At a nationwide travel agency, newly hired employees are classified as beginners (B). Every six months the performance of each agent is reviewed. Past records indicate that transitions through the ranks to intermediate (I) and qualified (Q) are according to the following Markov chain, where F indicates workers that were fired BIQ F B 45.4 0.15 Q 0 0 1 0 F 0 0 0 1 (a) What fraction eventually become qualified? (b) What is the expected time until a beginner is fired or becomes qualified?

Answer:

a) 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 true mean is not significantly different from 8.17 at 5% of signficance.  

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

[tex]p_v =2*P(t_{(49)}<-1.525)=0.067[/tex]  

Step-by-step explanation:

1) Data given and notation  

[tex]\bar X=8.159[/tex] represent the mean weight for the sample  

[tex]s=0.051[/tex] represent the sample standard deviation

[tex]n=50[/tex] sample size  

[tex]\mu_o =8.17[/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)  

Part a

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is different from 8.17, the system of hypothesis would be:  

Null hypothesis:[tex]\mu = 8.57[/tex]  

Alternative hypothesis:[tex]\mu \neq 8.57[/tex]  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

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

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

Calculate the statistic

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

[tex]t=\frac{8.159-8.17}{\frac{0.051}{\sqrt{50}}}=-1.525[/tex]    

P-value

The first step is calculate the degrees of freedom, on this case:  

[tex]df=n-1=50-1=49[/tex]  

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

[tex]p_v =2*P(t_{(49)}<-1.525)=0.067[/tex]  

Conclusion  

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significantly different from 8.17 at 5% of signficance.  

To fill out the normal distribution curve, use the average weight and standard deviation. To find the percentage of people who would receive a bag with a weight greater than a certain amount, calculate the Z-score and use a Z-table.

To fill out the normal distribution curve for this situation, we can use the given information. The bags have an average weight of 1,040 grams with a standard deviation of 25 grams. This means that the mean of the distribution is 1,040 and the standard deviation is 25. We can plot the normal distribution curve using these values.

To find the percentage of people who would receive a bag that had a weight greater than 1,115 grams, we need to find the area under the normal distribution curve to the right of 1,115. We can calculate this using a Z-score calculator or a Z-table. The Z-score for 1,115 is (1,115 - 1,040) / 25 = 3. From the Z-table, we can find that the area to the right of Z-score 3 is approximately 0.0013. This means that approximately 0.13% of people would receive a bag that weighs more than 1,115 grams.

Answer:

[tex]z=\frac{7.25-7.5}{\frac{1.4}{\sqrt{10}}}=-0.565[/tex]  

[tex]p_v =P(Z<-0.565)=0.286[/tex]  

a) 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 true average is not significantly less than 7.5.  

b) [tex]\chi^2 =\frac{10-1}{1.96} 4 =18.367[/tex]  

[tex]p_v =P(\chi^2 >18.367)=0.0311[/tex]

If we compare the p value and the significance level provided we see that [tex]p_v >\alpha[/tex] so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population variance is not significantly higher than 1.96.

Step-by-step explanation:

Assuming this info: "Suppose birth weights follow a normal distribution with mean 7.5 pounds and standard deviation 1.4 pounds"

1) Data given and notation  

[tex]\bar X=7.25[/tex] represent the sample mean  

[tex]s=1.2[/tex] represent the sample standard deviation

[tex]\sigma=1.4[/tex] represent the population standard deviation

[tex]n=10[/tex] sample size  

[tex]\mu_o =7.5[/tex] represent the value that we want to test  

[tex]\alpha=0.05,0.01[/tex] represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

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

2) State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is less than 7.5, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \geq 7.5[/tex]  

Alternative hypothesis:[tex]\mu < 7.5[/tex]  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

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

z-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]z=\frac{7.25-7.5}{\frac{1.4}{\sqrt{10}}}=-0.565[/tex]  

4)P-value  

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

[tex]p_v =P(Z<-0.565)=0.286[/tex]  

5) Conclusion  

Part a

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 true average is not significantly less than 7.5.  

