A sample of birth weights of 48 girls was taken. Below are theresults (in g):2724.53340.13243.63442.12996.9304235053161.84022.83362.93768.92951.93229.72997.53758.13350.23210.73684.33530.13132.72776.93492.72655.13790.93064.63744.63294.52654.73536.62781.13756.33531.53654.23505.14018.630383202.330063701.63451.42986.53541.23232.9296631863550.33056.63467.1* = 3314.6 gs = 347.35 gUse a 5% significance level to test the claim that thestandard deviation of birth weights of girls is less than thestandard deviation of birth weights of boys, which is 530 g.Round all answers to 3 decimal places if possible. Step 4. Testing Procedure:Provide the critical value(s) for the Rejection Region.For a one-tailed test, use DNE for the unneededcritical value.• Left CV :• Right CV =The p-value isLink E to Chi-Square Table:https://www.itl.nist.gov/div898/handbook/eda/section3/edaStep 5. DecisionIs the test statistic in the rejection region? ?Is the p-value less than the significance level? ?Conclusion:Select an answerStep 6. Interpretation:At a 5% significance level we Select an answer O havesufficient evidence to reject the null hypothesis infavor of the alternative hypothesis.

From the given data Perform Chi square test for one population variance

Answered: Step 4. Testing Procedure: Provide the…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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H0 : =150.00 kPa H1 :>150.00 kPa b) α= 0.01; sample variance : 1300 (freq. table ) sample deviation : 36.0555 (freq. table ) Calculation of Data Mean of the results= = 6700/50 =134 Mid Range of the results= (217.32+46.32)/2= 131.82 Range = (217.32-46.32) = 171 Modal Class= Class 121-145 with the highest frequency of 15 Median of the sorted set of data = 132.25 Mode = there are not clear modes In the samples. Sample variance S o= 1300 Sample deviation: S =36.55
What is the predicted total number of wins in a regular season for a team that is averaging 75 points per game with a relative skill level of 1350?
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edu/ultra/courses/_313006_1/cl/outline * Question Completion Status: QUESTION 1 EPA fuel economy estimates for automobile models tested recently predicts a mean of 24 mpg and a standard deviation of 6 mpg. Assume that a Normal model applies, N(24, 6). What proportion of automobiles would you expect to have gas mileage of more than 27 mpg? a. 0.6915 b. 0.0013 Oc C. 0.5000 O d. 0.3085 QUESTION 2
An advertising agency is studying social media campaigns and how they impact website traffic. The advertising agency states the average increase in website traffic after a social media campaign is greater than 20 percent. If we would like to test the advertising agency's claim with a hypothesis test using a significance level of α = 0.015, what is the rejection region? Use the graph below to show the rejection region. First, move the blue dot to choose the appropriate test (left-, right, or two- tailed). Then, slide the shaded region(s) to represent the rejection region. Provide your answer below:
Please don't use any kind of software to solve this problem. Please do all calculations showing formulas used, also complete the ANOVA table, and draw effects plots if necessary so I can understand how to do this problem. Thank you
Pls answer Activity 2 , table 2
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True or false
Use the information given in the table below to find a 95% Prediction Interval for y when x = 10. n = 20 x_bar = 15 %3D beta0_hat = 53 SSXX = 62 beta1_hat = - 4.5 S = 9 (- 14.79, 30.79) (-0.57, 16.57) (-10.25, 26.25) O (- 4.73, 20.73)
• The value of the test statistic is given by 1= • The p-value is the area under the curve to the right of the value of the test statistic. Student's t Distribution 0,4 + Step 1: Enter the number of degrees of freedom. 17 Step 2: Select one-tailed or two-tailed. 0.3+ O One-tailed O Two-tailed Step 3: Enter the test statistic. (Round to 3 decimal places.) 0.2+ 1.928 Step 4: Shade the area represented by the p-value. 0.1+ Step 5: Enter the p-value. (Round to 3 decimal places.) 0.0354 1= 1.928 (c) Based on your answer to part (b), choose what can be concluded, at the 0.05 level of significance, about the claim made by Ashley. Since the p-value is less than (or equal to) the level of significance, the null hypothesis is rejected. So, there is enough evidence to support the claim that the mean number of cases directly generated ? by one previous case is greater than 2.6. Since the p-value is less than (or equal to) the level of significance, the null hypothesis is not rejected. So, there is…
use the date form the second picture to answer question 3 please help me with question 3

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