_{2}=5*cos150=-4.33[/tex]

[tex]s_{1}=(6.88,9.82); s_{2}=(-4.33,2,5)[/tex]

we must find the distance S1-S2

[tex]d_{s1-s2}=\sqrt{(y_{2}-y_{1})^{2}+(x_{2}-x_{1})^{2}}=\sqrt{(2.5-9.8)^{2}+(-4.33-6.88)^{2}}\\=\sqrt{(24.5)^{2}+(-11.21)^{2}}=26.94m[/tex]

but the units are requested to be miles

d=26.94m*mille/1,609.34m=0.0167mille

Answer:

The required probability is 0.988.

Step-by-step explanation:

Consider the provided information.

Based on a​ poll, 67​% of Internet users are more careful about personal information when using a public​ Wi-Fi hotspot.

That means the probability of more careful is 0.67

The probability of not careful is: 1-0.67 = 0.33

We have selected four random Internet​ users. we need to find the probability that at least one is more careful about personal information.

P(At least one careful) = 1 - P(None of them careful)

P(At least one careful) = 1 - (0.33×0.33×0.33×0.33)

P(At least one careful) = 1 - 0.012

P(At least one careful) = 0.988

Hence, the required probability is 0.988.

The result may be higher because of the convenience bias in retrieving the sample. Because the survey subjects volunteered to​ respond not random.

The probability that at least one out of four randomly selected Internet users is more careful about personal information when using a public Wi-Fi hotspot is approximately 98.81%, assuming that each event is independent and the probability of an individual being careful is 67%. However, voluntary responses to the survey might introduce a non-response bias, affecting the accuracy of this probability.

The question asks for the probability that among four randomly selected Internet​ users, at least one is more careful about personal information when using a public​ Wi-Fi hotspot, given that 67% of Internet users are like this. To find the probability of 'at least one', it is easier to calculate the complement—that is, the probability that none of the four users are careful—and subtract it from 1 (the total probability of any outcome).

The probability that a randomly selected Internet user is not more careful is 1 - 0.67 = 0.33. Since we are considering four independent events, we raise the single-event probability to the fourth power:

(0.33)^4 = 0.33 * 0.33 * 0.33 * 0.33

The calculation results in approximately 0.0119. Now, subtract this from 1 to get the probability of at least one person being more careful:

1 - 0.0119 = 0.9881

Therefore, the probability that at least one out of four randomly selected Internet users is more careful about personal information when using a public Wi-Fi hotspot is about 98.81%.

However, the accuracy of the result could be influenced by the fact that the survey subjects volunteered to respond, which can result in a non-response bias. Volunteers might have different behaviors or opinions compared to the general Internet user population, potentially skewing the results of the survey and, consequently, the estimated probability.

Answer:

Z= -1.429

Step-by-step explanation:

Hello!

The owner thinks that the number of customers decreased. The hystorical average per day is 800 customers.

So the variable X: " Customers per day" X~N(μ;σ²)

σ= 350

Symbolically the test hypothesis are:

H₀: μ ≥ 800

H₁: μ < 800

The statistic value is:

Since you are studying the population sample, the study variable has a normal distribution and you know the value of the population variance, the propper statistic to make the test is:

Z= X[bar] - μ = 750 - 800 = -1.4285 ≅ -1.429

     σ/√n           350/√100

where X[bar] is the sample mean

μ is the population mean under the null hypothesis

σ is the population standard  

n is the sample taken

The value is Z= -1.429

I hope it helps!

Answer:

4.) Step 2; it should be m∠o + m∠p = 180 degrees (supplementary angles)

Step-by-step explanation:

o and p are supplementary angles, and therefore add up to 180 degrees.

Answer:

solution is in the image below

Step-by-step explanation:

Final answer:

To construct confidence intervals for the mean compressive strength of concrete with known variance, formulas involving the z-score are employed. The 95% confidence interval is narrower than the 99% interval, illustrating that higher confidence requires a wider interval to encapsulate the true mean with more assurance.

Explanation:

To construct a confidence interval for the mean compressive strength of concrete, we utilize the formula for the confidence interval of the mean when the population variance (σ2) is known. Given that the variance is 1000 psi2 and the mean (μ) is 3250 psi for a sample of n=12 specimens, the z-score corresponding to a 95% confidence level is approximately 1.96, and for a 99% confidence level, the z-score is approximately 2.576.