Part b

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

[tex]n=10[/tex] represent the sample size

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

[tex]s^2 =4 [/tex] represent the sample variance obtained

[tex]\sigma^2_0 =1.96[/tex] represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population variance increase, so the system of hypothesis would be:

Null Hypothesis: [tex]\sigma^2 \leq 1.96[/tex]

Alternative hypothesis: [tex]\sigma^2 >1.96[/tex]

Calculate the statistic  

For this test we can use the following statistic:

[tex]\chi^2 =\frac{n-1}{\sigma^2_0} s^2[/tex]

And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.

[tex]\chi^2 =\frac{10-1}{1.96} 4 =18.367[/tex]

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case 9. And since is a right tailed test the p value would be given by:

[tex]p_v =P(\chi^2 >18.367)=0.0311[/tex]

In order to find the p value we can use the following code in excel:

"=1-CHISQ.DIST(18.367,9,TRUE)"

Conclusion

If we compare the p value and the significance level provided we see that [tex]p_v >\alpha[/tex] so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population variance is not significantly higher than 1.96.

Answer:

There should be at most 24 lucky numbers in the third bag.

Step-by-step explanation:

Initially, there are 200 numbers. Two bags with 100 each. There are 31+18 = 49 lucky numbers. So there is a 49/200 = 0.245 probability that a randomly selected number from a random bag is the lucky number.

Now with 300 numbers, we want this probability to be lower than 24.5%. So we should solve the following rule of three:

200 - 49

300 - x

[tex]200x = 300*49[/tex]

[tex]x = 1.5*49[/tex]

[tex]x = 73.5[/tex]

With the third bag, the probability will be the same if 73.5-49 = 24.5 lucky numbers are added. So there should be at most 24 lucky numbers in the third bag.

Answer:

give me a second to research the answer

Step-by-step explanation:

Answer:

[tex]p_v = P(\chi^2_{29}>42.963)=1-0.954=0.0459[/tex]

b. between .025 and .05

Step-by-step explanation:

Previous concepts and notation

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

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

n = 30 sample size

s= 0.2 represent the sample deviation

[tex]\sigma_o =\sqrt{0.027}=0.164[/tex] the value that we want to test

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

t represent the statistic

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

State the null and alternative hypothesis

On this case we want to check if the population standard deviation is less than 0.027, so the system of hypothesis are:

H0: [tex]\sigma \leq 0.027[/tex]

H1: [tex]\sigma >0.027[/tex]

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

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

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

What is the value of your test statistic?

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

[tex] t=(30-1) [\frac{0.2}{0.164}]^2 =42.963[/tex]

What is the approximate p-value of the test?

The degrees of freedom are given by:

[tex] df=n-1= 30-1=29[/tex]

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

[tex]p_v = P(\chi^2_{29}>42.963)=1-0.954=0.0459[/tex]

And the best option would be:

b. between .025 and .05

The 95% confidence interval for the population proportion of college students who work to pay for tuition is (0.4228, 0.5172) and the 99% confidence interval is (0.4086, 0.5314).

To calculate a confidence interval for a population proportion, we use the formula p ± Z*(√((p(1-p))/n)), where p is the sample proportion, Z is the Z-score associated with our desired confidence level, and n is the sample size.

In this case, p = 0.47 (47%), n = 450, and the Z-scores are 1.96 for a 95% confidence interval and 2.58 for a 99% confidence interval.

For the 95% confidence interval, we calculate:

0.47 ± 1.96*(√((0.47*(1-0.47))/450)) = 0.47 ± 0.0472, so the 95% confidence interval is (0.4228, 0.5172).

