A nationwide job recruiting firm wants to compare the annual incomes forchildcare workers in New York and Rhode Island. Due to recent trends inthe childcare industry, the firm suspects that the that the mean annualincome of childcare workers in the state of New York is greater than themean annual income of childcare workers in Rhode Island. To see if this istrue, the firm selected a random sample of 10 childcare workers from NewYork and an independent random sample of 10 childcare workers fromRhode Island and asked them to report their mean annual income. The dataobtained were as follows.Annual income in dollarsNew York 52865, 45781, 48933, 44123, 52450, 54008, 46219, 44862, 54254, 39329Rhode Island 52347, 47038, 49732, 47329, 34151, 51204, 39798, 47921, 44854, 39128Send data to calculatorSend data to ExcelThe population standard deviations for the annual incomes of childcareworkers in New York and in Rhode Island are estimated as 6400 and 6600,respectively. It is also known that both populations are approximatelynormally distributed. At the 0.10 level of significance, is there sufficientevidence to support the claim that the mean annual income, ₁, of childcareworkers in New York is greater than the mean annual income, ₂,ofchildcare workers in Rhode Island? Perform a one-tailed test. Thencomplete the parts below.Carry your intermediate computations to at least three decimal places. (Ifnecessary, consult a list of formulas.)(a) State the null hypothesis H and the alternative hypothesis H₁.HOH₁ : 0(b) Determine the type of test statistic to use.Chi-SquareDegrees of freedom:(c) Find the value of the test statistic. (Round to three or more decimal places.)(d) Find the p-value. (Round to three or more decimal places.)0(e) Can we support the claim that the mean annual income of childcare workersin New York is greater than the mean annual income of childcare workers inRhode Island?OYes NoHXI00OXS0=0 OSO00OOOE

Given, n1=10, n2=10, σ1=6400 and σ2=6600 Level of significance=0.10

Answered: (a) State the null hypothesis H and 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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What are the hypotheses for the test? A. H0: β1≠0 and Ha: β1=0 B. H0: β1=0 and Ha: β1<0 C. H0: β1=0 and Ha: β1>0 Your answer is not correct. D. H0: β1=0 and Ha: β1≠0 This is the correct answer. What is the test statistic? t= (Round to two decimal places as needed.) Find the P-value. P-value= (Round to three decimal places as needed.) Part 4 Which of the following is the correct conclusion for the hypothesis test? A. Do not reject H0; the data provide sufficient evidence to conclude that x is useful for predicting y. B. Reject H0; the data do not provide sufficient evidence to conclude that x is useful for predicting y. C. Do not reject H0; the data do not provide sufficient evidence to conclude that x is useful for predicting y. Your answer is correct. D. Reject H0; the data provide sufficient evidence to conclude that x is useful for predicting y. Part 5 b. Find a 95%…
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Use technology to find the P-value for the hypothesis test described below. The claim is that for a smartphone carrier's data speeds at airports, the mean is μ=13.00 Mbps. The sample size is n=28 and the test statistic is t=−1.288.
Test Scores An instructor who taught an online statistics course and a classroom course feels that the variance of the final exam scores for the students who took the online course is greater than the variance of the final exam scores of the students who took the classroom final exam. The following data were obtained. At =α0.05 is there enough evidence to support the claim? Use σ21 for the variance of the online scores. Use the critical value method and tables. State the Hypothesis and Identify the claim: H0: Claim or not a claim H1: Claim or not a claim Find the critical Value: Round the answer to at least two decimal places. Compute the test value: Make the decision reject or accept the null hypothesis. Summarize the results. Is there enough evidence to support the claim? Online Course Classroom Course =s13.4 =s22.2 =n121 =n231
Use technology to find the P-value for the hypothesis test described below. The claim is that for a smartphone carrier's data speeds at airports, the mean is μ = 11.00 Mbps. The sample size is n = 28 and the test statistic is t = 2.358.
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(...) Dairy cows - The USDA collects and distributes a wide variety of data about agriculture in the United States. One statistic reported each year is the number of milk cows (in thousands) in each state. A random sample of 10 states is selected, with the number of milk cows reported in both 2011 and 2015. 1. Which hypotheses should be used to conduct the test? diff=0 vs. H₁ : 0 vs. Ha diff 5.7 20.202 mean sd 2015 19 47 125 195 14 1279 14 408 2011 19 45 119 188 14 1265 18 366 329 323 98 94 246.1 251.8 379.881 385.58 Difference 0 2 6 7 0 14 -4 42 -6 -4 5.7 14.111 3. Based on the p value we have Little evidence that the null model is not a good fit for our observed data. 4. Construct a 99% confidence interval for the mean difference in the number of milk cows (in thousands) between 2011 and 2015.
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Is there a statistically significant difference between males and females in terms of newspaper readership? The proportion of each sex that says they read a newspaper daily is reported below. Results are from a nationally representative random sample.Use α = .01 and the five-step model to conduct the hypothesis test. Write a sentence or two interpreting your results. Males Ps1 = .56 N1 =786 Females Ps2 = .59 N2 =965
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AGE MINDTAP hapter 7 and 9 Homework Consider the following hypothesis test: Hạ: p2 0.75 H: p< 0.75 A sample of 300 items was selected. Compute the p-value and state your conclusion for each of the following sample results. Use a =.05. Round your answers to four decimal places. a. p=0.64 p-value Conclusion: p-value less than or equal to 0.05, reject На b. p 0.75 P-value Conclusion: p-value greater thán 0.0s, do nat reject На Cp-0.7 p-value Conclusion: pvalue less than or equal to 0.05, reject d. p 0.77 Pvalue Conclusion: value oreater than O.05, de not reject Partially Correct

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(a) State the null hypothesis H and the alternative hypothesis H₁. Ho :O H₁:0 (b) Determine the type of test statistic to use. Chi-Square Degrees of freedom: (c) Find the value of the test statistic. (Round to three or more decimal places.) (d) Find the p-value (Bound to ro decimal places