The high price of medicines is a source of major expense for those seniors in the United States who have to pay for these medicines themselves. A random sample of 1800 seniors who pay for their medicines showed that they spent an average of $ 4600 last year on medicines with a standard deviation of $ 800 . Make a 90 % confidence interval for the corresponding population mean.

Answers

Answer 1

Answer:

[tex]4600-1.646\frac{800}{\sqrt{1800}}=4568.96[/tex]    

[tex]4600+1.646\frac{800}{\sqrt{1800}}=4631.04[/tex]    

We are confident at 90% that the true mean for the amount spent on medicines is between (4568.96 and 4631.04)

Step-by-step explanation:

Information given

[tex]\bar X=4600[/tex] represent the sample mean for the amount spent on medicines

[tex]\mu[/tex] population mean

s=800 represent the sample standard deviation

n=1800 represent the sample size  

Solution

The confidence interval for the true population mean is given by:

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

We can calculate the degrees of freedom with:

[tex]df=n-1=1800-1=1799[/tex]

We know that the Confidence level is 0.90 or 90%, the value of significance is [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: "=-T.INV(0.05,1799)".And we see that [tex]t_{\alpha/2}=1.646[/tex]

Replcing in the formula for the confidence interval we got:

[tex]4600-1.646\frac{800}{\sqrt{1800}}=4568.96[/tex]    

[tex]4600+1.646\frac{800}{\sqrt{1800}}=4631.04[/tex]    

We are confident at 90% that the true mean for the amount spent on medicines is between (4568.96 and 4631.04)


Related Questions

Suppose a recent plant journal indicated that the mean height of mature plants of a certain species of sunflower is 10.4 ft. A

biologist observing a huge field of mature plants of this particular species of sunflower thinks that the mean height is shorter

than reported. He measures the heights of 45 randomly selected sunflowers and finds the mean height to be 10.1 ft.


The hypotheses of interest are :


(A) H0 : µ = 10.1 vs Ha : µ < 10.1


(B) H0 : µ > 10.4 vs Ha : µ = 10.4


(C) H0 : ¯x = 10.1 vs Ha : ¯x > 10.1


(D) H0 : ¯x = 10.1 vs Ha : ¯x < 10.1


(E) H0 : µ = 10.4 vs Ha : µ < 10.4

Answers

Answer: (E) H0 : µ = 10.4 vs Ha : µ < 10.4

Step-by-step explanation:

The null hypothesis is the hypothesis that is assumed to be true. It is an expression that is the opposite of what the researcher predicts.

The alternative hypothesis is what the researcher expects or predicts. It is the statement that is believed to be true if the null hypothesis is rejected.

From the given situation,

The recent plant journal indicated that the mean height of mature plants of a certain species of sunflower is 10.4 ft. This is the null hypothesis.

The biologist sunflower thinks that the mean height is shorter than reported. This is the alternative hypothesis.

Therefore, the correct null and alternative hypotheses are

(E) H0 : µ = 10.4 vs Ha : µ < 10.4

A process that fills packages is stopped whenever a package is detected whose weight falls outside the specification. Assume that each package has probability 0.01 of falling outside the specification and that the weights of the packages are independent. Find the mean number of packages that will be filled before the process is stopped.

Answers

The mean number of packages that will be filled before the process is stopped  is 100

Step-by-step explanation:

Step 1

Given that each package has probability 0.01 of falling outside the specification (Probability of Failure)

The probability of Success is (100-probability of failure)=(100-0.01)=0.99

It is important to note that the weight of the packages are independent

Using the data given in the  question we get  the following:

P(fail) =Probability of Failure

P(success)=Probability of Success

P(fail) = 0.01 and P(success) = 0.99

Step 2

The mean of the Geometric distribution is:

P = 0.01;

μx = 1/p = 1/0.01 = 100

Step 3

Thus, we can say that the mean number of packages that will be filled before the process is stopped is 100

A teacher is experimenting with computer-based instruction. In which situation could the teacher use a hypothesis test for a population mean?

A) She gives each student a pretest. Then she teaches a lesson using a computer program. Afterwards, she gives each student a posttest. The teacher wants to see if the difference in scores will show an improvement.

B) She randomly divides the class into two groups. One groups receives computer-based instruction. The other group receives traditional instruction without computers. After instruction, each student has to solve a single problem. The teachers wants to compare the proportion of each group who can solve the problem.