For the 99% confidence interval, we calculate:

0.47 ± 2.58*(√((0.47*(1-0.47))/450)) = 0.47 ± 0.0614, so the 99% confidence interval is (0.4086, 0.5314).

https://brainly.com/question/34700241

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Answer:

Step-by-step explanation:

Given

mean [tex]\mu =12.9 Pounds[/tex]

Standard deviation [tex]\sigma =1\ Pound[/tex]

[tex]P(x<11)=P\left ( \frac{x-\mu }{\sigma }< \frac{11-12.9}{1}\right )[/tex]

[tex]P(x<11)=P\left ( z< -1.9\right )[/tex]

From z table

[tex]P(x<11)=0.0289\approx 0.029[/tex]

Answer:

a. none of these answers

Data provides sufficient evidence, at 1% significance level, to support the expert's claim. In addition the p-value (or the observed significance level) is equal to P( T >3.281)

Step-by-step explanation:

Data given and notation  

[tex]\bar X=13.5[/tex] represent the mean height for the sample  

[tex]s=3.2[/tex] represent the sample standard deviation for the sample  

[tex]n=9[/tex] sample size  

[tex]\mu_o =10[/tex] represent the value that we want to test

[tex]\alpha=0.01[/tex] represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

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

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is higher than 10 years, the system of hypothesis would be:  

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

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

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

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

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{13.5-10}{\frac{3.2}{\sqrt{9}}}=3.28[/tex]    

P-value

The first step is calculate the degrees of freedom, on this case:  

[tex]df=n-1=9-1=8[/tex]  

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

[tex]p_v =P(t_{(8)}>3.281)=0.00558[/tex]  

Conclusion  

If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can't conclude that we have a mean higher than 10 years at 1% of significance.  

a. none of these answers

Answer:

1. [tex]t_{0.01/2,(10)}=3.169[/tex]

2. Confidence interval = [15.2, 25.2]

Step-by-step explanation:

X = 20.2

S = 5.19

n = 11

Step 1 of 2: Critical value

α = 1 - 0.99 = 0.01

df = 11 - 1 = 10

Looking at t distribution table with α = 0.01 and df = 10, we find

[tex]t_{0.01/2,(10)}=3.169[/tex]

Step 2 of 2: 99% confidence interval

[tex]std-err=\frac{S}{\sqrt{n}}=\frac{5.19}{\sqrt{11}}=1.5648[/tex]

[tex]X+t_{0.01/2,10}*std-err[/tex]

20.2 + 3.169*1.5648

20.2 + 4.9595

25.2

[tex]X-t_{0.01/2,10}*std-err[/tex]

20.2 - 3.169*1.5648

20.2 - 4.9595

15.2

Confidence interval = [15.2, 25.2]

Hope this helps!

Step-by-step explanation:

Since we have given that

Sample size = 25

Mean = 46 years

Standard deviation = 5 years

We need to find the margin of error of a 98% confidence interval for the true mean client age.

So, Margin of error is given by

[tex]z\times \dfrac{\sigma}{\sqrt{n}}\\\\=2.33\times \dfrac{5}{\sqrt{25}}\\\\=2.33\times \dfrac{5}{5}\\\\=2.33[/tex]

Hence, margin of error is 2.33.

Answer:

B

Step-by-step explanation:

Type I error is basically rejection of the null hypothesis when the null hypothesis is true. In the given scenario the null hypothesis consists of the percentage of students who own cars is 35%. Hence the type I error would be rejection of null hypothesis that the percentage of students who own cars is 35% while the percentage is 35%.

Final answer:

A Type I error is rejecting the null hypothesis when it's true, and a Type II error is failing to reject the null hypothesis when it's false. In this case, Type I error is option B and Type II error is option C.

Explanation:

When performing hypothesis testing, there is potential to make two types of errors: Type I error and Type II error. A Type I error occurs when we reject the null hypothesis even though the null hypothesis is actually true. In the context of the question, the Type I error would correspond to option B: Reject the null hypothesis that the percentage of college students who own cars is equal to 35% when that percentage is actually equal to 35%.

In contrast, a Type II error happens when we fail to reject the null hypothesis whereas the null hypothesis is false. It is correctly identified by option C: Fail to reject the null hypothesis that the percentage of college students who own cars is equal to 35% when that percentage is actually different from 35%.