95% Confidence Interval:

CI = μ ± z(σ/√n)
= 3250 ± 1.96(/1000/12)
= 3250 ± (1.96)(28.8675)
= 3250 ± 56.61
= (3193.39, 3306.61) psi

99% Confidence Interval:

CI = μ ± z(σ/√n)
= 3250 ± 2.576(/1000/12)
= 3250 ± (2.576)(28.8675)
= 3250 ± 74.36
= (3175.64, 3324.36) psi

Comparing the widths, the 99% confidence interval is wider than the 95% confidence interval, which is a reflection of increased certainty (or confidence level) requiring a wider interval.

To determine the angles of diffraction maxima for a helium-neon laser light incident on a diffraction grating, use the formula d sin(θ) = m λ with d the grating spacing and λ the wavelength. Calculate for each order by substituting m=1, 2, etc., and solve for θ using inverse sine function.

To find the angles at which one would observe the first order maximum, second-order maximum, and so forth for monochromatic light incident on a diffraction grating, we use the diffraction grating equation: d × sin(θ) = m × λ, where d is the grating spacing, λ is the wavelength of light, m is the order of maximum, and θ is the diffraction angle.

For a diffraction grating with 6000 lines/cm, the spacing d is 1/6000 cm, or 1.67 x 10^-4 cm (since 1 cm = 10^-2 m, d = 1.67 x 10^-6 m). Given the wavelength λ = 632.8 nm or 6.328 x 10^-7 m, we can substitute these values into the equation to solve for θ for any order m.

To find the actual angles, use the inverse sine function (arcsin), keeping in mind the units.

Answer:

we reject H₀

Step-by-step explanation:

Normal Distribution

sample size      n  =  25        degees of fredom     =  25 - 1  df = 24

sample standard deviation = s = 0,24

sample mean     11.88

We have a one tail test (left) investigation

1.-Test hypothesis

H₀          ⇒              null hypothesis         μ₀  =  12

Hₐ          ⇒     Alternative hypothesis     μ₀  <  12  

2.-Significance level   0,05     t(c)  = - 1.7109

3.-Compute of t(s)

t(s)  = (  μ  -   μ₀ )/s/√n          ⇒  t(s)  =[ ( 11.88 - 12 )*√25 ]/0.24

t(s)  = - 0.12*5/0.24

t(s)  = - 2.5

4.-We compare  t(s) with  t(c)

   In this case   t(s) < t(c)        - 2.5  <  -1.71

5.-t(s) is in the rejection region, we reject H₀

The machine is not adjusted

Answer:

∠NMP, so m∠1 = m∠2.

Step-by-step explanation:

Final Answer:

A. approximately-normal; μp = 0.35, σp = 0.030


1. Identify conditions for a normal sampling distribution:

Random sampling: The problem states it's a simple random sample.

Large sample size: n = 250 is greater than 10% of the population (assumed to be large), satisfying the condition.

Success-failure condition: np = 250 * 0.35 = 87.5 and n(1-p) = 250 * 0.65 = 162.5 are both greater than 10, meeting the condition.

2. Calculate the mean and standard deviation of the sampling distribution:

Mean (μp): μp = p = 0.35 (equal to the population proportion)

Standard deviation (σp): σp = sqrt(p(1-p)/n) = sqrt(0.35*0.65/250) ≈ 0.030

Final Answer:

A. approximately-normal; μp = 0.35, σp = 0.030

Explanation:

Answer:

16 goes in the blank

Step-by-step explanation:

V(c) = 2*π*r*h

Differentiating boh sides

DV(c)/Dt   =  2πr Dh/Dt    now radius is 8 m

DV(c)/Dt   =  8π Dh/Dt

That expression gives the relation of changes in V and h

DV(c)/Dt  is the speed of growing of the volume

Dh/Dt  is the speed of increase in height

so if the cylinder is filling at a rate of  2 m³/min   the height will increase at a rate of 16π m/min  

Answer:

A) As α increases, β decreases

Step-by-step explanation:

I am assuming that the probabilities of both kind of error are positive, otherwise there is no reason to make an hypothesis test.

