An insurance company sets up a statistical test with a null hypothesis that the average time for processing a claim is 7 days, and an alternative hypothesis that the average time for processing a claim is greater than 7 days.
After completing the statistical test, it is concluded that the average time exceeds 7 days.

However, it is eventually learned that the mean process time is really 9 days.

What type of error occurred in the statistical test?

Answer:

c) Is not a property (hence (d) is not either)

Step-by-step explanation:

Remember that the chi square distribution with k degrees of freedom has this formula

[tex]\chi_k^2 = \matchal{N}_1^2 +  \matchal{N}_2^2 + ... + \, \matchal{N}_{k-1}^2 +  \matchal{N}_k^2[/tex]

Where N₁ , N₂m .... [tex] N_k [/tex] are independent random variables with standard normal distribution. Since it is a sum of squares, then the chi square distribution cant take negative values, thus (c) is not true as property. Therefore, (d) cant be true either.

Since the chi square is a sum of squares of a symmetrical random variable, it is skewed to the right (values with big absolute value, either positive or negative, will represent a big weight for the graph that is not compensated with values near 0). This shows that (a) is true

The more degrees of freedom the chi square has, the less skewed to the right it is, up to the point of being almost symmetrical for high values of k. In fact, the Central Limit Theorem states that a chi sqare with n degrees of freedom, with n big, will have a distribution approximate to a Normal distribution, therefore, it is not very skewed for high values of n. As a conclusion, the shape of the distribution changes when the degrees of freedom increase, because the distribution is more symmetrical the higher the degrees of freedom are. Thus, (b) is true.

Answer:

35$

Step-by-step explanation:

Let the shirt be = X

And the tie =Y

X + Y= 48$

X = 22 + Y (The shirt costs $22 more than the tie)

22 + 2y = 48

2y = 26

y = 13

X= 48 – 13

X = 35

therefore, the cost of the shirt is $35

and the cost of the tie is $13

The cost of the shirt is 35$ such the shirt and tie together cost $48.

Determine the known quantities and designate the unknown quantity as a variable while trying to set up or construct a linear equation to fit a real-world application.

In other words, an equation is a set of variables that are constrained through a situation or case.

Let's say the shirt cost is S while the tie is T

Together;

S + T = 48

And,

S = 22 + T

By substituting

22 + T + T = 48

2T = 26

T = 13

So,

S = 48 - 13 = 35

Hence "The cost of the shirt is 35$ such the shirt and tie together cost $48".

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Answer:

We reject the null hypothesis that the population means are equal and accept the alternative hypothesis that the population means are different.

Step-by-step explanation:

We have large sample sizes [tex]n_{1} = 49[/tex] and [tex]n_{2} = 64[/tex], the unbiased point estimate for [tex]\mu_{1}-\mu_{2}[/tex] is [tex]\bar{x}_{1} - \bar{x}_{2}[/tex], i.e., 12-14 = -2.

The standard error is given by [tex]\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}[/tex], i.e.,

[tex]\sqrt{\frac{(3)^{2}}{49}+\frac{(4)^{2}}{64}}[/tex] = 0.6585.

We want to test [tex]H_{0}: \mu_{1}-\mu_{2} = 0[/tex] vs [tex]H_{1}: \mu_{1}-\mu_{2} \neq 0[/tex] (two-tailed alternative). The rejection region is given by RR = {z | z < -2.5758 or z > 2.5758} where -2.5758 and 2.5758 are the 0.5th and 99.5th quantiles of the standard normal distribution respectively. The test statistic is [tex]Z = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}}[/tex] and the observed value is [tex]z_{0} = \frac{-2}{0.6585} = -3.0372[/tex]. Because -3.0372  fall inside RR, we reject the null hypothesis.

The test statistic follow a standard normal distribution because we are dealing with large sample sizes.

In this scenario of comparing two independent samples and given that the sample sizes are large, the sample test statistic follows the Standard Normal distribution or Z-distribution. The Z-test statistic representing the difference in sample means (in units of standard error) is compared with critical values for a two-tailed test at 0.01 significance level to determine if there's sufficient evidence to reject the null hypothesis that the two population means are equal.