Answer:

A .55

Step-by-step explanation:

Given that Leroy and Fred play chess at a club every Wednesday. The probability that Leroy will lose is .3, that he will stalemate is .5, and that he will win is .2. The probability that Fred will lose is .25, that he will stalemate is .4, and that he will win is .35.

lose stale win

Leroy 0.3 0.5 0.2

Fred 0.25 0.4 0.35

Probability that atleast one wins = Prob Leroy wins + Prob fred wins - prob both wins

(using addition theorem on probability)

But prob of both winning is 0

Hence required prob = 0.2 + 0.35 = 0.55

Option a is right

Answer:

0.48

Step-by-step explanation:

Answer:

(0,-6)

Step-by-step explanation:

recall that the y-intercept is simply the point where the line crosses the y-axis at x = 0.

to find the y intercept, we simply substitute x=0 into the equation and solve for y

12x - 10y = 60, when x = 0,

12(0) - 10y = 60

-10y = 60

y = 60 / (-10)

y = -6

hence the coordinate of the y - intercept is (0,-6)

Answer:

a) The 99% confidence interval would be given by (589.588;731.038)

b) If we see the confidence interval the 620 is included on the interval we don't have enough evidence to reject the rating is 620. But since we need a score that at least 620 and our lower limit is 589.588 we cant conclude that all the clients would have a score that at least of 620.  

Step-by-step explanation:

Previous concepts

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

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

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The data is:

661 595 548 730 791 678 672 491 492 583 762 624 769 729 734 706

Part a

Compute the sample mean and sample standard deviation.  

In order to calculate the mean and the sample deviation we need to have on mind the following formulas:  

[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex]  

[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex]  

=AVERAGE(661, 595, 548, 730, 791, 678, 672, 491 ,492, 583, 762 ,624 ,769, 729, 734, 706)

On this case the average is [tex]\bar X= 660.313[/tex]

=STDEV.S(661, 595, 548, 730, 791, 678, 672, 491 ,492, 583, 762 ,624 ,769, 729, 734, 706)

The sample standard deviation obtained was s=95.898

Find the critical value t* Use the formula for a CI to find upper and lower endpoints

In order to find the critical value we need to take in count that our sample size n =16<30 and on this case we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 99% of confidence, our significance level would be given by [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2 =0.005[/tex]. The degrees of freedom are given by:

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

We can find the critical values in excel using the following formulas:

"=T.INV(0.005,15)" for [tex]t_{\alpha/2}=-2.95[/tex]

"=T.INV(1-0.005,15)" for [tex]t_{1-\alpha/2}=2.95[/tex]

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

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

And if we find the limits we got:

[tex]660.313- 2.95\frac{95.898}{\sqrt{16}}=589.588[/tex]  

[tex]660.313+ 2.95\frac{95.898}{\sqrt{16}}=731.038/tex]  

So the 99% confidence interval would be given by (589.588;731.038)

Part b

If we see the confidence interval the 620 is included on the interval we don't have enough evidence to reject the rating is 620. But since we need a score that at least 620 and our lower limit is 589.588 we cant conclude that all the clients would have a score that at least of 620.  

Answer:

[tex]t=\frac{39-40}{\frac{8.8}{\sqrt{70}}}=-0.9508[/tex]      

[tex]p_v =P(t_{69}<-0.9508)=0.1725[/tex]  

If we compare the p value and the significance level given for example [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significantly less than 40 months.

Step-by-step explanation:

1) Data given and notation      

[tex]\bar X=39[/tex] represent the sample mean

[tex]s=8.8[/tex] represent the standard deviation for the sample      

[tex]n=70[/tex] sample size      

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

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

t would represent the statistic (variable of interest)      

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

State the null and alternative hypotheses.      

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

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

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

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:      

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

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

Calculate the statistic      

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

[tex]t=\frac{39-40}{\frac{8.8}{\sqrt{70}}}=-0.9508[/tex]      

Calculate the P-value      

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

[tex]df=n-1=70-1=69[/tex]  

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

[tex]p_v =P(t_{69}<-0.9508)=0.1725[/tex]  

Conclusion      

If we compare the p value and the significance level given for example [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significantly less than 40 months.      

