Diana_Cuevas_An_Unexpected_Guest_Workbook

N -.5 0 2 What does the red area indicate? p(2>Z>-.5) O p(-.5>Z>2) Op(2> Z) or p(Z > -.5) p(-2> Z) or P(Z > .5)

It was found that a certain type of battery supercharger takes an average of15.4minutes to charge the battery of a standard electric vehicle. Recently, the firmware for this type of supercharger was updated but the update has received negative reviews from users. A researcher suspects that the average time it takes for upgraded superchargers to charge a standard EV battery is longer than 15.4minutes. To test his claim, choosefiftferandomly upgraded superchargers and measures the time each takes to charge a standard EV battery. Find that superchargers take a sample mean of 16.1 minutes to charge a standard EV battery, with a sample standard deviation of 1.8minutes. Assume that the population of the amounts of time that upgraded superchargers take to charge a standard EV battery approximately follows a normal distribution. Complete the parts below to perform a hypothesis test to determine if there is sufficient evidence, at the significance level of0.05, to support the claim thatμ,the…

Use the given information to find the number of degrees of freedom, the critical values x? and y%, and the confidence interval estimate of o. It is reasonable to assume that a simple random sample has been selected from a population with a normal distribution. Nicotine in menthol cigarettes 90% confidence; n= 22, s= 0.23 mg. E Click the icon to view the table of Chi-Square critical values. df =(Type a whole number.)

Area to the Right of the Critical Value Degrees of Freedom 0.995 0.99 0.975 0.95 0.90 0.10 0.05 0.025 0.01 0.005 Degrees of Freedom 1 - - 0.001 0.004 0.016 2.706 3.841 5.024 6.635 7.879 1 2 0.010 0.020 0.051 0.103 0.211 4.605 5.991 7.378 9.210 10.597 2 3 0.072 0.115 0.216 0.352 0.584 6.251 7.815 9.348 11.345 12.838 3 4 0.207 0.297 0.484 0.711 1.064 7.779 9.488 11.143 13.277 14.860 4 5 0.412 0.554 0.831 1.145 1.610 9.236 11.071 12.833 15.086 16.750 5 6 0.676 0.872 1.237 1.635 2.204 10.645 12.592 14.449 16.812 18.548 6 7 0.989 1.239 1.690 2.167 2.833 12.017 14.067 16.013 18.475 20.278 7 8 1.344 1.646 2.180 2.733 3.490 13.362 15.507 17.535 20.090 21.955 8 9 1.735 2.088 2.700 3.325 4.168 14.684 16.919 19.023 21.666 23.589 9 10 2.156 2.558 3.247 3.940 4.865 15.987 18.307 20.483 23.209 25.188 10 11 2.603 3.053 3.816 4.575 5.578 17.275 19.675 21.920 24.725 26.757 11 12 3.074 3.571 4.404 5.226 6.304 18.549 21.026 23.337 26.217 28.299 12 13 3.565 4.107 5.009 5.892 7.042 19.812 22.362 24.736…

Area to the Right of the Critical Value Degrees of Freedom 0.995 0.99 0.975 0.95 0.90 0.10 0.05 0.025 0.01 0.005 Degrees of Freedom 1 - - 0.001 0.004 0.016 2.706 3.841 5.024 6.635 7.879 1 2 0.010 0.020 0.051 0.103 0.211 4.605 5.991 7.378 9.210 10.597 2 3 0.072 0.115 0.216 0.352 0.584 6.251 7.815 9.348 11.345 12.838 3 4 0.207 0.297 0.484 0.711 1.064 7.779 9.488 11.143 13.277 14.860 4 5 0.412 0.554 0.831 1.145 1.610 9.236 11.071 12.833 15.086 16.750 5 6 0.676 0.872 1.237 1.635 2.204 10.645 12.592 14.449 16.812 18.548 6 7 0.989 1.239 1.690 2.167 2.833 12.017 14.067 16.013 18.475 20.278 7 8 1.344 1.646 2.180 2.733 3.490 13.362 15.507 17.535 20.090 21.955 8 9 1.735 2.088 2.700 3.325 4.168 14.684 16.919 19.023 21.666 23.589 9 10 2.156 2.558 3.247 3.940 4.865 15.987 18.307 20.483 23.209 25.188 10 11 2.603 3.053 3.816 4.575 5.578 17.275 19.675 21.920 24.725 26.757 11 12 3.074 3.571 4.404 5.226 6.304 18.549 21.026 23.337 26.217 28.299 12 13 3.565 4.107 5.009 5.892 7.042 19.812 22.362 24.736…