C) The teacher uses a combination of traditional methods and computer-based instruction. She asks students which they liked better. She wants to determine if the majority prefer the computer-based instruction.

Answers

Final answer:

A teacher can use a hypothesis test for a population mean in situation A, where pretest scores are compared to posttest scores to determine if there is a significant improvement.

Explanation:

A hypothesis test for a population mean can be used in situation A. The teacher can use the pretest scores as the population mean and then compare it to the posttest scores to see if there is a statistically significant improvement. The hypothesis test will determine if the difference in scores is due to chance or if it is a result of the computer-based instruction.

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An 8-foot by 5-foot section of wall is to be covered by square tiles that measure 4 inches on each side. If the tiles are not cut, how many of them will needed to cover the wall ?

Answers

Answer:

30 tiles

Step-by-step explanation:

SO we need to find the area of the wall so we do

8 feet * 5 feet = 40 feet squared

then we find that area in inches

40*12=480

Now we find the area of the tiles

4*4=16

So now we divide to find how many tiles are needed

480/16=30

Our answer is 30

Which two events are most likely to be independent? A) Being a senior; going to homeroom. B) Registering to vote; being left-handed. C) Having a car accident; having a junior license. D) Doing the Statistics homework; getting an A on the test.

Answers

Answer:

It is B) Registering to vote; being left-handed.

Step-by-step explanation:

These two events would be the only situation where the occurrence of one has no effect on the other, and each one of the other examples contain an event that affects the other probability of the other event.

Registering to vote; being left-handed are independent events so option (B) will be correct.

What are dependent and independent events?

If the probability of an event affects the other then that will call a dependent event.

For example, getting two red balls from a bag without replacement.

While independent events are events, where the probability of the first event doesn't affect another.

For example tossing a coin.

Given the two events in all options.

In option (A) if you are a senior then it is possible that you can go to homeroom.

In option (B) if you are left-handed then also you can get a mark on the right finger it doesn't matter so that will be an independent event.

Hence "Registering to vote; being left-handed are independent events".

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Solve the equation, 4 - (x + 1) = 6.

Answers

Answer:

X=-3

Step-by-step explanation

First of all, remove the parenthesis by distributing the value of the negative sign so it would be 4-x-1=6 , next find the like terms which is 4-1 is 3. so 3-x=6.

transpose the 3 to the other side so it would be -x=6-3

-x=3 divide both sides by -1. X=-3

The capacity of n elevator is 12 people or 1968 pounds. The capacity will be exceeded if 12 people have weights with a mean greater than 1968/12=164 pounds. Suppose the people have weights that are normally distributed with a mean of 171 lb and a standard deviation of 34 lb.




a. find the probability that if a person is randomly selected, his weight will be greater than 164 pounds.




The probability is approximately ___




b. Find the probability that 12 randomly selected people will have a neam that is greater than 164 pounds.

Answers

Answer:

a) 58.32% probability that his weight will be greater than 164 pounds.

b) 76.11% probability that 12 randomly selected people will have a neam that is greater than 164 pounds.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central Limit Theorem

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

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

[tex]\mu = 171, \sigma = 34[/tex]

a. find the probability that if a person is randomly selected, his weight will be greater than 164 pounds.

This is 1 subtracted by the pvalue of Z when X = 164. So

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

[tex]Z = \frac{164 - 171}{34}[/tex]

[tex]Z = -0.21[/tex]

[tex]Z = -0.21[/tex] has a pvalue of 0.4168

1 - 0.4168 = 0.5832

58.32% probability that his weight will be greater than 164 pounds.

b. Find the probability that 12 randomly selected people will have a neam that is greater than 164 pounds.

Now [tex]n = 12, s = \frac{34}{\sqrt{12}} = 9.81[/tex]

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

By the Central Limit Theorem

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

[tex]Z = \frac{164 - 171}{9.81}[/tex]

[tex]Z = -0.71[/tex]

[tex]Z = -0.71[/tex] has a pvalue of 0.2389

1 - 0.2389 = 0.7611

76.11% probability that 12 randomly selected people will have a neam that is greater than 164 pounds.

Angle C is inscribed in circle O.
AB is a diameter of circle O.
What is the measure of angle B?

Answers

Answer:

The measure of m∠B = 19°.