Answer:

a) If we see the alternative hypothesis we see that we are conducting a bilateral test or two tail.

b) (-1.9886, 1.9886)

c) [tex]t=\frac{(100 -98.4)-(0)}{\sqrt{\frac{3.4^2}{41}}+\frac{5.6^2}{45}}=1.617[/tex]

d) [tex]p_v =2*P(t_{84}>1.617) =0.1096[/tex]

So with the p value obtained and using the significance level given [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 mean of the group 1 is NOT significantly different than the mean for the group 2.  

e) [tex]p_v =2*P(t_{84}>1.617) =0.1096[/tex]

We can use the following excel code to calculate it: "=2*(1-T.DIST(1.617;84;TRUE))"

Step-by-step explanation:

The system of hypothesis on this case are:

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

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

Or equivalently:

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

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

Our notation on this case :

[tex]n_1 =41[/tex] represent the sample size for group 1

[tex]n_2 =45[/tex] represent the sample size for group 2

[tex]\bar X_1 =100[/tex] represent the sample mean for the group 1

[tex]\bar X_2 =98.4[/tex] represent the sample mean for the group 2

[tex]s_1=3.4[/tex] represent the sample standard deviation for group 1

[tex]s_2=5.6[/tex] represent the sample standard deviation for group 2

Part a

If we see the alternative hypothesis we see that we are conducting a bilateral test or two tail.

Part b

On this case since the significance level is 0.05 and we are conducting a bilateral test we have two critical values, and we need on each tail of the distribution [tex]\alpha/2 = 0.025[/tex] of the area.

The distribution on this cas since we don't know the population deviation for both samples is the t distribution with [tex]df=n_1+n_2 -2= 41+45-2=84[/tex] degrees of freedom.

We can use the following excel codes in order to find the critical values:

"=T.INV(0.025,84)", "=T.INV(1-0.025,84)"

And we got: (-1.9886, 1.9886)

Part c

The statistic is given by this formula:

[tex]t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{S^2_2}{n_2}}[/tex]

And now we can calculate the statistic:

[tex]t=\frac{(100 -98.4)-(0)}{\sqrt{\frac{3.4^2}{41}}+\frac{5.6^2}{45}}=1.617[/tex]

The degrees of freedom are given by:

[tex]df=41+45-2=84[/tex]

Part d

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

[tex]p_v =2*P(t_{84}>1.617) =0.1096[/tex]

So with the p value obtained and using the significance level given [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 mean of the group 1 is NOT significantly different than the mean for the group 2.  

Part e

[tex]p_v =2*P(t_{84}>1.617) =0.1096[/tex]

We can use the following excel code to calculate it: "=2*(1-T.DIST(1.617;84;TRUE))"

Answer:

Step-by-step explanation:

Given that a curve in polar coordinates is given by:

r=9+3cosθ

a) At point P, we have

[tex]θ=\frac{21\pi}{18}[/tex]

Substitute to get

[tex]r=9+cos \frac{21\pi}{18}\\=9.3932[/tex]

b) Cartesian coordinate is

[tex]x= rcos \theta=3.6934\\y =r sin \tjeta =8.6365[/tex]

c) At the origin r =0

when r =0

we have

[tex]9+3cos\theta=0\\cos\theta =-3[/tex]

Since cos cannot take values as -3 it doe snot pass through origin.

In summary, the polar coordinate r for point P is approximately 9.59. The Cartesian coordinates of P are approximately (0.99, -9.43). The given curve passes through the origin twice in the interval 0<θ<2π.

This question is about a curve in polar coordinates. The curve is given by r=9+3cosθ. Let's explore the three sub-questions in order:

a. Find polar coordinate r for P

Because Point P is at θ=21π/18, we can plug this into the equation r=9+3cosθ. This gives r = 9 + 3cos(21π/18) = 9.59.

b. Find Cartesian coordinates for point P.