Area to the Right of the Critical Value Degrees of Freedom 0.995 0.99 0.975 0.95 0.90 0.10 0.05 0.025 0.01 0.005 1 ​- ​- 0.001 0.004 0.016 2.706 3.841 5.024 6.635 7.879 2 0.010 0.020 0.051 0.103 0.211 4.605 5.991 7.378 9.210 10.597 3 0.072 0.115 0.216 0.352 0.584 6.251 7.815 9.348 11.345 12.838 4 0.207 0.297 0.484 0.711 1.064 7.779 9.488 11.143 13.277 14.860 5 0.412 0.554 0.831 1.145 1.610 9.236 11.071 12.833 15.086 16.750 6 0.676 0.872 1.237 1.635 2.204 10.645 12.592 14.449 16.812 18.548 7 0.989 1.239 1.690 2.167 2.833 12.017 14.067 16.013 18.475 20.278 8 1.344 1.646 2.180 2.733 3.490 13.362 15.507 17.535 20.090 21.955 9 1.735 2.088 2.700 3.325 4.168 14.684 16.919 19.023 21.666 23.589 10 2.156 2.558 3.247 3.940 4.865 15.987 18.307…

Area to the Right of the Critical Value Degrees of Freedom 0.995 0.99 0.975 0.95 0.90 0.10 0.05 0.025 0.01 0.005 1 ​- ​- 0.001 0.004 0.016 2.706 3.841 5.024 6.635 7.879 2 0.010 0.020 0.051 0.103 0.211 4.605 5.991 7.378 9.210 10.597 3 0.072 0.115 0.216 0.352 0.584 6.251 7.815 9.348 11.345 12.838 4 0.207 0.297 0.484 0.711 1.064 7.779 9.488 11.143 13.277 14.860 5 0.412 0.554 0.831 1.145 1.610 9.236 11.071 12.833 15.086 16.750 6 0.676 0.872 1.237 1.635 2.204 10.645 12.592 14.449 16.812 18.548 7 0.989 1.239 1.690 2.167 2.833 12.017 14.067 16.013 18.475 20.278 8 1.344 1.646 2.180 2.733 3.490 13.362 15.507 17.535 20.090 21.955 9 1.735 2.088 2.700 3.325 4.168 14.684 16.919 19.023 21.666 23.589 10 2.156 2.558 3.247 3.940 4.865 15.987 18.307…

Use the technology display to make a decision to reject or fail to reject the null hypothesis at a=0.01. Make the decision using the standardized test statistic and using the P-value. Assume the sample sizes are equal Click the icon to view the technology display. Choose the correct answer below. O A. Reject the null hypothesis because the P-value is less than the level of significance. OB. Reject the null hypothesis because the P-value is greater than the level of significance OC. Fail to reject the null hypothesis because the P-value is greater than the level of significance OD. Fail to reject the null hypothesis because the P-value is less than the level of significance CHI Technology Output μ1> μ2 z 2.609496821 p=0.0045337744 x1 = 47 x2 = 44 In1=45 2-SampZTest Print Done - X

ΜΑΤΉ SELFIE Find the area under the normal curve in each of the following cases: 1. Between z = 0 and z= 1.63 2. Between z = 0 and z=-1.78 3. Between 2 = 1.56 and z = 2.51 4. Between z = -1.36 and z=2.55 5. Between z = -2.46 and z= 1.55 6. Between z== -2.76 and z=-1.25 2.35 z = 7. Between z= 0.86 and z = 8. Between 2 =-0.76 and z = 1.35 9. Between z= 1.26 and z = 1.78 10. Between z = -1.78 and z= -0.26 11. To the right of z = 2.35 12. To the right of z = -1.31 13. To the left of z = 0.35 14. To the left of z= 1.85 15. To the right of z== -0.95 16. To the right of z = -0.75 17. To the right of z = -1.54 18. To the right of z = 0.89 19. To the right of z = -2.85 20. To the right of z = 0.3

Histogram #1 (N=4) 450 400 350 300 250 200 150 100 50 [0, 0.2] (0.2, 0.4] (0.4, 0.6) (0.6, 0.8] Type of Gas Particles (0.8, ) 350 Histogram #4 (N=8) 300 250 200 150 100 50 (0, 0.111111111) ,. (0.111111111, (0.222222222,. (0.333333333,. (0.444444444 (0.555555556, (0.666666667,. (0.777777778, Type of Gas Particles (0.888888889, 1] 180 Histogram #3 (N=702) 160 140 120 100 80 60 40 20 Type of Gas Particles W of Gas Particles Nof Gas Particles l of Gas Particles [0.438746439,. 0.524216524, (0.545584046, (0.552706553, (0.55982906,.. (0.566951567, (0.574074074

Random Variable X has PMF given by the following. What is the PMF of Y = X^2  + 1.   x              -1          1         2 P(X=x)    0.4       0.1      0.5

0.00 100.00 0.02 81.86 0.04 67.03 0.06 54.89 0.08 44.91 0.10 36.77 21 (microooulombs) 100 90 70 60 50 B. 0.02 0,04 0.06 0.08 0.1 I (seconds) mPR (0.00, 100.00)| -824.25 (0.02,81.86) -741.50 (0.06, 54.89) -607.00 (0.08, 44.91) -553.00 (0.10, 36.77) 504,33 R. 8822 8

Find the specified areas for a normal density. (a) The area above 7 on a N (5, 1.4) distribution Round your answer to three decimal places. Area= the absolute tolerance is +/-0.001 (b) The area below 15 on a N (20, 3.1) distribution Round your answer to three decimal places. Area= the absolute tolerance is +/-0.001 (c) The area between 86 and 100 on a N (100, 6.1) distribution Round your answer to three decimal places. Area= the absolute tolerance is +/-0.001