Both events arent possible simultaneously, so they cant be independent because the probability of both kind of errors is greater than 0.

If the p-value increases, then by definition, the probability of type 1 error increases, in this case, you are more likely to reject the null hypothesis, therefore you are less likely to make the mistake of not rejecting the null hypothesis when you have to. Thus, the probability of type 2 error decreases.

On the other way around, if you are less likely to reject the null hypothesis (hence, the probability of type 1 error decreases), then you will be more likely to not reject the null hypothesis even on weird scenarios, therefore the probability of type 2 error increases.

There is no formula to determine the relatioship between both errors, but the errors are related: when the probability of type 1 error increases, then the probability of type 2 error decreases. The correct answer is A)

Answer:

2.5

Step-by-step explanation:

We are given that

Mean=[tex]\mu=3.2[/tex] pounds

Standard deviation=[tex]\sigma=0.8[/tex] pounds

n=25

We have to find the Z-score if the mean of  a sample of 25 roasts is 3.6 pounds.

We know that Z-score formula

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

We have [tex]\bar X=3.6[/tex]

Substitute the values then we get

Z- score=[tex]\frac{3.6-3.2}{\frac{0.8}{\sqrt{25}}}[/tex]

Z- score=[tex]\frac{0.4}{\frac{0.8}{5}}=\frac{0.4\times 5}{0.8}=2.5[/tex]

Hence, the Z-score value for the given sample mean=2.5

The Z-score for a sample mean of 3.6 pounds for 25 roasts, given a population mean of 3.2 pounds and a population standard deviation of 0.8 pounds, is calculated through the formula 'Z = (X - μ) / (σ / √n)' and is determined to be 2.5. The fact that the Z-score is positive indicates that the sample mean is above the population mean.

The question presented involves calculation of a Z-score in a situation involving weights of roasts. The Z-score is a measure used in statistics that describes a value's relationship to the mean of a group of values. It is measured in terms of standard deviations from the mean. Here, we can use the formula for calculating Z-score from a sample mean:

Z = (X - μ) / (σ / √n)

where 'X' is the sample mean, 'μ' is the population mean, 'σ' is the population standard deviation, and 'n' is the size of the sample. Plugging in the values given in the problem, we get:

Z = (3.6 - 3.2) / (0.8 / √25) = 0.4 / 0.16 = 2.5

So, the Z-score for the sample mean of 25 roasts weighing 3.6 pounds each is 2.5.

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Answer:

36x^2

Step-by-step explanation:

Since area is base*height:

8*4.5=36

x*x=x^2

The monomial is 36x^2

To construct a confidence interval for the population mean with an unknown standard deviation, we use Student's t-distribution. The 99% confidence interval for the given sample mean (x = 880) and sample standard deviation (s = 5) is (877.371, 882.629). Assuming a different sample standard deviation, the 99% confidence intervals are (874.24, 885.76) for s = 10 and (868.48, 891.52) for s = 20. The confidence interval width increases as the sample standard deviation increases.

To construct a confidence interval for the population mean with an unknown standard deviation, we use Student's t-distribution. First, we find the value of the t-statistic for the given confidence level and degrees of freedom. For a 99% confidence level and a sample size of 24, the degrees of freedom is 23. Using Appendix D or a t-table, we find the critical t-value to be approximately 2.819.

(a) To construct a confidence interval with the given sample mean (x = 880) and sample standard deviation (s = 5), the margin of error is calculated as: E = t * (s / sqrt(n)) = 2.819 * (5 / sqrt(24)) = 2.819 * 1.021 = 2.88 (rounded to 3 decimal places). The 99% confidence interval is then calculated as: (x - E, x + E) = (880 - 2.88, 880 + 2.88) = (877.12, 882.88), rounded to 3 decimal places as (877.371, 882.629).

(b) Using the same method, but assuming a sample standard deviation of s = 10, the margin of error is calculated as: E = 2.819 * (10 / sqrt(24)) = 5.76 (rounded to 3 decimal places). The 99% confidence interval is then (874.24, 885.76).