To convert from polar to Cartesian coordinates, we use the equations x = rcosθ and y = rsinθ. Plugging in our values, we get x = 9.59cos(21π/18) and y = 9.59sin(21π/18). Calculating these gives us x approximately equal to 0.99 and y approximately equal to -9.43. So the Cartesian coordinates for point P are (0.99, -9.43).

c. How many times does the curve pass through the origin when 0<θ<2π?

The curve reaches the origin when r is 0. From the curve equation r = 9 + 3cosθ = 0, we can see the curve crosses the origin twice, at θ = 2π/3 and θ = 4π/3.

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Answer:

12 6/8 or 12 3/4

Step-by-step explanation:

Answer:the total amount that they recycled is 12 3/4 boxes

Step-by-step explanation:

Maria's class recycled 2 2/8 boxes of paper in a month. Converting 2 2/8 boxes of paper in a month to improper fraction, it becomes

18/8 boxes of paper in a month.

They recycled another 10 4/8 boxes the next month. Converting 10 4/8 boxes to improper fraction, it becomes 84/8 boxes.

Therefore, the total amount of boxes that they recycled would be

18/8 + 84/8 = 102/8 boxes

Converting to whole number, it becomes

12 3/4 boxes

Answer:

[tex]p_v =P(t_{74}<-1.227)=0.112[/tex]    

If we compare the p value and a significance level for example [tex]\alpha=0.1[/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 true mean it's not significantly less than 20 min.

Step-by-step explanation:

Data given and notation    

[tex]\bar X=18.81[/tex] represent the average lateral recumbency for the sample    

[tex]s=8.4[/tex] represent the sample standard deviation    

[tex]n=75[/tex] sample size    

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

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

t would represent the statistic (variable of interest)    

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

State the null and alternative hypotheses.    

We need to apply a left tailed  test.  

What are H0 and Ha for this study?    

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

Alternative hypothesis :[tex]\mu < 20[/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{18.81-20}{\frac{8.4}{\sqrt{75}}}=-1.227[/tex]

The degrees of freedom are given by:

[tex]df=n-1=75-1=74[/tex]    

Give the appropriate conclusion for the test  

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

[tex]p_v =P(t_{74}<-1.227)=0.112[/tex]    

Conclusion    

If we compare the p value and a significance level for example [tex]\alpha=0.1[/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 true mean it's not significantly less than 20 min.

Answer: B

Step-by-step explanation:

Answer:

the real answer is2 or d

Answer:Angle AEC is 139 degrees.

Step-by-step explanation:

Since line EC bisects angle BED, it divides angle BED equally into 2. This means that

Angle BEC = angle CED

If angle CED = 4x + 1

Therefore,

Angle BED = 2 × angle CED

= 2(4x + 1) = 8x + 2

The sum of the angles in a straight line is 180 degrees. Therefore

Angle AEB + angle BED = 180

Angle AEB = 11x - 12. Therefore

11x - 12 + 8x + 2 = 180

19x - 10 = 180

19x = 180 + 10 = 190

x = 190/19 = 10

Angle BEC = 4x + 1 = 4×10 + 1 = 41

Angle AEB = 11x - 12 = 11×10 - 12 = 98

Angle AEC = 41 + 98 = 139

Answer: AEC = 139

Step-by-step explanation:

You first have to find x.

To find x, we need to add all the angles together to get 180°.

Since EC bisects BED, we know angle BEC and CED equal the same measure.

AEB + BEC + CED = 180°

(11x - 12) + (4x + 1) + (4x + 1) = 180°

Add like terms and solve

19x - 10 = 180

19x = 190

x = 10

Now, we substitute 'x' in AEB and BEC

AEB = 11x - 12

11 (10) - 12

AEB = 98

BEC = 4x + 1

4 (10) + 1

BEC = 41

98 + 41 = 139

AEC = 139

Answer:

The mean number favoring the substation is 12 citizens.

Step-by-step explanation:

For each citizen surveyed, there are only two possible outcomes. EIther they favored building a police substation in their neighborhood, or they opposed. This means that we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, with p probability, and X can only have two outcomes.