Step-by-step explanation:

Given

Angle C is inscribed in circle O.m∠A = 71°

As we know that an angle inscribed in a semicircle is always a right angle.

So from the given diagram,

m∠C = 90°

As we know that the sum of the three interior angles is equal to 180.

so

m∠A + m∠B + m∠C = 180°

71° + m∠B + 90° = 180°

m∠B = 180° - 71° - 90°

m∠B = 19°

Therefore, the measure of m∠B = 19°.

Answer:

Step-by-step explanation: 19

write a quadratic function f whose zeros are 6 and 1

Answers

Answer:

x^2-7x+6

Step-by-step explanation:

(x-6)(x-1)

x^2-1x-6x+6

x^2-7x+6

The quadratic function whose zeros are 6 and 1 is: f(x) = x² - 7x + 6.

How do we find the quadratic function with zero's at x = 6 and x = 1?

To write a quadratic function f(x) with zeros at x=6 and x=1, we can start by representing it in factored form:

f(x) = a(x−6)(x−1)

Where a is a constant. Depending on the desired leading coefficient, you can choose any value for a. For simplicity, let's choose a=1.

Therefore, the quadratic function with zeros at 6 and 1 is:

f(x) = (x−6)(x−1)

f(x) = x² − 7x + 6

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An adult African elephant weighs 8.25 tons. How much do 27 elephants weigh in tons? Answer without units.

Answers

Answer:

222.75 tons

Step-by-step explanation:

1 elephant = 8.25 tons

x 27               x 27

27 elephants = 222.75 tons

I hope this helps.

Final answer:

To find the weight of 27 elephants, multiply the weight of one elephant by 27. The total weight of 27 elephants, each weighing 8.25 tons, is 222.75 tons.

Explanation:

To calculate the total weight of 27 elephants each weighing 8.25 tons, we simply multiply the weight of one elephant by the number of elephants:

8.25 tons/elephant × 27 elephants = 222.75 tons

It's clear we need to multiply when converting from a single elephant's weight to the combined weight of multiple elephants. We use the fact that 1 ton = 2,000 pounds for conversion between units if needed, but here we keep the weight in tons as the question requested.

When drawn in standard position, an angle B has a terminal ray that lies in the second quadrant and sin(B)=5/13. Find the value of cos(B), (show your answer as a fraction). Please show all your work

Please a help here.
Thank you

Answers

Answer:

[tex]\cos{B} = -\frac{12}{13}[/tex]

Step-by-step explanation:

For any angle B, we have the following trigonometric identity:

[tex]\sin^{2}{B} + \cos^{2}{B} = 1[/tex]

In this problem, we have that:

[tex]\sin{B} = \frac{5}{13}[/tex]

So, applying the trigonometric identity:

[tex]\sin^{2}{B} + \cos^{2}{B} = 1[/tex]

[tex](\frac{5}{13})^{2}) + \cos^{2}{B} = 1[/tex]

[tex]\cos^{2}{B} = 1 - (\frac{5}{13})^{2})[/tex]

[tex]\cos^{2}{B} = 1 - \frac{25}{169}[/tex]

[tex]\cos^{2}{B} = \frac{169}{169} - \frac{25}{169}[/tex]

[tex]\cos^{2}{B} = \frac{144}{169}[/tex]

[tex]\cos{B} = \pm \sqrt{\frac{144}{169}}[/tex]

[tex]\cos{B} = \pm \frac{12}{13}[/tex]

In the second quadrant, the cosine is negative. So

[tex]\cos{B} = -\frac{12}{13}[/tex]

Final answer:

The cosine in the second quadrant is negative, giving cos(B) = -12/13.

Explanation:

To find the value of cos(B) when an angle B is in standard position with its terminal ray in the second quadrant and given that sin(B) = 5/13, we can use the Pythagorean identity:

sin²(B) + cos²(B) = 1.

First, we square the sine value to get sin²(B):
sin²(B) = (5/13)² = 25/169.

Now we use the Pythagorean identity to find cos²(B):
cos²(B) = 1 - sin²(B) = 1 - 25/169 = 144/169.

Since angle B is in the second quadrant, the cosine value must be negative.

Therefore, cos(B) = -√(cos²(B)) = -√(144/169).

The square root of 144/169 is 12/13, and thus:
cos(B) = -12/13.