The test in your question pertains to a hypothesis testing scenario featuring two independent samples. This scenario typically involves two population means given that population standard deviations are known. The distribution followed by the sample test statistic in such cases is the Standard Normal distribution or Z-distribution, as the sample sizes (n1 = 49, n2 = 64) are sufficiently large. To test the claim that population means are different (at a significance level of 0.01), you'd typically construct a Z-test statistic that represents the difference in sample means (x1 - x2) in units of its standard error. The Z-test statistic is calculated as follows:  

Here, x1 and x2 are the sample means, σ1 and σ2 are the population standard deviations and n1 and n2 are the samples sizes. The resulting Z-score can be compared with critical Z-scores for a two-tailed test at the given level of significance (0.01) to determine whether or not the null hypothesis (two population means are equal) can be rejected.

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Answer: -204.8

Step-by-step explanation: if there are and more questions like this use m   a   t    h   w    a  y

Answer:

65%

Step-by-step explanation:

No, the events "drank coffee" and "drank tea" are not mutually exclusive, as there are 10% of employees drank both coffee and tea.

If there are 50% drank coffee and 10% of them enjoy both, then there are 40% of the employees enjoy only coffee.

Similarly, there are 15% of employees who only enjoy tea.

Then the probability of selecting a person who only enjoy tea or coffee is

40% + 15% = 65%

Answer:

No.

0.65 or 65%

Answer:

0.546 is the probability that a randomly selected smartphone users in the age range 11 to 55+ is between 30 and 54 years old.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  34.8 years

Standard Deviation, σ = 14.1 years

We are given that the distribution of ages of smartphone is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

P( age range is between 30 and 54 years old)

[tex]P(30 \leq x \leq 54) = P(\displaystyle\frac{30 - 34.8}{14.1} \leq z \leq \displaystyle\frac{54-34.8}{14.1}) = P(-0.3404 \leq z \leq 1.3617)\\\\= P(z \leq 1.3617) - P(z < -0.3404)\\= 0.913 - 0.367 = 0.546 = 54.6\%[/tex]

[tex]P(30 \leq x \leq 54) = 54.6\%[/tex]

0.546 is the probability that a randomly selected smartphone users in the age range 11 to 55+ is between 30 and 54 years old.

The 90% confidence interval using the same sample values is E. 65 to 85.

The 90% confidence interval would be narrower than the 95% confidence interval but the middle point always remains the same.

The middle point there should be (60 + 90)/2 = 75 for the confidence interval. The confidence interval width for a 95% confidence interval width is 30.

For the 65 to 85 confidence interval, the width is 20, therefore this can be true because 20 < 30, therefore 65 to 85 could be the possible confidence interval required here.

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Answer:

21 I believe

Step-by-step explanation:

3 x 5 = 15

15 + 1 = 16

because its one for every three blueberry then there are 5 strawberries and 16 blueberries so 5 + 16 = 21

Answer: there are 12 blueberries and 4 strawberries in the bowl

Step-by-step explanation:

The total number of blueberries and strawberries contained in the bowl is 16. The ratio of blueberries to strawberries is 3:1

Total ratio will be sum of the proportion the blueberries to the proportion of strawberries.

Total ratio = 3+1 = 4

To determine how many of each berry are in the bowl,

Number of blueberries in the bowl will be

(Proportion of blueberries / total ratio ) × 100

This becomes

3/4 × 16 = 12 blueberries

Number of strawberries in the bowl will be

(Proportion of strawberries / total ratio ) × 100

This becomes

1/4 × 16 = 4 strawberries

Answer:

The proof is given below.

Step-by-step explanation:

this theorem says that, if a point is on the angular bisector of an angle, then it is equidistant from the sides of the angle.

(the proof of  angle bisector theorem can be explained, but it is difficult to type the whole thing. so watch this video for the proof of this theorem :  https://youtu.be/6GS4lS4btNI    )

Answer:

250

Step-by-step explanation:

36% are rabbits, and 28% are dogs.  That means the percent that are hamsters is:

100% − 36% − 28% = 36%

There are 160 dogs and hamsters, which are 28% + 36% = 64% of the pets.  Therefore, we can write and solve a proportion:

160 / 64% = x / 100%

x = 250

There are a total of 250 pets.

Answer: 250 pets

Step-by-step explanation:

Let the total number of pets in the pet shop be x

36% of them are rabbits. This means that the number of pets that are rabbit 36/100 × x = 0.36×x = 0.36x

28% of them are dogs. This means that the number of pets that are dogs are 28/100 × x = 0.28×x = 0.28x

The rest are hamsters. This means that total number of hamsters is total number of pets minus sum of rabbits and dogs.

