Q2: A hypotheses regarding newborn infants weight at a community hospital is that the mean is 33 pounds. A sample of 51 infants is randomly selected and their average sample weight at a birth are 12 and the standard deviation is 11. The null hypotheses: HO: u =
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- You would like to estimate the variance of GPA’s at SDSU. You randomly select 14 students and get a standard deviation of .981. Construct a 90% confidence interval of the population variance of GPA’s for SDSU. Using the previous information from SDSU GPA’s: You would like to do a hypothesis test to see if the variability of GPA’s at SDSU has increased from its historical average variance of .80. Use a level of significance of .05 and after doing the hypothesis test, what critical value would be used for this test? If the test statistic was equal to 35, what would be the probability associated with this test statistic?A chemical engineer claims that the average daily yield of a chemical manufactured in a chemical plant is 880. A sample of 50 daily yields is taken, and the sample mean is determined to be 871, with a standard deviation of 21. What would be the null hypothesis?A certain test preparation course is designed to improve students' SAT Math scores. The students who took the prep course have a mean SAT Math score of 550, while the students who did not take the prep course have a mean SAT Math score of 541. Assume that the population standard deviation of the SAT Math scores for students who took the prep course is 35.8 and for students who did not take the prep course is 31.6 The SAT Math scores are taken for a sample of 77 students who took the prep course and a sample of 87students who did not take the prep course. Conduct a hypothesis test of the claim that the SAT Math scores for students who took the prep course is higher than the SAT Math scores for students who did not take the prep course. Let μ1 be the true mean SAT Math score for students who took the prep course and μ2 be the true mean SAT Math score for students who did not take the prep course. Use a 0.01 level of significance. Step 1 of 5 : State the null and alternative…
- A researcher is interested in seeing if the average income of rural families is less than that of urban families. To see if his claim is correct he randomly selects 23 families from a rural area and finds that they have an average income of $69033 with a standard deviation of $591. He then selects 24 families from a urban area and finds that they have an average income of $65076 with a standard deviation of $846. Perform a hypothesis test using a significance level of 0.01 to help him decide. Let the rural families be sample 1 and the urban families be sample 2. The correct hypotheses are: Ho:µ1 µ2(claim) Ho: µ1 2 µ2 HA: H1 < H2(claim) Ho: µ1 = µ2 HA:µ1 # µ2(claim) Since the level of significance is 0.01 the critical value is -2.42 The test statistic is: (round to 3 places) The p-value is: (round to 3 places)A sample of 100 observations from a normal population has a mean of 50 and a standard deviation of 10. The null hypothesis is that the population mean is equal to 48, and the alternative hypothesis is that the population mean is greater than 48. Using a significance level of 0.05, what is the decision and conclusion of the hypothesis test?A random sample of 46 adult coyotes from Northern Minnesota showed the average age to be sample mean of 2.05 years. The population standard deviation is known to be o = 1.13 years. However, it is thought that the average age of coyotes is u = 1.75 years. Test to see if the coyotes in Northern Minnesota live longer than the national average of 1.75 years. Use 0.03 level of significance. What are the hypotheses? A: Ho µ= 1.75 years vs HA µ> 1.75 years B: Ho µ= 1.75 years vs HA H 2.05 years D: Ho µ = 2.05 years vs HA u< 2.05 years O A D O O
- When 40 people used the Weight Watchers dies for one year, their mean weight loss was 3.0 lb and the standard deviation was 4.9 lb. Use a 0.01 significance level to test the claim that the mean weight loss is greater than 0. (Show the solution) A. The null hypothesis is i. The mean weight loss is equal to 0. ii. The mean weight loss is not equal to 0. iii. The mean weight loss is greater than 0. iv. The mean weight loss is less than 0. B. The computed t-test statistic is i. 4.468 ii. 4.977 iii. 3.872 iv. 3.391 C. The critical/tabular t-test statistic is i. 2.426 ii. 1.684 iii. 1.685 iv. 2.423 D. Decision rule: i. Do not reject the null hypothesis. ii. Reject the null hypothesis. E. The final conclusion is i. The mean weight loss is not equal to 0 therefore Weight Watchers is effective. ii. The mean weight loss is less than 0 therefore Weight Watchers is not effective. iii. The mean weight loss is equal to 0 therefore Weight Watchers…The average life of light bulbs produced by SABA Electric Co. is expected to be normally distributed with the mean service life of 950 hours and standard deviation of 100 hours. A consumer agency tests a random sample of 100 bulbs and finds mean service life of 910 hours. If the agency conducts hypothesis testing, what would be the computed p-value for the test? 4 10 0% 3.49 none of the above.When 50 people used a certain diet for one year, their weight losses had a standard deviation of 5.4lb. Use a 0.05 significance level to test the claim that the amounts of weight loss have a standard deviation equal to 6.0lb, which appears to be the standard deviation for the amounts of weight loss with a different diet. Assume that a simple random sample is selected from a normally distributed population. What are the correct hypotheses for this test? H0: H1: Calculate the value of the test statistic. (Round to two decimal places as needed.) Use technology to determine the P-value for the test statistic. The P-value is (Round to three decimal places as needed.) What is the conclusion at the 0.05 level of significance? Since the P-value is ......(greater than/less than/equal to)the level of significance,......(do not reject/ reject) the null hypothesis. There ....(is/ is not) sufficient evidence to conclude that the sample is from a population with a…
- Among private universities in the United States, the mean ratio of students to professors is 35.2 (i.e., 35.2 students for each professor) with a standard deviation of 8.8. Suppose a random sample of 50 universities is selected and the observed mean student-to-professor ratio is 38. Is there evidence that the reported mean ratio actually exceeds 35.2? Use = 0.05 and conduct the appropriate hypothesis test. ( See attached picture )An engineer is comparing voltages for two types of batteries (K and Q) using a sample of 3939 type K batteries and a sample of 5757 type Q batteries. The mean voltage is measured as 8.558.55 for the type K batteries with a standard deviation of 0.6830.683, and the mean voltage is 8.828.82 for type Q batteries with a standard deviation of 0.7910.791. Conduct a hypothesis test for the conjecture that the mean voltage for these two types of batteries is different. Let μ1μ1 be the true mean voltage for type K batteries and μ2μ2 be the true mean voltage for type Q batteries. Use a 0.020.02 level of significance. Step 3 of 4 : Determine the decision rule for rejecting the null hypothesis H0H0. Round the numerical portion of your answer to two decimal placesSixty gym members were randomly selected and their weights were recorded and inputted into MINITAB. The results are summarized in Exhibit 1 below. One-Sample Z: Weight Test of mu = 195 vs < 195 The assumed standard deviation = 22.11 Variable = N Weight = 60 Mean = 193.13 Standard Deviation =22.11 SE Mean = *What is the value of test statistic for this test.?