The number P, in hundreds of bacteria in a sample, can be modeled by the Equation P= T^4+ 5T^3 + 5T^2 + 6t where t is measured in weeks. explain how to determine the number of weeks which the population would be greater than 10,000

Answers

Answer:

The values of T which satisfies the inequality are: [tex](8.8, \infty)[/tex]

Step-by-step explanation:

In the equation [tex]P= T^4+ 5T^3 + 5T^2 + 6T[/tex]

Represent P by 10,000Write an inequality where the expression is greater than 10000: [tex]T^4+ 5T^3 + 5T^2 + 6T>10000[/tex]Get 0 on one side of the inequality.

        [tex]T^4+ 5T^3 + 5T^2 + 6T-10000>0[/tex]

Graph the polynomial function. We have real x-intercepts of 8.84 and -11.38.Determine intervals where the graph is above the x-axis. Since negative values of x in this situation are irrelevant, the values of T which satisfies the inequality are: [tex](8.8, \infty)[/tex]We now test a value from the set of solution to see if it is valid.Let T=9

[tex]T^4+ 5T^3 + 5T^2 + 6T>10000\\9^4+ 5(9)^3 + 5(9)^2 + 6(9)>10000\\10665>10000[/tex]

Since 10665 is grater than 10000, the result is reasonable.

A courier service company wishes to estimate the proportion of people in various states that will use its services. Suppose the true proportion is 0.04. If 469 are sampled, what is the probability that the sample proportion will be less than 0.03

Answers

Using the normal distribution and the central limit theorem, it is found that there is a 0.1335 = 13.35% probability that the sample proportion will be less than 0.03.

Normal Probability Distribution

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

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

It 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, which is the percentile of X.By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1 - p)}{n}}[/tex], as long as [tex]np \geq 10[/tex] and [tex]n(1 - p) \geq 10[/tex].

In this problem:

The true proportion is of p = 0.04.469 people are sampled, hence n = 469.

The mean and the standard error are given by:

[tex]\mu = p = 0.04[/tex]

[tex]s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.04(0.96)}{469}} = 0.009[/tex]

The probability that the sample proportion will be less than 0.03 is the p-value of Z when X = 0.03, hence:

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

By the Central Limit Theorem

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

[tex]Z = \frac{0.03 - 0.04}{0.009}[/tex]

[tex]Z = -1.11[/tex]

[tex]Z = -1.11[/tex] has a p-value of 0.1335.

0.1335 = 13.35% probability that the sample proportion will be less than 0.03.

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(1 point) Each of the following statements is an attempt to show that a given series is convergent or divergent not using the Comparison Test (NOT the Limit Comparison Test.) For each statement, enter C (for "correct") if the argument is valid, or enter I (for "incorrect") if any part of the argument is flawed. (Note: if the conclusion is true but the argument that led to it was wrong, you must enter I.) I equation editorEquation Editor 1. For all n>2, nn3−2<2n2, and the series 2∑1n2 converges, so by the Comparison Test, the series ∑nn3−2 converges. equation editorEquation Editor 2. For all n>2, 1n2−1<1n2, and the series ∑1n2 converges, so by the Comparison Test, the series ∑1n2−1 converges. I equation editorEquation Editor 3. For all n>2, n+1−−−−−√n>1n, and the series ∑1n diverges, so by the Comparison Test, the series ∑n+1−−−−−√n diverges. I equation editorEquation Editor 4. For all n>2, ln(n)n2>1n2, and the series ∑1n2 converges, so by the Comparison Test, the series ∑ln(n)n2 converges. I equation editorEquation Editor 5. For all n>1, 1nln(n)<2n, and the series 2∑1n diverges, so by the Comparison Test, the series ∑1nln(n) diverges. I equation editorEquation Editor 6. For all n>1, n7−n3<1n2, and the series ∑1n2 converges, so by the Comparison Test, the series ∑n7−n3 converges.

Answers

Answer:

I.

CORRECT.

II.

CORRECT.

III.

CORRECT.

IV.

CORRECT.

V.

INCORRECT.

VI.

CORRECT

Step-by-step explanation:

To understand let us restate the comparison test in simple terms.