Number of hamsters = x -(0.28x + 0.36x) = 0.36x

If there are 160 dogs and hamsters,it means that

0.28x + 0.36x = 160

0.64x =160

x = 160/0.64

x = 250

The equation modeling the displacement d as a function of time t for an object in simple harmonic motion with a period of 7 seconds and amplitude of 3cm is d = -3cos(2pi/7 * t + pi).

The displacement d of an object moving in simple harmonic motion can be modeled by the equation d=Acos(wt).

Given that the period T is 7 seconds, we can use the formula T=2pi/w to solve for the angular frequency w. Rearranging the equation, we have w = 2pi/T. Plugging in the given period T=7, we get w = 2pi/7.

Since the object initially moves in a negative direction, we would have a phase shift of pi in the cosine function. Therefore, the equation modeling the displacement d as a function of time t is d = -3cos(2pi/7 * t + pi).

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Answer:

The expected value of the safe bet equal $0

Step-by-step explanation:

If  

[tex]S=\left\{s_1,s_2,...,s_n\right\}[/tex]

is a finite numeric sample space and

[tex]P(X=s_k)=p_k[/tex] for k=1, 2,..., n

is its probability distribution, then the expected value of the distribution is defined as

[tex]E(X)=s_1P(X=s_1)+s_2P(X=s_2)+...+s_nP(X=s_n)X) [/tex]

What is the expected value of the safe bet?

In the safe bet we have only two possible outcomes: head or tail. Woodrow wins $100 with head and “wins” $-100 with tail So the sample space of incomes in one bet is

S = {100,-100}

Since the coin is supposed to be fair,  

P(X=100)=0.5

P(X=-100)=0.5

and the expected value is

E(X) = 100*0.5 - 100*0.5 = 0

The expected value of the safe bet is $0, which means that you would neither gain nor lose money on average if you played this game repeatedly.

The expected value of the safe bet can be calculated by multiplying the possible outcomes by their respective probabilities and summing them up. In this case, the safe bet has two possible outcomes: winning $100 with a 50% probability and losing $100 with a 50% probability. So, the expected value can be calculated as follows:

Expected value = (100 * 0.5) + (-100 * 0.5) = $0

Therefore, the expected value of the safe bet is $0, which means that you would neither gain nor lose money on average if you played this game repeatedly.

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Answer:

i know im late but it's 2 crates.

Step-by-step explanation:

To find the number of crates, divide the total number of cookies by the number of cookies in each box.

To find the number of crates that can be filled with the 3,258 cookies, we divide the total number of cookies by the number of cookies in each box. In this case, there are 12 cookies in each box.

So, we divide 3,258 by 12:

3,258 ÷ 12 = 271.5

Since we can't have half of a crate, we round down to the nearest whole number:

271.5 ≈ 271

Therefore, 271 crates can be filled from the total of 3,258 cookies.

Answer:

a) (0.72, 0.78)

Step-by-step explanation:

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".  

Description in words of the parameter p

[tex]p[/tex] represent the real population proportion of all residents in that county who think their local government did a good job

[tex]\hat p[/tex] represent the estimated proportion of all residents in that county who think their local government did a good job

n=1000 is the sample size required  

[tex]z_{\alpha/2}[/tex] represent the critical value for the margin of error  

The population proportion have the following distribution  

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]

Numerical estimate for p

In order to estimate a proportion we use this formula:

[tex]\hat p =\frac{X}{n}[/tex] where X represent the number of people with a characteristic and n the total sample size selected.

[tex]\hat p=\frac{750}{1000}=0.75[/tex] represent the estimated proportion of all residents in that county who think their local government did a good job

Confidence interval

The confidence interval for a proportion is given by this formula  

[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]  

For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.  

[tex]z_{\alpha/2}=1.96[/tex]  

And replacing into the confidence interval formula we got:  

[tex]0.750 - 1.96 \sqrt{\frac{0.75(1-0.75)}{1000}}=0.72[/tex]  

[tex]0.750 + 1.96 \sqrt{\frac{0.75(1-0.75)}{1000}}=0.78[/tex]  

And the 95% confidence interval would be given (0.72;0.78).  

We are confident at 95% that the true proportion of people who think their local government did a good job is between (0.72;0.78).

a) (0.72, 0.78)

Answer:

(B) Fail to reject the null hypothesis that the true population mean battery life of the smartphone is equal to 20 hours.

(C) There is not enough evidence at the α=0.05 level of significance to suggest that the true population mean battery life of the smartphone is less than 20 hours.

