If you receive a p-value of 0.233 when performing a statistical test. Using the normal alpha level, what does this result mean? no conclusions can be made reject the null fail to reject the null
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- The correlation coefficient r is a sample statistic. What does it tell us about the value of the population correlation coefficient ?(Greek letter rho)? You do not know how to build the formal structure of hypothesis tests of ? yet. However, there is a quick way to determine if the sample evidence based on ? is strong enough to conclude that there is some population correlation between the variables. In other words, we can use the value of r to determine if ? ≠ 0. We do this by comparing the value |r| to an entry in the correlation table. The value of ? in the table gives us the probability of concluding that ? ≠ 0 when, in fact, ? = 0 and there is no population correlation. We have two choices for ?: ? = 0.05 or ? = 0.01. (a) Look at the data below regarding the variables x = age of a Shetland pony and y = weight of that pony. Is the value of |r| large enough to conclude that weight and age of Shetland ponies are correlated? Use ? = 0.05. (Round your answer for r to four decimal…A researcher is testing a hypothesis of a single mean. The critical t value for a = .05 and a one-tailed test is 2.0639. The observed t value from sample data is 1.742. The decision made by the researcher based on this information is to the null hypothesis. reject not reject redefine change the alternate hypothesis into restateYou are part of a trivia team and have tracked your team's performance since you started playing, so you know that your scores are normally distributed with u = 88 and a = 15 Recently, a new person joined the team, and you think the scores have changed Sample data is: n = 28, sample mean = 72 Assuming a two tailed test, What is the test statistic (Z)?
- A sample mean is 94, sample size is 27, and population standard deviation is 14 are given. Ho: p=89, Hl: μ#89 and a significance level is 0.10. Use the P-value approach to test the hypothesis. O a z=186; P-value = 0.0314; reject null hypothesis O b. z 186, P-value = 0.0628; reject null hypothesis O c. z=0.36; P-value = 0.7188, do not reject null hypothesis Od. z=0.36, P-value = 0.3594; do not reject null hypothesisA sample of 33 lights in Austin gives an average of 46 cars/min while the city SD=3.4 A sample of 34 Dallas lights give a mean of 44 cars/mim with the city SD=1.4 At the 91% level, do the towns differ? 1. Is the test (A-Left Tailed, B-Right Tailed, C-2-Tailed) (separate with comma if there are 2) 2. Critical Value 3. Test statistic 4. 3-decimal p-ValueProfessor Nord stated that the mean score on the final exam from all the years he has been teaching is a 79%. Colby was in his most recent class, and his class’s mean score on the final exam was 82%. Colby decided to run a hypothesis test to determine if the mean score of his class was significantly greater than the mean score of the population. α = .01. What is the mean score of the population? What is the mean score of the sample? Is this test one-tailed or two-tailed? Why?
- Do men talk less than women? The table shows results from a study of the words spoken in a day by men and women. Assume that the two samples are randomly selected, independent, the population standard deviations are not know and not considered equal. At the 0.05 significance level, test the claim that the mean number of words spoken by men is less than the mean number of words spoken by women. Men Women n1 = 212 n2 = 218 xˉx̄1 = 15706.4 words xˉx̄2 = 15683.2 words s1 = 1595.44 words s2 = 1597.54 words What are the correct hypotheses? (Select the correct symbols and use decimal values not percentages.)H0: Select an answer μ₁ μ σ₁² x̄₂ p x̄₁ s₁² p̂₁ μ(men) p₂ p₁ μ₂ ? ≤ > ≥ = ≠ < Select an answer μ₁ p₁ μ₂ x̄₂ σ₁² p̂₁ μ s₁² μ(women) p x̄₁ p₂ H1: Select an answer p̂₂ p s₂² σ₂² μ(men) μ μ₂ x̄₂ x̄₁ p₂ μ₁ p₁ ? < > = ≠ ≥ ≤ Select an answer μ₂ μ p₂ s₁² p₁ x̄₁ σ₁² p̂₁ p μ₁ x̄₂ μ(women) Original Claim = Select an answer H₁ H₀ df = Based on the hypotheses, find the…The correlation coefficient r is a sample statistic. What does it tell us about the value of the population correlation coefficient ? (Greek letter rho)? You do not know how to build the formal structure of hypothesis tests of ? yet. However, there is a quick way to determine if the sample evidence based on ? is strong enough to conclude that there is some population correlation between the variables. In other words, we can use the value of r to determine if ? ≠ 0. We do this by comparing the value |r| to an entry in the correlation table. The value of ? in the table gives us the probability of concluding that ? ≠ 0 when, in fact, ? = 0 and there is no population correlation. We have two choices for ?: ? = 0.05 or ? = 0.01.The average salary for American college graduates is $48,600. You suspect that the average is less for graduates from your college. The 49 randomly selected graduates from your college had an average salary of $46,486 and a standard deviation of $9,200. What can be concluded at the αα = 0.01 level of significance? For this study, we should use Select an answer t-test for a population mean z-test for a population proportion The null and alternative hypotheses would be: H0:H0: ? p μ Select an answer = ≠ > < H1:H1: ? μ p Select an answer > ≠ = < The test statistic ? z t = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is ? > ≤ αα Based on this, we should Select an answer reject accept fail to reject the null hypothesis. Thus, the final conclusion is that .
- My answer is as follows: T critical = 2.70 T value= 7.775 P< a (0.01) Therefore we reject the null. Is this correct?Please sir or ma’am, please help me with hypothesis, critical region, and t statistic.What is the name of this hypothesis test? Do male and female servers at Applebee's work the same number of hours? A random sample of 45 female servers worked an average of 36 hours per week, with a standard deviation of 2. A random sample of 62 male servers worked an average of 29.5 hours per week, with a standard deviation of 4. O Chi-Square Test for Independence O One Sample z Test OTwo Proportion z Test Z. OLinear Regression t Test O Chi-Square Goodness of Fit Test O Paired Samples t Test OTwo Independent Samples t Test O One Sample t Test O One Proportion z Test OANOVA