Comparison test :

Given     [tex]\text{Series}_A[/tex]   and [tex]\text{Series}_B[/tex] such that    [tex]\text{Series}_A < \text{Series}_B[/tex] , then

1.   If    [tex]\text{Series}_B[/tex] converges then [tex]\text{Series}_A[/tex]  converges as well.

2.  If  [tex]\text{Series}_A[/tex]  diverges then [tex]\text{Series}_B[/tex]  diverges as well.

Now to give you a more intuitive idea of what is going on, think about it like this.  When   the series on top converges it is like an "upper bound" for what you have on the bottom, therefore what you have on the bottom has to converge as well.

Similarly if what you have on the bottom explotes, then what you have on top will explote as well.

That's how I like to think about that intuitively.

Now, using those results let us examine the statements.

I.

[tex]\frac{1}{n} < \frac{ln(n)}{n}[/tex]

Since the infinite sum of 1/n  diverges in fact the infinite sum of ln(n)/n does not converge.

Therefore, CORRECT.

II.

[tex]\frac{\arctan(n)}{n^3} < \frac{\pi}{2}\frac{1}{n^3}[/tex]  

Since the infinite sum of    [tex]\frac{\pi}{2}\frac{1}{n^3}[/tex]   is in fact convergent then  [tex]\frac{\arctan(n)}{n^3}[/tex] converges as well using the comparison theorem. Therefore

CORRECT.

III.

[tex]\frac{n}{2-n^3} < \frac{1}{n^2}[/tex]

Once again   [tex]1/n^2[/tex]  does converge so what you have on the bottom converges as well. Therefore

CORRECT.

IV.

[tex]\frac{\ln(n)}{n^2} < \frac{1}{n^{1.5}}[/tex]

Once again   [tex]\frac{1}{n^{1.5}}[/tex]  converges therefore since it is on top what is on the bottom converges as well. Therefore.

CORRECT.

V.

[tex]\frac{\ln(n)}{n} < \frac{2}{n}[/tex]

Now the fact that  [tex]\frac{2}{n}[/tex]    diverges does not necessarily imply that what you have on the bottom diverges. Therefore

INCORRECT.

VI.

That is correct as well since what you have on top converges therefore what you have on the bottom converges as well.

 

the sum 27 and a number g

Answers

27 and g can not be added together therefore it is 27+g

Answer:

27 + g is how you write if u want a solution tell me what G is = to

Step-by-step explanation:

Find the greatest common factor of 4c and 10c".

Answers

Answer:

Step-by-step explanation:

One day, Donnie observes that the wind is blowing at 6 miles per hour. A unladen swallow nesting near Donnie’s house flies three quarters of a mile down the road (in the direction of the wind), turns around, and returns exactly 4 minutes later. What is the airspeed of the unladen swallow? (Here, ‘airspeed’ is the speed that the swallow can fly in still air.) socratic.org

Answers

Answer: 11.24 miler per hour

Step-by-step explanation:

The package of a particular brand of rubber band says that the bands can hold a weight of 7 lbs. Suppose that we suspect this might be an overstatement of the breaking weight. So we decide to take a random sample of 36 of these rubber bands and record the weight required to break each of them. The mean breaking weight of our sample of 36 rubber bands is 6.6 lbs. Assume that the standard deviation of the breaking weight for the entire population of these rubber bands is 2 lbs. True or false

Answers

Answer:

The statement is True.

Step-by-step explanation:

In this case we need to determine whether the rubber bands in a package of a particular brand of rubber band can hold a weight of 7 lbs or less.

A one-sample test can be used to perform the analysis.

The hypothesis can be defined as follows:

H₀: The mean weight the rubber bands can hold is 7 lbs, i.e. μ = 7.

Hₐ: The mean weight the rubber bands can hold is less than 7 lbs, i.e. μ < 7.

The information provided is:

 [tex]n=36\\\bar x=6.6\ \text{lbs}\\\sigma=2\ \text{lbs}[/tex]

As the population standard deviation is provided, we will use a z-test for single mean.

Compute the test statistic value as follows:

[tex]z=\frac{\bar x-\mu}{\sigma/\sqrt{n}}=\frac{6.6-7}{2/\sqrt{36}}=-1.20[/tex]  

The test statistic value is -1.20.

Decision rule:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected.

Compute the p-value for the two-tailed test as follows:

 [tex]p-value=P(Z<-1.20)=0.1151[/tex]

*Use a z-table for the probability.

The p-value of the test is 0.1151.