Has an expected value of:

[tex]E(X) = np[/tex]

In this problem, we have that:

[tex]n = 15, p = 0.8[/tex]

E(X) = 15*0.8 = 12.

The mean number favoring the substation is 12 citizens.

Final answer:

The mean number of citizens out of 15 who favor the construction of a police substation is 12, calculated by multiplying the sample size (15) by the probability of favoring the substation (0.80).

Explanation:

To find the mean number favoring the substation, we'll use the probability of a citizen favoring the substation to calculate the expected value in a sample of 15 citizens. Since 80% of the community favored building a police substation, the mean number favoring the substation can be computed by multiplying the sample size by the probability of favoring the substation.

Mean = Sample Size × Probability of Favoring the Substation

Mean = 15 × 0.80

Mean = 12

Therefore, the mean number of citizens out of 15 who favor the construction of the police substation is 12.

Answer:

4.57ft  by 1.57 ft

Step-by-step explanation:

We are given that

Emma's square patio has been area=31 sq.ft

One dimension decrease by 1 foot and other dimension decrease by 4 feet.

We have to find the new dimensions of Emma's patio.

Let x be the side of Emma's square patio

We know that

Area of square=[tex]x^2[/tex]

[tex]x^2=31[/tex]

[tex]x=\sqrt{31}=5.57[/tex] ft

One dimension=[tex]5.57-1=4.57[/tex] ft

other dimension=[tex]5.57-4=1.57 ft [/tex]

Hence, the new dimension of Emma's patio is given by

4.57ft  by 1.57 ft

Answer:

a) Statement is true

b) Statement incorrect. Company goal is to attend customer at 190 second at the most. So test should be one tail test  (right)

c) We have to reject H₀, and differences between sample mean and the hypothesized population mean should not be attributed only to sampling error

Step-by-step explanation:

a)  as the significance level is = 0,02  that means  α = 0,02 (the chance of error type I )  and the β (chance of type error II   is

1  -  0.02  =  0,98

b) The company establishe in 190 seconds time for customer at its ticket counter (if this time is smaller is excellent ) company is concerned about bigger time because that could be an issue for customers. Therefore the test should be a one test-tail to the right

c) test statistic

Hypothesis test should be:

null hypothesis                H₀   =  190

alternative hypothesis    H₀   >  190

t(s)  =  ( μ  -  μ₀ ) / s/√n      ⇒   t(s)  = ( 202 - 190 )/(28/√100 )

t(s)  = 12*10/28

t(s)  = 4.286

That value is far away of any of the values found for 99 degree of fredom and between  α  ( 0,025 and 0,01 ). We have to reject H₀, and differences between sample mean and the hypothesized population mean should not be attributed only to sampling error

If we look table  t-student we will find that for 99 degree of freedom and α = 0.02.

Answer:  a. 0.8

Step-by-step explanation:

When we test whether the variances of the two populations are equal we use F- test.

Test statistic : [tex]F=\dfrac{s_1^2}{s_2^2}[/tex]

, where [tex]s_1^2[/tex]= sample variance from population 1.

[tex]s_2^2[/tex]= sample variance from population 2.

As per given , we have

[tex]s_1^2=8[/tex] and  [tex]s_2^2=10[/tex]

Then,  If we test whether the variances of the two populations are equal, the test statistic will be [tex]F=\dfrac{8}{10}=0.8[/tex]

Hence, the test statistic will have a value of 0.8 .

Thus , the correct answer is  a. 0.8 .

Answer:

So, the interval is : (7.206,7.794)

Step-by-step explanation:

The mean μ = 7.5

Standard deviation σ=3

n = 400

At 95% confidence interval, the z score is 1.96

[tex]7.5+1.96(\frac{3}{\sqrt{400} } )[/tex]

And [tex]7.5-1.96(\frac{3}{\sqrt{400} } )[/tex]

7.5+0.294 and 7.5-0.294

So, the interval is : (7.206,7.794)

Answer:

a) [tex]df= k-1[/tex]

b) [tex]df= (k_1-1)(k_2-1)[/tex]

Step-by-step explanation:

Previous concepts

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

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

Solution to the problem

Let's define some notation:

N= Total number of individuals for a population

n = Total of individuals selected from a sample

n1= number of objects/individuals with characteristic 1

n2= number of objects/individuals with characteristic 2

k1= number of levels for the variable or factor 1

k2= number of levels for the variable or factor 2

Part a

For a chi square goodness of fit test the degrees of freedom are given by:

[tex]df= k-1[/tex]

Where k represent the total number of categories on the godness of fit test.

