Preliminary data analysis indicate that it is reasonable to use a t-test to carry out the specified hypothesis test. Perform the r-test using the critical-valueapproach.EspañolUse a significance level of a=D0.05 to test the claim that U 32.6. The sample data consists of 15 scores for which x =42.8 and s=7.2.oloH:42.8H:p=42.8Test statistic: t=5.49. Critical values: t=±2.145. Reject H :p=42.8.There is sufficient evidence to support the claim that the mean is different from 42.8.H:p= 32.6H:pz 32.6Test statistic:t%=5.49. Critical values: t= ±2,145. Reject H:p=32.6.There is sufficient évidence to support the claim that the mean is different from 32.6.H:pz32.6H:p=32.6Test statistic: t=5.49. Critical values: t= ±2.145. Reject H u=32.6.There is sufficient evidence to support the claim that the mean is different from 32.6.H:=42.8H:pz42.8Test statistic:t%3D5.49. Critical values: t= +2.145. Reject H u=42.8.There is sufficient evidence to support the claim that the mean is different from19

Answered: Preliminary data analysis indicate that…

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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You are conducting a study to see if the proportion of men over 50 who regularly have their prostate examined is significantly less than 0.15. You use a significance level of a =0.02. Ho:p= 0.15 H1:p < 0.15 You obtain a sample of sizen= 504 in which there are 60 successes. What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic = What is the p-value for this sample? (Report answer accurate to four decimal places.) p-value = The p-value is. O less than (or equal to) a O greater than a This test statistic leads to a decision to... O reject the null O accept the null O fail to reject the null As such, the final conclusion is that.. O There is sufficient evidence to warrant rejection of the claim that the proportion of men oven 50 who regularly have their prostate examined is less than 0.15. There is not sufficient evidence to warrant rejection of the claim that the proportion of men over 50 who regularly have their prostate examined is…
The display provided from technology available below results from using data for a smartphone carrier's data speeds at airports to test the claim that they are from a population having a mean less than 4.00 Mbps. Conduct the hypothesis test using these results. Use a 0.05 significance level. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. LOADING... Click the icon to view the display from technology. Question content area bottom Part 1 Assuming all conditions for conducting a hypothesis test are met, what are the null and alternative hypotheses? A. H0: μ=4.00 Mbps H1: μ>4.00 Mbps B. H0: μ<4.00 Mbps H1: μ=4.00 Mbps C. H0: μ=4.00 Mbps H1: μ≠4.00 Mbps T-Test μ<4.00 t=−2.869485 p=0.003303 x=3.43 Sx=1.256322 n=40 Identify the test statistic.
A sample mean, sample size, and sample standard deviation are provided below. Use the one-mean t-test to perform the required hypothesis test at the 10% significance level. x=25, s=10, n=24, H0: μ=24, Ha: μ>24 The test statistic is t= (round to two decimals as needed)
I need help with this problem Conduct a hypothesis test about the mean cost per square foot at a chosen significance level. Reference an internet source for the current average (or median) cost per square foot of homes in Orange County. Discuss/explain your choice of significance level for your hypothesis test in #1 as it relates to the power of a test and Type I/II errors.
K The display provided from technology available below results from using data for a smartphone carrier's data speeds at airports to test the claim that they are from a population having a mean less than 5.00 Mbps. Conduct the hypothesis test using these results. Use a 0.05 significance level. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. Click the icon to view the display from technology. D iew an example Get more help. 42 W S Assuming all conditions for conducting a hypothesis test are met, what are the null and alternative hypotheses? OA. Ho: 5.00 Mbps J D. Ho: -5.00 Mbps H₁: - Clear all : - ; X option F11 { [ ? + = 1 I Check answer F12 } ] delete retu
What is the test statistic?(3 decimal places) What is the critical value(s)?(3 decimal places) Reject or don’t reject?
Please help me with this question.
Identify test statistic Identify P value Identify a conclusion Fail to reject H0... Reject H0....
Report an appropriate hypothesis test and use a 5% significance level. The alternative hypothesis for the One-way Anova F-test would be Group of answer choices Ha: Density affects firing temperature. Ha: Firing temperature affects density. Ha: Firing temperature does not affect density. Ha: Density does not affect firing temperature.
Salma claimed that the gas price in Blue Park neighborhood (mean of $3.65) is higher than that in Green Spring (mean of $3.59). She obtained the t-value of 2.223 after calculating data from 6 gas stations in Blue Park and 5 gas stations in Green Spring. Use the chart below and determine if Salma's claim is supported! Table 10-3. Critical values of "t" significance level p= 0.05 significance level p= 0.01 TP 1 12.706 63.657 2 4.303 9.925 3 3.182 5.841 2.776 4.604 5 2.571 4.032 6 2.447 3.707 7 2.365 3.499 2.306 3.355 9 2.262 3.250 10 2.228 3.169 11 2.201 3.106 Copy the following into your text box, fill in the blanks, and answer the questions A. The degree of freedom is: B. The t-calculated value is: and the t-critical value is : C. Comparison: t-calc t-crit (answer with ()) D. Is the null hypothesis accepted or rejected? Explain how you determined that! Also, explain what it means to reject or accept the hypothesis! E. Explain if Salma's claim is supported based on your t-value…
The summary statistics are n=85, x=3.93, s=0.63. Use a 0.01 significance level to test the claim that the population of student course evaluations has a mean equal to 4.00. identify null/ alt hypothesis, test statistic, P- value, and state final conclusion that addresses the original claim
A data set includes data from student evaluations of courses. The summary statistics are n= 95, x over bar = 3.44, s= 0.64. Use a 0.10 significance level to test the claim that the population of student course evaluations has a mean equal to 3.50. Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. What is the hypothesis? What is the test statistic? What is the p-value? What is the final conclusion that addresses the original claim? Is there suficient evidence to conclude that the original claim that the mean of the population of student course evaluations is equal to 3.50 is or is not correct?

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