The p-value of the test is very large for all the commonly used significance level. The null hypothesis will not be rejected.

Thus, it can be concluded that the mean weight the rubber bands can hold is 7 lbs.

Hence, the statement is True.

The sodium content of a popular sports drink is listed as 205 mg in a 32-oz bottle. Analysis of 20 bottles indicates a sample mean of 219.2 mg with a sample standard deviation of 18.0 mg. (a) State the hypotheses for a two-tailed test of the claimed sodium content. H0: μ ≥ 205 vs. H1: μ < 205 H0: μ ≤ 205 vs. H1: μ > 205 H0: μ = 205 vs. H1: μ ≠ 205

Answers

Answer:

[tex]t=\frac{219.2-205}{\frac{18}{\sqrt{20}}}=3.528[/tex]    

[tex]p_v =2*P(t_{19}>3.528)=0.0022[/tex]    

If we compare the p value and a significance level for example [tex]\alpha=0.05[/tex] we see that [tex]p_v <\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the lifetime is signficantly different from 205 hours.    

Step-by-step explanation:

Data given and notation    

[tex]\bar X=219.2[/tex] represent the sample mean

[tex]s=18[/tex] represent the sample standard deviation    

[tex]n=20[/tex] sample size    

[tex]\mu_o =205[/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 apply a two tailed  test.  

What are H0 and Ha for this study?    

Null hypothesis:  [tex]\mu = 205[/tex]  

Alternative hypothesis :[tex]\mu \neq 205[/tex]  

Compute the test statistic  

The statistic for this case is given by:  

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)    

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

Calculate the statistic    

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

[tex]t=\frac{219.2-205}{\frac{18}{\sqrt{20}}}=3.528[/tex]    

Give the appropriate conclusion for the test  

The degreed of freedom are:

[tex] df = n-1= 19[/tex]

Since is a two sided test the p value would be:    

[tex]p_v =2*P(t_{19}>3.528)=0.0022[/tex]    

Conclusion    

If we compare the p value and a significance level for example [tex]\alpha=0.05[/tex] we see that [tex]p_v <\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the lifetime is signficantly different from 205 hours.    

Answer:

H0: mu equals 205 vs. H1: mu not equals 205.

Step-by-step explanation:

A null hypothesis (H0) is a statement from a population parameter which is either rejected or accepted (fail to reject) upon testing. It expresses equality.

An alternate hypothesis (H1) is also a statement from the population parameter which negates the null hypothesis and is accepted if the null hypothesis is true. It expresses inequality.

A two-tailed test is one in which the alternate hypothesis is expressed using any of the inequality signs below:

not equal to, less than or equal to, greater than or equal to.

A manufacturer claims that the thickness of the spearmint gum it produces is 7.5 one-hundredths of an inch. A quality control specialist regularly checks this claim. From his experience, the specialist knows the distribution is right skewed and that the standard deviation is 0.4 one-hundredths of an inch. On one production run, he took a random sample of n = 100 pieces of gum and measured their thickness. Which hypothesis test would be most appropriate for this task?

Answers

Final answer:

The most appropriate hypothesis test for this task would be a one-sample t-test.

Explanation:

For this task, the most appropriate hypothesis test to use would be a one-sample t-test. A one-sample t-test is used to determine whether the mean of a sample is significantly different from a population mean.

In this case, the quality control specialist wants to test whether the mean thickness of the gum produced by the manufacturer is significantly different from the claimed value of 7.5 one-hundredths of an inch. The specialist takes a random sample of n = 100 pieces of gum and measures their thickness.

The one-sample t-test is appropriate because the population distribution is assumed to be right skewed and the standard deviation is known. The t-test allows for the use of a smaller sample size and accommodates the assumption of a non-normal distribution.

Learn more about Hypothesis testing here:

https://brainly.com/question/34171008

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A scientist claims that 7%7% of viruses are airborne. If the scientist is accurate, what is the probability that the proportion of airborne viruses in a sample of 418418 viruses would be greater than 6%6%? Round your answer to four decimal places.

Answers

Final answer:

The probability that the proportion of airborne viruses in the sample would be greater than 6% is approximately 0.5596.

Explanation:

To find the probability that the proportion of airborne viruses in a sample of 418 viruses would be greater than 6%, we can use the normal distribution. First, let's find the mean and standard deviation of the proportion of airborne viruses. The mean is 7% (0.07) and the standard deviation can be calculated using the formula: sqrt((p(1-p))/n), where p is the proportion and n is the sample size.