(c) Assuming s = 20, the margin of error is calculated as: E = 2.819 * (20 / sqrt(24)) = 11.52 (rounded to 3 decimal places). The 99% confidence interval is then (868.48, 891.52).

(d) As the sample standard deviation (s) increases, the margin of error (E) and confidence interval width will also increase. This means that as s increases, we become less certain about the true population mean, resulting in a wider range of values in the confidence interval.

Answer:

Independent Sampling

Step-by-step explanation:

There are two scenarios for independent sampling .

Testing the mean we get the differences between samples from each population. When both samples are randomly  inferences about the populations. can get.

Independent sampling are sample that are selected randomly. Observation does not depend upon value.Many analysis assume that sample are independent.

In this statement 90 dash are divides into two groups Group 1 and Group 2 . Both are standardized that mean both are randomly selected. Means are observed. Observation doesn't depend upon value.  So this style of sampling is independent Sample.

Answer: a) 0.23, b) No, c) No.

Step-by-step explanation:

Since we have given that

Probability of homes for sale have garages = 61%

Probability of homes having swimming pool = 39%

Probability of homes having both = 14%

a) If a home for sale has a​ garage, what's the probability that it has a​ pool, too?

Using "Conditional probability", we get that

[tex]P(P|G)=\dfrac{P(P\cap G)}{P(G)}=\dfrac{0.14}{0.61}=0.23[/tex]

​b) Are having a garage and having a pool independent​ events? Explain.

P(P).P(S) = 0.61×0.39=0.24≠P(P∩S)=0.14

Hence, they are not independent events.

c) Are having a garage and having a pool mutually​ exclusive? Explain.

Since P(P∩S)≠0

So, they are not mutually exclusive.

Hence, a) 0.23, b) No, c) No.

Final answer:

To calculate the probability of a home having a pool given it has a garage, one uses the given percentages to find that it's approximately 23%. Examining whether having a garage and having a pool are independent or mutually exclusive reveals they are neither independent (because the probabilities differ when one condition is known) nor mutually exclusive (because homes can and do have both features).

Explanation:

Given information implies that 61% of homes have garages (G), 39% have pools (P), and 14% have both a garage and a pool (G∩P).

a) The probability that a home for sale has a pool given it has a garage can be calculated using the formula P(P|G) = P(G∩P) / P(G). Substituting the given values, we get P(P|G) = 0.14 / 0.61, which approximately equals 0.23 or 23%. This means if a home has a garage, there's a 23% chance it also has a pool.

b) Two events are independent if the occurrence of one does not affect the probability of the occurrence of the other. To check if G and P are independent, we calculate P(P) and compare it to P(P|G). Since P(P) = 0.39 and P(P|G) = 0.23, they are not equal; thus, having a garage and having a pool are not independent events. This is evident because knowing that a home has a garage changes the probability that it also has a pool.

c) Two events are mutually exclusive if they cannot occur at the same time. Given that 14% of homes have both a garage and a pool, it's clear that having a garage and having a pool are not mutually exclusive, as a home can have both features simultaneously.

Answer:

[tex]D_{up}(t)=3t[/tex]

[tex]D_{down}(t)=8t[/tex]

[tex]D_{flat}(t)=5t[/tex]

Explanation:

To construct a linear model of their distance, in miles, as a function of time in hours, since we were given their traveling speed in miles per hour, simply multiply the traveling speed in mph by the time, t, in hours.

When traveling up a hill:

[tex]D_{up}(t)=3t[/tex]

When traveling down a hill:

[tex]D_{down}(t)=8t[/tex]

When traveling along a flat portion:

[tex]D_{flat}(t)=5t[/tex]

Final answer:

To construct linear models that describe the distance traveled on different portions of the path in terms of time, we can use the formula for distance: D = V * T, where D is the distance, V is the velocity, and T is the time.

Explanation:

To construct linear models that describe the distance traveled on different portions of the path in terms of time, we can use the formula for distance:

D = V * T

where D is the distance, V is the velocity, and T is the time.

For the uphill portion, the speed is 3 mph, so the linear model would be:

D = 3T

For the downhill portion, the speed is 8 mph, so the linear model would be:

D = 8T

For the flat portion, the speed is 5 mph, so the linear model would be:

D = 5T

The probability of drawing a spade from a 52-card deck is 0.25. If it's known that the card is black, the probability that it's a spade increases to 0.5.