Step-by-step explanation:

1) Data given and notation    

[tex]\bar X=19.5[/tex] represent the battery life sample mean  

[tex]s=1.9[/tex] represent the sample standard deviation    

[tex]n=33[/tex] sample size    

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

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

t would represent the statistic (variable of interest)    

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

2) State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean battery life is less than 20 :    

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

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

Since we don't know the population deviation, 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".    

3) Calculate the statistic    

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

[tex]t=\frac{19.5-20}{\frac{1.9}{\sqrt{33}}}=-1.51[/tex]    

4) P-value    

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

[tex]df=n-1=33-1=32[/tex]

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

[tex]p_v =P(t_{(32)}<-1.51)=0.0704[/tex]    

5) 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, so we can conclude that the average battery life it's not significantly different less than 20 hours at 5% of signficance. If we analyze the options given we have:

(A) Reject the null hypothesis that the true population mean battery life of the smartphone is equal to 20 hours. FALSE, we FAIL to reject the null hypothesis.

(B) Fail to reject the null hypothesis that the true population mean battery life of the smartphone is equal to 20 hours. TRUE, we fail to reject the null hypothesis that the mean would be 20 or higher .

(C) There is not enough evidence at the α=0.05 level of significance to suggest that the true population mean battery life of the smartphone is less than 20 hours.  TRUE, we FAIL to reject the null hypothesis that the mean is greater or equal to 20 hours, so we reject the alternative hypothesis that the mean is less than 20 hours.

(D) There is enough evidence at the α=0.05 level of significance to support the claim that the true population mean battery life of the smartphone is not equal to 20 hours. FALSE the claim is not that the mean is different from 20. The real claim is: "Javier believes the mean battery life is less than 20 hours".

Javier's t-test with a test statistic of −1.51 and 32 degrees of freedom has a p-value greater than 0.05, hence we fail to reject the null hypothesis, indicating not enough evidence to suggest the true mean is less than the claimed 20 hours. Thus, the correct answer is option (C).

This problem involves conducting a one-sample t-test to determine whether the true population mean battery life is different from the claimed 20 hours. The null hypothesis here is H0: [tex](mu = 20)[/tex] hours, and the alternative hypothesis is Ha: [tex](mu < 20)[/tex] hours. With 32 degrees of freedom, Javier's calculated test statistic is t ≈ −1.51. To decide whether to reject or fail to reject the null hypothesis, we must compare the p-value to the significance level, [tex]\(\alpha = 0.05\)[/tex].

If the p-value is less than [tex]\(\alpha = 0.05\)[/tex], then we reject the null hypothesis. If the p-value is higher, we fail to reject. In this exercise, the p-value associated with Javier's test statistic of t ≈ −1.51 for a one-tailed test is greater than 0.05, thus we should fail to reject the null hypothesis (Option B). This implies that there is not enough evidence at the 0.05 level of significance to support the claim that the true population mean battery life of the smartphone is less than 20 hours (Option C).

Answer:

p-value:  0.6527

Step-by-step explanation:

Hello!

You have two samples to study, from each sample the weight of each child was measured and counted the total of overweight kids (x: "success") in each group:

Sample 1 (Boys aged 6-11)

n₁= 546

x₁= 89

^p₁= x₁/n₁ = 89/546 ≅0.16

Sample 2 (girls aged 6-11)

n=508

x= 74

^p= x/n = 74/508 ≅ 0.15

If the hypothesis statement is "The proportion of boys that are overweight differs from the proportion of girls that are overweight", the test hypothesis is:

H₀: ρ₁ = ρ₂

H₁: ρ₁ ≠ ρ₂

This type of hypothesis leads to a two-tailed rejection region, then the p-value will also be two-tailed. To calculate the p-value you have to first calculate the value of the statistic under the null hypothesis, in this case, is a test for the difference between two proportions:

Z=      (^ρ₁ - ^ρ₂) - (ρ₁ - ρ₂)         ≈ N(0;1)

    √(ρ` * (1 - ρ`) * (1/n₁ + 1/n₂))

ρ`= x₁ + x₂   =  89+74     = 0.154 ≅ 0.15

     n₁ + n₂     546 + 508

Z⁰ᵇ =          (0.16-0.15) - (0)                    

       √(0.15 * (1 - 0.15) * (1/546 + 1/508))

Z⁰ᵇ = 0.45

I've mentioned before that in this test you have a two-tailed p-value. The value calculated (0.45) corresponds to the right or positive tail and the left tail is symmetrical to it concerning the distribution mean, in this case, is 0, so it is -0.45. To obtain the p-value you need to calculate the probability of both values and add them:

P(Z>0.45) + P(Z<-0.45) = (1- P(Z<0.45)) + P(Z<-0.45) = (1-0.67364) + 0.32636 = 0.65272 ≅ 0.6527

p-value:  0.6527

Since there is no signification level in the problem, I'll use the most common to reach a decision. α: 0.05

Since the p-value is greater than α, you do not reject the null Hypothesis, in other words, there is no significative difference between the proportion of overweight boys and the proportion of overweight girls.

