6. Suppose we want to test, at the 5% level of significance, whether there is a linear relationship between x and y. That iswe want to test the null hypothesis Ho : B1 = 0 against the alternative hypothesis Ha : B1 7 0. What is the value of theobserved test statistic for this test?O 2.673-2.6731.5240O -1.5240 Suppose the simple linear regression model, Y; = Bo + B1 a; + Ej, is used to explain the relationship between x and y.A random sample of n10 values for the explanatory variable (x) was selected and the corresponding values of theresponse variable (y) were observed. The summary statistics were:Σ.8, Συ- 10 , Σα,y 9.5,Σ-7.4, Σ-20

From the given data Find test statistic ( t ) =?

Answered: 6. Suppose we want to test, at the 5%…

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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Suppose ?1 and ?2 are true mean stopping distances at 50 mph for cars of a certain type equipped with two different types of braking systems. Use the two-sample t test at significance level 0.01 to test H0: ?1 − ?2 = −10 versus Ha: ?1 − ?2 < −10 for the following data: m = 8, x = 114.8, s1 = 5.01, n = 8, y = 129.9, and s2 = 5.35. Calculate the test statistic and determine the P-value. (Round your test statistic to one decimal place and your P-value to three decimal places.)
Test the claim that the mean GPA of night students is significantly different than the mean GPA of day students at the 0.1 significance level.The null and alternative hypothesis would be: H0:μN=μDH0:μN=μDH1:μN≠μDH1:μN≠μD H0:μN=μDH0:μN=μDH1:μN>μDH1:μN>μD H0:pN=pDH0:pN=pDH1:pN≠pDH1:pN≠pD H0:pN=pDH0:pN=pDH1:pN>pDH1:pN>pD H0:pN=pDH0:pN=pDH1:pN<pDH1:pN<pD H0:μN=μDH0:μN=μDH1:μN<μDH1:μN<μD The test is: two-tailed left-tailed right-tailed The sample consisted of 45 night students, with a sample mean GPA of 2.15 and a standard deviation of 0.06, and 35 day students, with a sample mean GPA of 2.16 and a standard deviation of 0.03.The test statistic is: ________(to 2 decimals)The positive critical value is: ________(to 2 decimals)Based on this we: Reject the null hypothesis Fail to reject the null hypothesis
The data below are stress-test scores from 6 nurses from the Emergency Care (EC) unit and 6 nurses from the Long Term (LT) care unit who were randomly selected for the study. The investigator wants to test, at the 5% level, whether there is a difference in stress test scores in the EC group and the LT group. EC LT 5 2 9 3 8 5 10 6 7 4 6 2 State the hypotheses. What statistical test should be used to determine if there is a difference in stress scores between EC and LT nurses? Determine the appropriate critical value and set up the decision rule. Compute the test statistic. Interpret your results. Give your conclusion.
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Which of the following is the correct way to report the results of a hypothesis test and a measure of effect size using a tstatistic? a. r2 = 0.42, t(19) = 2.30 , p < .05 b. t(19) = 2.30, p < .05, r2 = 0.42 c. t = 2.30, df = 19, p < .05, r2 = 0.42 d. t(19) = 2.30, r2 = 0.42, p < .05
Regression. A coach wants to see the relationship between the statistics of practice games and official games of a local soccer team. A sample of 24 players was used and the resulting (partial) Excel output is shown below. Assume both x and y form normal distributions. Regression Statistics Multiple R *** R Square 0.70524 Adjusted R Square 0.668395 Standard Error 8.703633 Observations 24 Coefficients Standard t Stat P-value Lower Upper
HR department wants to make sure that their hiring practices are independent of gender. In other words, they want to ensure that both men and women have an equal chance of being hired. The observed data related to this survey is summarized in the following table: Observed frequency Male Female Total Hired 45 35 Not Hired 32 28 Total Given Ho = “HR department hired male and female employees equally”, decide whether to reject or fail to reject the null hypothesis (α = 0.05).
Test the claim that the mean GPA of night students is significantly different than 2.5 at the 0.1 significance level. The null and alternative hypotheses would be: Ho p=0.625 Hop=2.5 Ho: p=0.625 Ho: : H₁: p > 0.625 The test is: right-tailed left-tailed two-tailed о :p = 0.625 Hoμ = 2.5 H₁ μ = 2.5 o H₁:μ> 2.5 H₁: p < 0.625 H₁: p0.625 H₁: μ2.5 H₁: μ< 2.5 Based on a sample of 35 people, the sample mean GPA was 2.47 with a standard deviation of 0.06 The p-value is: -1.645 The positive significance level is: .10 Based on this we: OTT Reject the null hypothesis Fail to reject the null hypothesis X (to 3 decimals) o(to 2 decimals)
In each part, we have given the value obtained for the test statistic, z, in a one-mean z-test.We have also specified whether the test is two tailed, left tailed, or right tailed. Determine the P-value in each case and decide whether, at the 5% significance level, the data provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis. a. z = -1.25; left-tailed testb. z = 2.36; right-tailed testc. z = 1.83; two-tailed test
If a significance test gives p-value 0.005... A.. the effect of interest is practically significant B...We do have good evidence against the null hypothesis C...we do not have good evidence against the null hypothesis D...the null hypothesis is very likely to be true E.. The margin of error is 0.005
I
H1.

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