Gallons/person/year Because whole milk is high in saturated fat, replacing whinte milk with lower-fat milk decreases saturated fat intake, potentinity reducing the risk of heart disease. However, milk is the most import ant source of calcium in the North American diet, and calclum i needed for bone health, so a reduction in total milk consampon could contribute to an increase in the incidence of fractures dus o low bone density, a condition called osteoporosis.¹¹ 25 10 20 15 5 A nutrition scientist looking at this graph, however, would see not just changes in the amount of milk Americans drank but the nutritional and public health implications of these changes well. 35 30 0 1970 1 74 78 L 82 86 90 HET 94 98 Year - Total milk -Lower-fat milks - Whole milk 02 the person's nutritional status can be assessed (Figure 2.2). record of an individual's food intake is Obtain 06 10 1 14 18 Interpret the Data If 1 gallon of milk has 4.8 grams of calcium, how much less cal- cium per person was…

Attached

95% CI for α99% CI for β

CO Determine the area of under the Normal Curve for z = -2.3 2.3 area= [a]

To test Ho: µ = 100 versus H,: µ# 100, a simple random sample size ofn=22 is obtained from a population that is known to be normally distributed. Answer parts (a)-(d). Click here to view the t-Distribution Area in Right Tail. ..... (a) If x = 105.8 and s = 8.9, compute the test statistic. t = (Round to three decimal places as needed.) (b) If the researcher decides to test this hypothesis at the a = 0.01 level of significance, determine the critical values. The critical values are (Use a comma to separate answers as needed. Round to three decimal places as needed.) (c) Draw a t-distribution that depicts the critical region(s). Which of the following graphs shows the critical region(s) in the t-distribution? A. O B. OC. (d) Will the researcher reject the null hypothesis? A. There is not sufficient evidence for the researcher to reject the null hypothesis since the test statistic is between the critical values. B. The researcher will reject the null hypothesis since the test statistic is…

The values listed below are waiting times (in minutes) of customers at two different banks. At Bank A, customers enter a single waiting line that feeds three teller windows. At Bank B, customers may enter any one of three different lines that have formed at three teller windows. Answer the following questions. Bank A 6.4 6.6 6.7 6.8 7.1 6.7 7.3 7.6 7.8 7.9 7.9 7.9 Bank B 4.1 5.4 5.9 6.2 7.8 8.6 9.3 10.0 1= Click the icon to view the table of Chi-Square critical values. Construct a 99% confidence interval for the population standard deviation a at Bank A. Omin < aBank A < min (Round to two decimal places as needed.)

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Calculate the Banzhaf power index for each voter. Suppose the voters in the order listed are A, B, C, and D. {34: 22, 12, 9, 3} O a. BPI(A) = 0.5 BPI(B) = 0.3 BPI(C) = 0.2 BPI(D) = 0.1 O b. BPI(A) = 0.5 ВР/(В) —D 0.3 BPI(C) = 0.1 ВР[D) %3D 0.1 c. BPI(A) = 0.5 BPI(B) = 0.5 BPI(C) = 0 BPI(D) = 0

Empirical Rule.

I need to solve this please

Childhood participation in sports, cultural groups, and youth groups appears to be related to improved self-esteem for adolescents (McGee, Williams, Howden-Chapman, Martin, & Kawachi, 2006). In a representative study, a sample of n = 100 adolescents with a history of group participation is given a stan- dardized self-esteem questionnaire. For the general population of adolescents, scores on this questionnaire form a normal distribution with a mean of p %3D 50 and %3D 15. The sample of group- a standard deviation of o %3D participation adolescents had an average of M = 53.8.

A regression model relating x, number of salespersons at a branch office, to y, annual sales at the office (in thousands of dollars) provided the following regression output. Where n total = 38. (NEED ANSWER FOR D)

iid Let Y; Uniform(a0, b0), where 0 > 0 and 0 < a < b, i = 1, ... , n. (a) Consider Yn as an estimator of 0. Find a function of Yn that is unbiased for 0. (b) Find a sufficient statistic for 0. (c) the improved estimator) How would you improve the estimator from part (a)? (You do not need to calculate (d) Does the CR bound apply here?

℗ Do Homework - Section 3.11 - Profile 1 - Microsoft Edge https://mylab.pearson.com/Student/PlayerHomework.aspx?homeworkId=644835720&questionId=12&flushed=false&cid=7318436¢erwin=yes 2t The concentration C in milligrams per milliliter (mg/ml) of a certain drug in a person's blood-stream t hours after a pill is swallowed is modeled by C(t) = 7+ changes from 40 to 50 minutes. 1+1³ The change in concentration is about mg/ml. (Type an integer or decimal rounded to the nearest thousandth as needed.) 1 58°F Cloudy ▬▬ ■ Q Search U IN hulu -0.06t - e Estimate the change in concentration when t Clear all x A Check answer 9:07 PM 4/22/2023