The subject of this question is the calculation of probability in a card game scenario. When a single card is selected from a standard 52-card deck, the probability that the card drawn is a spade​ is calculated by dividing the total number of spades in the deck (13) by the total number of cards in the deck (52). This gives a probability of 13/52 or 0.25.

However, if a single card is drawn from a standard 52-card deck, and it is known that the card is black, then the total number of possible outcomes is reduced to 26 (the number of black cards in the deck). In this scenario, the probability that the card drawn is a spade becomes 13/26 or 0.5.

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The probability of drawing a spade from a 52-card deck is 1/4. If we already know the card is black, the probability increases to 1/2.

This question relates to the field of probability. To address part a, in a standard 52-card deck, there are 13 spades. We therefore have a 13 out of 52 chance to draw a spade, simplifying to a probability of 1/4.

For part b, being told that the card drawn is black means our sample space becomes 26 (13 spades and 13 clubs). Since there are still 13 spades, the probability of drawing a spade given that the card is black is 13 out of 26, or 1/2.

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Answer:

[tex]x_0=-5\\x_1=-2\\x_2=1\\x_3=4\\x_4=7[/tex]

Δn=3

Step-by-step explanation:

Remember, if we need to divide the interval (a,b) in n equal subinterval, then we need divide the distance (d) between the endpoints of the interval and divide it by n. Then the width Δn of each subinterval is d/n.

We have the interval [-5,7]. The distance between the endpoints of the interval is

[tex]d=7-(-5)=12[/tex].

Now, we divide d by 4 and obtain [tex]\frac{d}{4}=\frac{12}{4}=3[/tex]

Then, Δn=3.

Now, to find the endpoints of each sub-interval, we add 3 from the left end of the interval.

[tex]-5=x_0\\x_0+3=-5+3=-2=x_1\\x_1+3=1=x_2\\x_2+3=4=x_3\\x_3+3=7=x_4[/tex]

So,

[tex]x_0=-5\\x_1=-2\\x_2=1\\x_3=4\\x_4=7[/tex]

The width of each subinterval is 3. The endpoints in ascending order are -5, -2, 1, 4, 7.

To find the width of each subinterval, we need to divide the length of the interval by the number of subintervals. In this case, the interval is [-5,7] and we need to divide it into 4 equal subintervals. The length of the interval is 7 - (-5) = 12. So the width of each subinterval is 12/4 = 3.

Next, we can find the endpoints of the subintervals. Since we have 4 subintervals, we need 5 endpoints. The first endpoint is the left endpoint of the interval, which is -5. Then we add the width of each subinterval to find the next endpoints: -5 + 3 = -2, -2 + 3 = 1, 1 + 3 = 4, and finally 4 + 3 = 7, which is the right endpoint of the interval. Therefore, the endpoints in ascending order are -5, -2, 1, 4, 7.

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Answer:

[tex]z=\frac{0.686 -0.68}{\sqrt{\frac{0.68(1-0.68)}{86}}}=0.119[/tex]  

[tex]p_v =P(z<0.119)=0.547[/tex]  

The p value obtained was a very high value and using the significance level given for example [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, and we can said that at 5% of significance the proportion interest is not significantly lower than 0.68.  

Step-by-step explanation:

1) Data given and notation

n=86 represent the random sample taken

X=59 represent the adults that match the condition you are testing

[tex]\hat p=\frac{59}{86}=0.686[/tex] estimated proportion of adults that match the condition you are testing

[tex]p_o=0.68[/tex] is the value that we want to test

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

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 population proportion is less than 0.68.:  

Null hypothesis:[tex]p=0.68[/tex]  

Alternative hypothesis:[tex]p < 0.68[/tex]  

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

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

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

3) Calculate the statistic  

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

[tex]z=\frac{0.686 -0.68}{\sqrt{\frac{0.68(1-0.68)}{86}}}=0.119[/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 is not given [tex]\alpha[/tex]. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

[tex]p_v =P(z<0.119)=0.547[/tex]  

So the p value obtained was a very high value and using the significance level given for example [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, and we can said that at 5% of significance the proportion interest is not significantly lower than 0.68.