alue for a right-tailed test using a significance level of ?=0.02.α=0.02
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Find the critical ?z value for a right-tailed test using a significance level of ?=0.02.α=0.02.
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- A study is conducted to investigate the average number of hours of sleep per night for a sample of 40 college students. The sample mean is found to be x̄ = 6.5 hours with a standard deviation of s = 1.2 hours. Perform a hypothesis test to determine if the population mean number of hours of sleep per night is less than 7 hours. Use a significance level of α = 0.05.Total blood volume (in ml) per body weight (in kg) is important in medical research. For healthy adults, the red blood cell volume mean is about µ = 28 ml/kg with a standard deviation of 4.75ml/kg. Red blood cells volume that is too low or too high can indicate a medical problem. Suppose he does a hypothesis test with a significance level of 0.01. Roger has had seven blood tests and the red blood cell volumes were: 32 25 41 35 30 37 29 Roger’s null hypothesis is that his blood tests are normal and that his alternative hypothesis is that something is wrong. Symbolically, the null and hypotheses are as follows Ho: µ = 28 ml/kg and Hi: µ ≠ 28 ml/kg What value of the test statistic (z) should he report? (Sampling a Normal Distribution for the Mean µ) Group of answer choices a. 0.99 b.2.42 c.2.62 d.2.22You are conducting a study to see if the proportion of voters who prefer Candidate A is significantly more than 0.71. Thus you are performing a right-tailed test. Your sample data produce the test statistic z = 3.006. Find the p-value accurate to 4 decimal places. p-value =
- You wish to test the following claim (HaHa) at a significance level of α=0.05 Ho:μ=59.9 Ha:μ>59.9You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size n=93n=93 with mean M=62.6M=62.6 and a standard deviation of SD=9.4SD=9.4.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... less than (or equal to) αα greater than αα This test statistic leads to a decision to... reject the null accept the null fail to reject the null As such, the final conclusion is that... There is sufficient evidence to warrant rejection of the claim that the population mean is greater than 59.9. There is not sufficient evidence to warrant rejection of the claim that the population mean is greater than 59.9. The sample data support the claim that the…The mean test score for a simple random sample of n=100 students was =80. The population standard deviation of test scores is σ=15. Construct a 95% confidence interval for the population mean test score μ.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 hypothesis
- Professor 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 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 .
- Suppose the mean weight of King Penguins found in an Antarctic colony last year was 15.4 kg. In a sample of 35 penguins same time this year in the same colony, the mean penguin weight is 14.6 kg. Assume the population standard deviation is 2.5 kg. At .05 significance level, can we reject the null hypothesis that the mean penguin weight does not differ from last year? Use P-value approach.Do shoppers at the mall spend more money on average the day after Thanksgiving compared to the day after Christmas? The 52 randomly surveyed shoppers on the day after Thanksgiving spent an average of $122. Their standard deviation was $29. The 59 randomly surveyed shoppers on the day after Christmas spent an average of $117. Their standard deviation was $35. What can be concluded at the αα = 0.05 level of significance? For this study, we should use The null and alternative hypotheses would be: H0:H0: H1:H1: The test statistic = (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 the null hypothesis.The average salary for American college graduates is $45,100. You suspect that the average is less for graduates from your college. The 50 randomly selected graduates from your college had an average salary of $42,788 and a standard deviation of $5,940. What can be concluded at the αα = 0.05 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: ? μ p Select an answer > < = ≠ H1:H1: ? p μ 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 fail to reject reject accept the null hypothesis. Thus, the final conclusion is that ... The data suggest that the population mean is not significantly less than 45,100 at αα = 0.05, so there is…