I hope it helps!

Answer:

(a) 4.148 x 10^(12) ways

(b) 5,827,360 ways

Step-by-step explanation:

Number of Demonstrators (D)  = 44

Number of Repudiators (R) = 56

(a)

5 senate members must be Repudiators and 5 must be demonstrators, assuming that the order at which they are selected is irrelevant:

[tex]N= C^{D}_{5} * C^{R}_{5}\\N=\frac{56!}{5!(56-5)!} *\frac{44!}{5!(44-5)!} \\N=3,819,816*1,086,008\\N=4.148 *10^{12}[/tex]

(b)

Since there are two different positions, (speaker and vice speaker), order is important in this situation, and the total number of ways to select two senators from each party is:

[tex]N= P^{D}_{2} * P^{R}_{2}\\N=\frac{56!}{(56-2)!} *\frac{44!}{(44-2)!} \\N=3,080*1,892\\N=5,827,360[/tex]

The question involves applying the concept of combinations in mathematics to determine the number of ways committee members, speakers, and vice speakers can be selected from two different political parties.

The subject of the question involves two main components of mathematics: combinatorics and probability. This involves calculating the number of ways certain events can occur given a certain number of possibilities.

Let's first solve part (a) of your question. We are asked how many ways there are to select a committee of 10 senate members with the same number of Demonstrators and Repudiators. We want five senators from each party. Given there are 44 Demonstrators and 56 Repudiators, the number of ways we can pick a committee is the product of comb(44,5) and comb(56,5) which are the combinations of picking 5 out of 44 Demonstrators and 5 out of 56 Repudiators, respectively.

In part (b) of your question, we are asked how many ways there are for two speakers and two vice speakers to be selected, one from each party. This is simply comb(44,1) multiplied by comb(43,1) multiplied by comb(56,1) multiplied by comb(55,1). This is because we first choose 1 out of 44 Demonstrators for a speaker position, then 1 out of the remaining 43 Demonstrators for a vice speaker position, then 1 out of 56 Repudiators for a speaker position, and finally 1 out of the remaining 55 Repudiators for a vice speaker position.

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Answer:

We accept H₀, we dont have evidence to say that the family size container

of ketchup has smaller quantity

Step-by-step explanation:

Population mean  μ₀ =  20 ounces

sample size  n  =  10   df = n -1  df = 10-1   df= 9

n < 30   use of t-student distribution

sample mean  μ  = 19.86

sample standard deviation  s  =  0,22

One tail-test ( left tail)

1.-Test Hypothesis

H₀     null hypothesis                μ₀ =  20

Hₐ  Alternative hypothesis       μ₀ < 20

2.- α  =  0,01   and one test tail

3.- Compute

t(s)  =  [ ( μ  -   μ₀ ) ] / s/√n           t(s)  = [( 19.86  -  20 )* √10 ] / 0.22

t(s)  =  - ( 0,14 * 3.16) / 0,22  

t(s)  =  - 2.01

4.- We go to  t-student table  t(c) for df = 9  and 0,01 = α  

and find t(c) =  - 2.821

5.-Compare   t (s)   and  t (c)

t (c)  <   t (s)      -  2.821  < - 2.01

6. t(s)  is inside de acceptance region  we accept H₀

Answer:

Step-by-step explanation:

The null hypothesis is that the mean weekly food budget for all households in the community is equal to the national average, while the alternative hypothesis is that it is higher. The test statistic is calculated using the sample mean and the standard deviation of the population. To determine if the results are significant, we compare the test statistic to the critical value.

The null hypothesis for this test is that the mean weekly food budget for all households in the community is equal to the national average, which is $99. The alternative hypothesis is that the mean weekly food budget for all households in the community is higher than the national average.

The test statistic in this case is calculated using the sample mean and the standard deviation of the population. It is equal to (sample mean - population mean) / (standard deviation / sqrt(sample size)).