Using the given information, the standard deviation is sqrt((0.07(1-0.07))/418) ≈ 0.0064.

Next, we convert the 6% threshold to a z-score using the formula (x - mean) / standard deviation. So, the z-score is (0.06 - 0.07) / 0.0064 ≈ -0.156.

To find the probability that the proportion is greater than 6%, we look up the z-score (-0.156) in the standard normal distribution table and subtract the corresponding probability from 1. The probability is approximately 0.5596.

The graphs below have the same shape. Complete the equation of the red graph. Enter exponents using the caret (^); for example, enter x2 as x^2. Do not include "F(x) =" in your answer.

Answers

Answer:

f(x) = 1-x^2

Step-by-step explanation:

The graph of f(x) is just shifted down 3 units

A shift down is a subtraction

f(x) = g(x) -3

     = 4- x^2 -3

     =1-x^2

Answer:

1 - x^2

Step-by-step explanation:

F(x) is 3 units down. So,

G(x) - 2 = 4 - x² - 3

F(x) = 1 - x²

10 boys share 7 cereal bars equally what fraction of the cereal bars does each boy get? show the work.

Answers

Answer:

1 43/100

Step-by-step explanation:

10 boys divide 7 cereal bars equally or

10 / 7 = 1.428571428571429

Or 1.43 rounded Or 1 43/100

What is the whole number for 6/1?

Answers

Answer:

6

Step-by-step explanation:

Answer:

6

Step-by-step explanation:

Because 1/1 = 1 whole and you add 5 more 1/1 and you get 6/1 so their is 6 wholes

Rhada has a 6-pound bag of day . Her craft project requires 6 ounces of clay for each batch of 3 ornaments,

Answers

Answer:

Rhada can make 48 ornaments if she uses all of the clay.

Step-by-step explanation:

The complete question is:

Rhada has a 6-pound bag of day . Her craft project requires 6 ounces of clay for each batch of 3 ornaments. If she uses all of the clay, how many ornaments can Rhada make?

Solution:

The amount of clay Rhada has, X = 6-pound.

Convert the weight of clay bag into ounces as follows:

1 pound = 16 ounces

6 pound = 16 × 6

              = 96 ounces.

So, Rhada has 96 ounces of clay.

It is provided that her craft project requires 6 ounces of clay for each batch of ornaments.

Compute the number of batches that can be made by 96 ounces of clay as follows:

Number of batches = Total weight of clay ÷ Amount required for each batch

                                [tex]=\frac{96}{6}\\=16[/tex]

Thus, Rhada can make 16 batches of ornaments.

Now, it is also provided that each batch has 3 ornaments.

Compute the number of ornaments in 16 batches as follows:

Number of ornaments = Number of batches × No. of ornament in 1 batch

                                     [tex]=16\times 3\\=48[/tex]

Thus, Rhada can make 48 ornaments if she uses all of the clay.

Which two events are independent? A.) go sledding, snow B.) sleep,dream C.) count to 100, fix a bike D.) do chores, earn allowance

Answers

Answer:

A and C

Step-by-step explanation:

The meaning of an event being independent is that one event doesn't effect the probability of they other event happening. So B is wrong because whether yo sleep or not can affect whether you dream or not and D is wrong because you earning and allowance is based off whether you do your chores or not.

Prior Knowledge Questions (Do these BEFORE using the Gizmo.)
Recall that a percent is a ratio of a number to 100. For example 39% means 39 out of 100.

1.Last year, there were 20 students on the track team. Eight of them competed in the long
jump. What percent of the team members competed in the long jump?

Answers

Answer:

1.6% should be right.

