Determine the critical value of Chi-square test, using 10 degrees of freedom and alpha of 0.05: A. Right tail test B. Left tail test C. Two tail test, lower value upper value
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A: It is an important part of statistics . It is widely used.
Q: Find the t-value for a=0.005 one-tailed test with n=25.
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A: Solution
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Q: what is the critical t-value for a right-tailed test when alpha = .025 and df = 12
A:
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A: Solution: Given information: α=0.2 Level of significance Tow tailed test.
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Q: With alpha = .05, (two tail, nondirectional test), the critical t score boundaries =
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Q: What is tcrit for a one sample, one tailed t-test, with n=13 and α=0.05
A: given data, one- tailed t- test n=13df=13-1=12α=0.05t-critical value=?
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A: When Wilcoxon signed ranks test instead of paired t-test.
Q: A company manufactures tennis balls. When its tennis balls are dropped onto a concrete surface from…
A: GivenMean(μ)=54.7sample size(n)=25sample mean(x)=56.8standard deviation(s)=0.25
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- At the alpha = 0.01 level , what is the correct conclusion for this test? The daily temperatures in fall and winter months in Virginia have a mean of 62F. A meteorologist in southwest Virginia believes the mean temperature is colder in this area. The meteorologist takes a random sample of 30 daily temperatures from the fall and winter months over the last five years in southwest Virginia. The mean temperature for the sample is 59 degrees * F with a standard deviation of 6.21 degrees * F The meteorologist conducts a one -sample t-test for and calculates a P value of 0.007. The meteorologist should reject the null hypothesis since 0.007 < 0.01 . There is convincing evidence that the mean temperature in fall and winter months in southwest Virginia is less than 62 F. The meteorologist should reject the null hypothesis since 0.007 < 0.01 . There is not convincing evidence that the mean temperature in fall and winter months in southwest Virginia is less than 62 F. The meteorologist…The manager of a political party wants to find out the association between the campaign spending (Large, medium and small) and the vote obtained (Majority, Average and minority). Use Chi-squared hypothesis test for association to test that there is an association between the variable in campaign spending and the variable in vote obtained with five percent level of significance. The manager uses MS excel to calculate the Chi-squared stat value and the result is 8.874. Critical value: At 1% = 13.277 :At 5% = 9.488 :At 10% = 7.779 1. State the decision rule 2. Conclusion and decision 3. State the alternative hypothesis and the null hypothesisA company manufactures tennis balls. When its tennis balls are dropped onto a concrete surface from a height of 100 inches, the company wants the mean height the balls bounce upward to be 54.8 inches. This average is maintained by periodically testing random samples of 25 tennis balls. If the t-value falls between −t0.95 and t0.95, then the company will be satisfied that it is manufacturing acceptable tennis balls. A sample of 25 balls is randomly selected and tested. The mean bounce height of the sample is 56.3 inches and the standard deviation is 0.25 inch. Assume the bounce heights are approximately normally distributed. Is the company making acceptable tennis balls?
- A random sample of 100 male employees showed that 42 are working from home while a random sample of 100 female employees showed that 63 are working from home. Are the two population proportions equal? Test at alpha=0.05. What is Ho and Ha? And what is the test statistic value for this test? Based on the test statistic value, what is your decision?SOLVE FOR CRITICAL VALUE Two suppliers manufacture a plastic gear used in a laser printer. The impact strength of these gears measures in foot-pounds is an important characteristic. A random sample of 10 gears from supplier 1 results in x1 = 290 and s1 = 12 while another random sample of 16 gears from the second supplier results in x2 = 321 and s2 = 22. Is there evidence to support the claim that supplier 2 provides gears with higher mean impact strength. Use a =0.04, and assume that both populations are normally distributed with equal variances.Level of significance = 0.04What is tcrit for a paired, two tailed t-test, with 15 individuals examined pre and post and α=0.05?
- According to a CCH Unscheduled Absence survey 9 of small According to a CCH Unscheduled Absence survey, 9% of small businesses use telecommuting of workers in an effort to reduce unscheduled absenteeism. This proportion compares to 6% for all businesses. Is there really a significant difference between small businesses and all businesses on this issue? Use these data and an alpha of .10 to test this question. Assume that there were 780 small businesses and 915 other businesses in this survey. According to a CCH Unscheduled Absence survey 9 of smallA study of parental empathy for sensitivity cues and baby temperament (higher scores mean more empathy) was performed. Let x, be a random variable that represents the score of a mother on an empathy test (as regards her baby). Let x, be the empathy = 68.36. Another random sample of 31 fathers gave score of a father. A random sample of 37 mothers gave a sample mean of x, X, = 60.06. Assume that o, = 11.55 and o, = 11.45. (a) Let u, be the population mean of x, and let u, be the population mean of x,. Find a 99% confidence interval for u, - Hz. (Round your answers to two decimal places.) lower limit upper limit (b) Examine the confidence interval and explain what it means in the context of this problem. Does the confidence interval contain all positive, all negative, or both positive and negative numbers? What does this tell you about the relationship between average empathy scores for mothers compared with those for fathers at the 99% confidence level? Because the interval contains only…Using the z table, find the critical value (or values) for an α = .09 two-tailed test.
- Determine the cri tical z-score corresponding to a = 0.2 in a left-tail test. crititical z-score = [three decimal accuracy]A company manufactures tennis balls. When its tennis balls are dropped onto a concrete surface from a height of 100 inches, the company wants the mean height the bails bounce upward to be 55.4 inches. This average is maintained by periodically testing random samples of 25 tennis balls. If the t-value falls between - to 90 and to so. then the company will be satisfied that it is manufacturing acceptable tennis balls. A sample of 25 balls is randomly selected and tested. The mean bounce height of the sample is 56.2 inches and the standard deviation is 0.25 inch. Assume the bounce heights are approximately normally distributed. Is the company making acceptable tennis balls? Find - to so and to so -o s0 to so O (Round to three decimal places as needed.)Define the Independent t Test for Two Samples and explain what why the word "independence" is used. In addition, come up with two hypothetical examples demonstrating this strategy or come up with "real-world" application using your professional life (i.e. how could you apply this in your work).