To determine if the results are significant at the 5% level, we compare the test statistic to the critical value for a one-tailed test with a significance level of 0.05. If the test statistic is greater than the critical value, we reject the null hypothesis and conclude that the mean weekly food budget for all households in the community is higher than the national average.

Answer:C

Step-by-step explanation:

Answer:

False

Step-by-step explanation:

Confidence intervals provide a range for a population parameter at a given significance level. The parameter can be mean, standard deviation etc.

In this example population is the prices of the rents of all the unfurnished one-bedroom apartments in the Boston area

significance level is 95%. Thus, the chance being the true population parameter in the given interval is 95%.

But, "This interval describes the price of 95% of the rents of all the unfurnished one-bedroom apartments in the Boston area." statement is false because the population parameter is missing. Confidence interval may describe population mean for example but it does not describe the whole population.

Answer:

Hip breadths less than or equal to 16.1 in. includes 90% of the males.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 14.5

Standard Deviation, σ = 1.2

We are given that the distribution of hip breadths is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

We have to find the value of x such that the probability is 0.10.

P(X > x)  

[tex]P( X > x) = P( z > \displaystyle\frac{x - 14.5}{1.2})=0.10[/tex]  

[tex]= 1 -P( z \leq \displaystyle\frac{x - 14.5}{1.2})=0.10 [/tex]  

[tex]=P( z \leq \displaystyle\frac{x - 14.5}{1.2})=0.90 [/tex]  

Calculation the value from standard normal z table, we have,  

[tex]P(z < 1.282) = 0.90[/tex]

[tex]\displaystyle\frac{x - 14.5}{1.2} = 1.282\\x = 16.0384 \approx 16.1[/tex]  

Hence, hip breadth of 16.1 in. separates the smallest 90​% from the largest 10%.

That is hip breaths greater than 16.1 in. lies in the larger 10%.

Answer:

The answer is B.8.  

Step-by-step explanation:

This is the answer because 8 is the greatest factor that will go into both 48 and 56. 8x6=48, 8x7=56.

The greatest common factor (GCF) of 48 and 56 is 8. It is the highest number that divides both numbers without leaving a remainder. Thus, the correct answer is Option B. 8.

The greatest common factor (GCF) of two numbers is the largest number that divides both of them without leaving a remainder. To find the GCF of 48 and 56, follow these steps:

Therefore, the correct answer is Option B. 8. This makes 8 the highest number that can evenly divide both 48 and 56.

Answer:

C. Fail to reject Upper H 0. There is insufficient evidence to warrant rejection of the claim that the variation of maximal skull breadths in 4000 B.C. is the same as the variation in A.D. 150

Step-by-step explanation:

Hello!

You have two different independent samples and are asked to test if the population variances of both variables are the same.

Sample 1 (4000 B.C)

X₁: Breadth of a skull from 4000 B.C. (mm)

X₁~N(μ₁;σ₁²)

n₁= 30 skulls

X[bar]₁= 131.62 mm

S₁= 5.19 mm

Sample 2 (A.D. 150)

X₂: Breadth of a skull from 150 A.D. (mm)

X₂~N(μ₂;σ₂²)

n₂= 30 skulls

X[bar]₂= 136.07 mm

S₂= 5.35 mm

Since you want to test the variances, the proper test to do is an F-test for the population variance ratio. The hypothesis can be established as equality between variances or as a quotient between them.

The hypothesis is:

H₀: σ₁²/σ₂² = 1

H₁: σ₁²/σ₂² ≠ 1

Remember, when you express the hypothesis as a quotient of variances, if it's true that they are the same, the result will be 1, this is the number you'll use to replace in the F-statistic.

α: 0.05

F= (S₁²/S₂²) * (σ₁²/σ₂²) ~F[tex]_{n1-1;n2-1}[/tex]

F= (5.19/5.35)*1 = 0.97

The p-value = 0.5324

Since the p-value is greater than the level of significance, the decision is to not reject the null hypothesis.

Using critical values:

Left: F[tex]F_{n1-1;n2-1;\alpha /2} = \frac{1}{F_{n2-1;n1-1;1-\alpha /2} } = \frac{1}{F_{29;29;0.95} } = \frac{1}{2.10} } =0.47[/tex]

Right: [tex]F_{n1-1; n2-1; 1-\alpha /2} = F_{29; 29; 0.975} = 2.10[/tex]

The calculated F-value (0.97) is in the not rejection zone (0.47<F<2.10) ⇒ Don't reject the null hypothesis.