Part b

For a chi square test of independence the degrees of freedom are given by:

[tex]df= (k_1-1)(k_2-1)[/tex]

Where k1= number of levels for the variable or factor 1, k2= number of levels for the variable or factor 2 for the chi square test for independence.

In statistical analysis, degrees of freedom (df) are calculated differently based on the type of test. For a chi-square goodness-of-fit test, df equals the number of categories minus one. For a chi-square test for independence, df equals the product of subtracting one from the number of columns and rows.

The degrees of freedom (df) play an important role in many statistical tests, including the chi-square tests for goodness-of-fit and independence. These degrees of freedom are calculated differently based on the type of test. In a chi-square goodness-of-fit test, the degrees of freedom is calculated as df = k - 1, where k represents the number of categories or groups in the dataset.

On the other hand, for a chi-square test for independence, the degrees of freedom are calculated using the formula df = (number of columns - 1)(number of rows - 1), where columns and rows represent the different categories or groups present in a contingency table used in the test of independence. In this context, the null hypothesis for this test states that the two factors being considered are independent.

It's important to remember that degrees of freedom often represent the quantity of the 'free' or varying elements in your dataset and directly impact the test results and conclusions. So, having an accurate calculation is crucial in your statistical analysis.

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Answer:

140

Step-by-step explanation:

To construct a subset of S with said property, we have two choices, include 3 in the subset or include four in the subset. These events are mutually exclusive because 3 and 4 can not both be elements of the subset.

First, let's count the number of subsets that contain the element 3.

Any of such subsets has five elements, but since 3 is already an element, we only have to select four elements to complete it. The four elements must be different from 3 and 4 (3 cannot be selected twice and the condition does not allow to select 4), so there are eight elements to select from. The number of ways of doing this is [tex]{}_8C_4=70[/tex].

Now, let's count the number of subsets that contain the element 4.

4 is already an element thus we have to select other four elements . The four elements must be different from 3 and 4 (4 cannot be selected twice and the condition does not allow to select 3), so there are eight elements to select from, so this can be done in [tex]{}_8C_4=70[/tex] ways.

We conclude that there are 70+70=140 required subsets of S.

Answer:  The required ordered pair (p, q) is (-7, -12).

Step-by-step explanation:  Given that (x+2)(x-1) divides the following polynomial f(x) :

[tex]f(x)=x^5-x^4+x^3-px^2+qx+4.[/tex]

We are to find the ordered pair (p,q).

We have the following theorem :

Factor theorem : If (x-a) divides a polynomial h(x), then h(a) = 0.

According to the given information, we can say that (x+2) divides f(x). So, we get

[tex]f(-2)=0\\\\\Rightarrow (-2)^5-(-2)^4+(-2)^3-p(-2)^2+q(-2)+4=0\\\\\Rightarrow -32-16-8-4p-2q+4=0\\\\\Rightarrow 2p+q=-26~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]

Also, (x-1) is a factor of f(x). So,

[tex]f(1)=0\\\\\Rightarrow (1)^5-(1)^4+(1)^3-p(1)^2+q(1)+4=0\\\\\Rightarrow 1-1+1-p+q+4=0\\\\\Rightarrow p-q=5~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)[/tex]

Adding equations (i) and (ii), we get

[tex]3p=-21\\\\\Rightarrow p=-7.[/tex]

From equation (ii), we get

[tex]-7-q=5\\\\\Rightarrow q=-12.[/tex]

Thus, the required ordered pair (p, q) is (-7, -12).