lable data, indicate the lower and upper 80% CI for the ratio between the 2 1 Trt A 1.75 16 a. 0.5377 [0²A/0²B] 4.4566 Trt B 0.94 41 b. 1.0088 [0²A/0²B] 3.0949 c. 0.8748 [0²A/0²B] 5.3952
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![The following information was obtained from an experiment with two treatments (A and B). From the
available data, indicate the lower and upper 80% CI for the ratio between the two variances:
Trt A
1.75
n
16
a. 0.5377 [o²a/o²â] 4.4566
Trt B
0.94
41
S²
b. 1.0088 [0²A/O²B] 3.0949
c. 0.8748 [0²A/0²B] 5.3952
d. 0.76219 [0²A/0²B] 2.8380](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd32c0099-d35b-45af-b7af-24c06e9e82b9%2F1972faf7-dee0-457c-a0a9-c1644c16812f%2Fw3q93bx_processed.png&w=3840&q=75)
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- 1. ln [calcium intake (mg)] among 25 females, 12 to 14 years of age, below the poverty level is 6.56 ± 0.64. Similarly, the mean ± 1 sd of ln [calcium intake (mg)] among 40 females, 12 to 14 years of age, above the poverty level is 6.80 ± 0.76. a. Test for a significant difference between the variances of the two groups. b. What is the appropriate procedure to test for a significant difference in means between the two groups? c. Implement the procedure in (b) using the critical-value method. d. What is the p-value corresponding to your answer in (c)? e. Compute a 95% CI for the difference in means between the two groups.At age 9 the average weight for both boys and girls are exactly the same. A random sample of 9 year olds gave the following results.... Boys: Girls:Sample Size: 65 50Mean weight (lbs): 130 125Pop. Variance: 120 130 At alpha 0.05, do the data support the claim that there is a difference in heights?Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 4.8 parts/million (ppm). A researcher believes that the current ozone level is at an excess level. The mean of 242samples is 5.2 ppm with a variance of 1.001.00. Does the data support the claim at the 0.05 level? Assume the population distribution is approximately normal. Find the value of the test statistic. Round your answer to three decimal places. Specify if the test is one-tailed or two-tailed. Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places. Make the decision to reject or fail to reject the null hypothesis.
- The following information was obtained from an experiment with two treatments (A and B). From the available data, indicate the lower and upper 80% CI for the ratio between the two variances: Trt A 1.75 16 a. 0.5377 [0²A/0²B] 4.4566 S² n Trt B 0.94 41 b. 1.0088 [0²A/0²B] 3.0949 c. 0.8748 [0²A/0²B] 5.3952 d. 0.76219 [0²A/0²B] 2.8380Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 4.6 parts/million (ppm). A researcher believes that the current ozone level is at an excess level. The mean of 26 samples is 4.9 ppm with a variance of 1.00 . Does the data support the claim at the 0.05 level? Assume the population distribution is approximately normal. 1. State the null and alternative hypotheses. 2, Find the value of the test statistic. Round your answer to three decimal places. 3. Specify if the test is one-tailed or two-tailed . 4. Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places. 5. Make the decision to reject or fail to reject the null hypothesisThe pH of rain, measured at a weather station in Michigan, was observed for 39 consecutive rain storms. The sample mean is 4.6982 and the sample variance is 0.39623. Test the research hypothesis that the mean pH is less than 5.00, with α = 0.01.
- AUA wants to determine whether the mean cumulative GPA, CGPA is significantly different among undergraduate and graduate students (Sheet 31). They collect data on several undergraduate and graduate students. What can they conclude at alpha=0.05? Select one: a. t(stat)=2.56, which is greater than t(cr)=1.68, therefore, we can conclude that there is significant difference in mean CGPAs b. Assuming unequal variances, t(stat)=2.56, which is greater than t(cr)=2.02, therefore, we can conclude that there is significant difference in mean CGPAs c. t(stat)=1.15, which is less than t(cr)=2.12, therefore, we can conclude that there is no significant difference in mean CGPAs d. Assuming equal variances, t(stat)=2.56, which is greater than t(cr)=2.02, therefore, we can conclude that there is significant difference in mean CGPAsRecords for the last 15 years have shown that the average rainfall in a certain region of the country, for the month of March, to be 1.20 inches, with s = 0.45 inches. A second region had an average rainfall of 1.35 inches, with s = 0.54. estimate the difference of the true average rainfalls in those two regions as a 95% C.I. with the assumption of normal populations and unequal variances.An automobile manufacturer has given its car a 46.246.2 miles/gallon (MPG) rating. An independent testing firm has been contracted to test the actual MPG for this car since it is believed that the car has an incorrect manufacturer's MPG rating. After testing 240240 cars, they found a mean MPG of 46.446.4. Assume the population variance is known to be 2.562.56. A level of significance of 0.050.05 will be used. Find the P-value of the test statistic. You may write the P-value as a range using interval notation, or as a decimal value rounded to four decimal places.
- An auto manufacture claims that the variance of the gas mileages in a model of hybrid vehicle is 0.16. A random sample of 30 vehicles has a varience of 0.26. At a=0.05 is there enough evidence to reject the claim??Unfortunately, arsenic occurs naturally in some ground water. A mean arsenic level of u = 8 parts per billion (ppb) is considered safe for agricultural use. A well in Los Banos is used to water cotton crops. This well is tested on a regular basis for arsenic. A random sample of 37 tests gave a sample mean of = 7.3ppb arsenic. It is known that o = 1.9 ppb for this type of data. Does this information indicate that the mean level of arsenic in this well is less than 8 ppb? Use the classical approach. Use a = 0.01 What is the Decision (step 5) for this problem? There is not sufficient evidence at the 0.01 level of significance to show the mean arsenic level in the Los Banos well is less than 7.3 ppb. O There is sufficient evidence at the 0.01 level of significance to show the mean arsenic level in the Los Banos well is less than 8.0 ppb. There is not sufficient evidence at the 0.01 level of significance to show the mean arsenic level in the Los Banos well is less than 8.0 ppb. O There is…Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 7.6 parts/million (ppm). A researcher believes that the current ozone level is not at a normal level. The mean of 1313 samples is 7.2 ppm with a variance of 0.49 Assume the population is normally distributed. A level of significance of 0.02 will be used. Make the decision to reject or fail to reject the null hypothesis.
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