I hope this helps!

The hypotheses are that there is no significant difference (null) or that there is a significant difference (alternative) in skull breadths from 4000 B.C. and A.D. 150. An F-test is used to test these via the comparison of sample variances. The conclusion depends on the P-value: if it is greater than 5%, the null is accepted (option C), and if less, rejected (option D).

The null and alternative hypotheses for this question can be stated as follows:

There is no significant difference in the variation of maximal skull breadths in 4000 B.C. and A.D. 150.

There is a significant difference in the variation of maximal skull breadths in 4000 B.C. and A.D. 150.

To test these hypotheses, we use the F-test for the equality of two variances. The test statistic (F) is calculated by taking the ratio of the sample variances, which in this case would be (5.19^2) / (5.35^2).

The P-value associated with the F statistic is then used to determine the significance of the evidence against the null hypothesis. If the P-value is less than the significance level (0.05), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

The conclusion of the hypothesis test depends on the calculated P-value. If P-value is less than 0.05, we conclude that there is a significant difference in the variation of skull breadths, thereby rejecting the null hypothesis (option D). If the P-value is greater than 0.05, we fail to reject the null hypothesis, concluding that the skull variations are not significantly different (option C).

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Answer: x = 0

y = 2

z = -1

Step-by-step explanation:

The system of equations are

x+y+z=1 - - - - - - - - - - 1

-2x+4y+6z=2 - - - - - - - - - 2

-x+3y-5z=11 - - - - - - - - - 3

Step 1

We would eliminate x by adding equation 1 to equation 3. It becomes

4y -4z = 12 - - - - - - - - - 4

Step 2

We would multiply equation 1 by 2. It becomes

2x + 2y + 2z = 2 - - - - - - - - - 5

We would add equation 2 and equation 5. It becomes

6y + 8z = 4 - - - - - - - - - 6

Step 3

We would multiply equation 4 by 6 and equation 6 by 4. It becomes

24y - 24z = 72 - - - - - - - - 7

24y + 32z = 16 - - - - - - - - 8

We would subtract equation 8 from equation 7. It becomes

-56z = 56

z = -56/56 = -1

Substituting z = -1 into 7, it becomes

24y - 24×-1 = 72

24y + 24 = 72

24y = 72 - 24 = 48

y = 48/24 = 2

Substituting y = 2 and z = -1 into equation 1, it becomes

x + 2 - 1 = 1

x = 1 - 1 = 0

Answer:

14.488 amperes

Step-by-step explanation:

The amplitude I of the current is given by

[tex]\large I=\displaystyle\frac{E_m}{Z}[/tex]

where

[tex]\large E_m[/tex] = amplitude of the energy source E(t).

Z = Total impedance.

The amplitude of the energy source is 120, the maximum value of E(t)  

The total impedance is given by

[tex]\large Z=\sqrt{R^2+(X_L-X_C)^2}[/tex]

where

R= Resistance

L = Inductance

C = Capacitance

w = Angular frequency

[tex]\large X_L=wL[/tex] = inductive reactance

[tex]\large X_C=\displaystyle\frac{1}{wC}[/tex] = capacitive reactance

As E(t) = 120sin(12t), the angular frequency w=12

So

[tex]\large X_L=12*0.37=4.44\\\\X_C=1/(12*7)=0.012[/tex]

and

[tex]\large Z=\sqrt{7^2+(4.44-0.012)^2}=8.283[/tex]

Finally

[tex]\large I=\displaystyle\frac{E_m}{Z}=\frac{120}{8.283}=14.488\;amperes[/tex]

the 95% confidence interval for the difference in proportions is approximately (0.1562, 0.2686).

To find the 95% confidence interval for the difference in proportions, we can use the formula:

[tex]\[\text{CI} = (\hat{p}_1 - \hat{p}_2) \pm z \times \sqrt{\frac{{\hat{p}_1(1 - \hat{p}_1)}}{n_1} + \frac{{\hat{p}_2(1 - \hat{p}_2)}}{n_2}}\][/tex]

Where:

- [tex]\(\hat{p}_1\) and \(\hat{p}_2\)[/tex] are the sample proportions.

- [tex]\(n_1\) and \(n_2\)[/tex] are the sample sizes.

- z is the z-score corresponding to the desired level of confidence.