Step-by-step explanation:

A study by Consumer Reports showed that 64% of supermarket shoppers believe supermarket brands to be as good as national name brands. To investigate whether this result applies to its own product, consider the hypothesis testing with H subscript 0 colon space p space equals space 0.64 space space space v s. space H subscript a colon space p space not equal to 0.64If a sample of 100 shoppers showed 52 stating that the supermarket brand was as good as the national brand, what is the p-value

Answers

Answer:

[tex]z=\frac{0.52 -0.64}{\sqrt{\frac{0.64(1-0.64)}{100}}}=-2.5[/tex]  

The p value for this case is given by:

[tex]p_v =2*P(z<-2.5)=0.0124[/tex]  

Step-by-step explanation:

Data given and notatio  

n=100 represent the random sample selected

X=52 represent the shoppers stating that the supermarket brand was as good as the national brand

[tex]\hat p=\frac{52}{100}=0.52[/tex] estimated proportion of stating that the supermarket brand was as good as the national brand

[tex]p_o=0.64[/tex] is the value that we want to test

z would represent the statistic

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

System of hypothesis

We need to conduct a hypothesis in order to test the claim that the true proportion of shoppers stating that the supermarket brand was as good as the national brand is 0.64 or not, then the system of hypothesis are.:  

Null hypothesis:[tex]p=0.64[/tex]  

Alternative hypothesis:[tex]p \neq 0.64[/tex]  

The statistic is given by:

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

Calculate the statistic  

The statistic is given by:

[tex]z=\frac{0.52 -0.64}{\sqrt{\frac{0.64(1-0.64)}{100}}}=-2.5[/tex]  

Statistical decision  

The p value for this case is given by:

[tex]p_v =2*P(z<-2.5)=0.0124[/tex]  

5) The records of 100 SNHU students show the following courses taken: 53 students took History 41 students took Marketing 48 students took Writing 18 students took History and Marketing 21 students took Marketing and Writing 7 students took all 3 courses 9 students took none of these courses Answer the following questions. Show how you obtained your solution. a) How many students took Marketing and Writing, but not History? b) How many students took only History?.

Answers

Answer:

a) How many students took Marketing and Writing, but not History?

14 Students.

b) How many students took only History?

23 Students.

Step-by-step explanation:

We were given the following values in the question

Total number of students = 100

53 students took History = n(H)

41 students took Marketing = n(M)

48 students took Writing = n(W)

18 students took History and Marketing = n(H n M)

21 students took Marketing and Writing = n ( M n W)

7 students took all 3 courses = n ( H n M n W)

9 students took none of these courses

a) How many students took Marketing and Writing, but not History?

This is calculated as

= n(M n W) - n ( H n M n W)

= 21 - 7

= 14

Therefore, only 14 Students took Marketing and Writing but not History.

b) How many students took only History?

Step 1

Subtract the Total number of students from the number of students who did not take any course

100 - 9 = 91 students

Step 2

The number of students who took History and Writing (H n W) was not given, so we find it.

Therefore,

n(H) + n(M) + n(W) -n( H n M) - n( M n W) - n( H n W) + n( H n M n W) = 91

53 + 41 + 48 - 18 - 21 - n( H n W) + 7 = 91

n( H n W) = 149 - 39 -91

n( H n W) = 149 - 130

n( H n W) = 19

Step 3

To find the number of students taking History only, we calculate it as

Number of Students that took History only = n( H) - n(H n W) - n( H n M) + n( H n M n W)

= 53 - 19 - 18 + 7

= 53 + 7 -19 - 18

= 60 - 37

= 23

Therefore, the number of students that took only History is 23 Students.

b. Interpret the Rsquared value. Does the multiple regression equation help us predict the total golf score much better than we could without knowing that​ equation? A. No​, because Rsquared is close to 1. B. Yes​, because Rsquared is not close to 1. C. No​, because Rsquared is not close to 1. D. Yes​, because Rsquared is close to 1.

Answers

Answer:

Depending upon the golf score predictors we can conclude the answer from given Conditional statements

If Someone want Total golf score better then, following .condition

Yes, Because R squared is close to 1 .  ....(If Regression values are prefect to fit with model then only)

(But the data is not sufficient to say 100%, it depends on score given)

Step-by-step explanation:

Given :

Explanatory Question on r-squared in regression.

To Find :

Does regression equation help us to know total golf score much better.

Solution;

As  in Question there is no golf score given so we cant see "yes" or "no".

But we can be conditional here ,

1) If golf score to be accurate then , multiple regression predict points should fit the model prefect ,hence the R squared value "must be close to 1" .

2) If gold score is not accurate then multiple regression   suggest that model does  not explain any variables, no linear relationship.

then r squared value "must not be close to 1".

Depending upon the golf score predictors we can conclude the answer from given Conditional statements.

that  Yes, Because R squared is close to 1 .

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