Given:

- [tex]\(n_1 = 520\), \(n_2 = 460\)[/tex]

- [tex]\(\hat{p}_1 = \frac{318}{520}\), \(\hat{p}_2 = \frac{379}{460}\)[/tex]

- z = 1.96 for a 95% confidence interval

Let's plug in the values and calculate:

[tex]\[\hat{p}_1 = \frac{318}{520} \approx 0.6115\]\[\hat{p}_2 = \frac{379}{460} \approx 0.8239\]\[\text{CI} = (0.8239 - 0.6115) \pm 1.96 \times \sqrt{\frac{{0.6115 \times (1 - 0.6115)}}{520} + \frac{{0.8239 \times (1 - 0.8239)}}{460}}\]\[\text{CI} = (0.2124) \pm 1.96 \times \sqrt{\frac{{0.6115 \times 0.3885}}{520} + \frac{{0.8239 \times 0.1761}}{460}}\][/tex]

[tex]\[\text{CI} = (0.2124) \pm 1.96 \times \sqrt{0.000457 + 0.000368}\]\[\text{CI} = (0.2124) \pm 1.96 \times \sqrt{0.000825}\]\[\text{CI} = (0.2124) \pm 1.96 \times 0.0287\]\[\text{CI} = (0.2124) \pm 0.0562\][/tex]

Thus, the 95% confidence interval for the difference in proportions is approximately (0.1562, 0.2686).

Answer:

The total length can be calculated by doing thesum of all sides

Step-by-step explanation:

The total length of planting around each stepping stone is 15x - 2.

To find the total length of planting around each stepping stone, we need to add up the lengths of all the sides of the irregular polygon.

The sides of the polygon are (3x-1), (2x+1), (4x-2), (4x-4), and (2x+4).

To find the total length, we can add up the lengths of all the sides:

(3x-1) + (2x+1) + (4x-2) + (4x-4) + (2x+4)

Simplifying the expression, we get:

15x - 2

Therefore, The total length of planting around each stepping stone is 15x - 2.

The probable question may be:

Belinda placed stepping stones in the shape of the irregular polygon shown. She will plant thyme around the edge of each stepping stone. What is the total length of planting around each stepping stone ?​

The sides of the irregular polygon are (3x-1), (2x+1), (4x-2), (4x-4), (2x+4)

Answer:

0.0004498

Step-by-step explanation:

Let us define the events:

A = The test returns positive.

B = The accused was present.

Since everyone in the town has an equal probability of being at the murder scene, and the town has a population of 10,000

P(B) = 1/10000 = 0.0001

We have that the probability the test returning positive given that the accused was actually present at the scene is 0.9

P(A | B) = 0.9

and the probability that the test returns negative given that the accused was not present at the scene is 0.8

[tex]\large P(A^c|B^c)=0.8[/tex]

where

[tex]\large  A^c,\;B^c[/tex] are the complements of A and B respectively.

We want to determine the probability that the DNA test returned a positive result given that the accused was at the murder scene, that is, P(B | A).

We know that P(A | B) = 0.9, so

[tex]\large \frac{P(A\cap B)}{P(B)}=0.9\Rightarrow P(A\cap B)=0.9P(B)=0.9*0.0001\Rightarrow\\\\P(A\cap B)=0.00009[/tex]

Now, we have

[tex]\large P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{0.00009}{P(A)}[/tex]

So if we can determine P(A), the result will follow.

By De Morgan's Law

[tex]\large A^c\cap B^c=(A\cup B)^c[/tex]

so

[tex]\large 0.8=P(A^c|B^c)= \frac{P(A^c\cap B^c)}{P(B^c)}=\frac{P((A\cup B)^c)}{P(B^c)}=\frac{1-P(A\cup B)}{1-P(B)}\Rightarrow\\\\\frac{1-P(A\cup B)}{1-0.0001}=0.8\Rightarrow P(A\cup B)=1-0.8(1-0.0001)\Rightarrow\\\\P(A\cup B)=0.20008[/tex]

Using the formula

[tex]\large P(A\cup B)=P(A)+P(B)-P(A\cap  B)[/tex]

and replacing the values we have found

[tex]\large 0.20008=P(A)+0.0001-0.00009\Rightarrow\\\\P(A)=0.20007[/tex]

and finally, the desired result is

[tex]\large P(B|A)=\frac{0.00009}{P(A)}=\frac{0.00009}{0.20007}\Rightarrow\\\\\boxed{P(B|A)=0.0004498}[/tex]

Answer:

50

Step-by-step explanation:

Please see attachment .