Six random determinations of sulfur content in steel at a particular point in a process gave the values 3.05, 3.11, 3.14, 3.24, 3.18, and 3.08. Assume the values are normally distributed. A previous study based on a sample of 21 random observations gave an estimate of variance of 1.51 × 10^–3. Use the 5% level of significance. a. Compute for the new sample variance. b. Using the variance ratio, is the variance significantly higher now?
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- The data for a random sample of six paired observations are shown below. Pair Observation 1 Observation 2 1 1 3 2 2 4 3 3 5 4 4 6 5 5 7 6 6 8 a. Calculate the difference between each pair of observations by subtracting observation 2 from observation 1. Use the differences to calculate sd2. b. Calculate the standard deviations s12 and s22 of each column of observations. Then find pooled estimate of the variance sp2. c. Comparing sd2 and sp2, explain the benefit of a paired difference experiment.Data on the weights (lb) of the contents of cans of diet soda versus the contents of cans of the regular version of the soda is summarized to the right. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.05 significance level for both parts. Diet Regular μ μ1 μ2 n 28 28 x 0.79741 lb 0.81023 lb s 0.00442 lb 0.00749 lb a. Test the claim that the contents of cans of diet soda have weights with a mean that is less than the mean for the regular soda. What are the null and alternative hypotheses? A. H0: μ1=μ2 H1: μ1≠μ2 B. H0: μ1≠μ2 H1: μ1<μ2 C. H0: μ1=μ2 H1: μ1>μ2 D. H0: μ1=μ2 H1: μ1<μ2 The test statistic, t, is nothing. (Round to two decimal places as needed.) The P-value is nothing.…Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 6.7 parts/million (ppm). A researcher believes that the current ozone level is at an insufficient level. The mean of 810 samples is 6.6 ppm. Assume the variance is known to be 1.00. Does the data support the claim at the 0.02 level? Step 1 of 5: Enter the hypotheses: Answer 2 Points E Keypad Keyboard Shortcuts Họ: Prev Ne: © 2020 Hawkes Learning MacBook Air 80 888 DII DD F1 F2 F3 F4 F5 F6 F7 F8 F9 F10 F11 ! @ %23 & 1 2 3 4 %3D Q W E T Y { P A S D F G H J K C M ? on command command option .. .- V - * 00 B.
- 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?(5). Calculate SS, variance and standard deviation for the following sample of n=7 scores: 8, 6, 5, 2, 6, 3, 5.A researcher takes sample temperatures in Fahrenheit of 16 days from San Francisco (Group 1) and 22 days from Atlanta (Group 2). Test the claim that the mean temperature in San Francisco is less than the mean temperature in Atlanta. Use a significance level of a = 0.10. Assume the populations are approximately normally distributed with unequal variances. You obtain the following two samples of data. Round answers to 4 decimal places. San Francisco 77.4 94.8 77 80.9 84 Ho: M₁ 73.4 84.9 92 60.6 83.6 92.5 76.3 93.4 87.2 81.5 70.4 What are the correct hypotheses? Note this may view better in full screen mode. Select the correct symbols for each of the 6 spaces. Atlanta 85.4 92.8 80.2 87.9 80.6 77.9 72.9 75.9 72.1 76.7 80.4 92.5 75.9 72.1 97.7 89.2 88.1 80 92.2 86 96.2 91.9 p-value H₁: M₁ Based on the hypotheses, find the following: Test Statistic = -0.635 X = 0.2494 × OH₂ The p-value is greater than alpha OF H₂ to alpha. The correct decision is to fail to reject the null hypothesis or The…
- The accompanying data table contains chest deceleration measurements (in g, whereg is the force of gravity) from samples of small, midsize, and large cars. Shown are the technology results for analysis of variance of this data table. Assume that a researcher plans to use a 0.05 significance level to test the claim that the different size categories have the same mean chest deceleration in the standard crash test. Complete parts (a) and (b) below. Click the icon to view the table of chest deceleration measurements A Click the icon to view the table of analysis of variance results. ..... a. What characteristic of the data specifically indicates that one-way analysis of variance should be used? O A. There are three samples of measurements. O B. The population means are approximately normal. O C. The measurements are categorized according to the one characteristic of size. O D. Nothing specifically indicates that one-way analysis of variance should be used. b. If the objective is to test…Data on the weights (Ib) of the contents of cans of diet soda versus the contents of cans of the regular version of the soda is summarized to the right. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.01 significance level for both parts. Diet Regular H2 27 27 0.79037 lb 0.80399 lb 0.00449 lb 0.00756 lb a. Test the claim that the contents of cans of diet soda have weights with a mean that is less than the mean for the regular soda. What are the null and alternative hypotheses? O A. Ho: H1 = H2 OB. Ho: H1#H2 Hq: HyOur environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 7.97.9 parts/million (ppm). A researcher believes that the current ozone level is at an excess level. The mean of 2424 samples is 8.18.1 ppm with a variance of 0.250.25. Does the data support the claim at the 0.10.1 level? Assume the population distribution is approximately normal. Step 3 of 5: Specify if the test is one-tailed or two-tailed.Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 4.4 parts/million (ppm). A researcher believes that the current ozone level is not at the normal level. The mean of 1090 samples is 4.3 ppm. Assume the variance is known to be 1.44. Does the data support the claim at the 0.05 level? Step 1 of 5: Enter the hypotheses: Step 2 of 5: Enter the value of the z test statistic. Round your answer to two decimal places. Step 3 of 5: Specify if the test is one-tailed or two-tailed. Step 4 of 5: Enter the decision rule. Step 5 of 5: Enter the conclusion.A researcher takes sample temperatures in Fahrenheit of 17 days from New York City and 18 days from Phoenix. Test the claim that the mean temperature in New York City is different from the mean temperature in Phoenix. Use a significance level of α=0.05. Assume the populations are approximately normally distributed with unequal variances. You obtain the following two samples of data. New York City Phoenix 99 94.2 95.5 72 93.2 86.8 102 122.1 85.4 114.4 80 94.7 86.4 89.7 75.4 104.7 79.5 77.6 83.4 106.8 64.3 98.6 65.5 91.5 87.7 82 104 97.7 74.3 64.9 59.5 82 82.8 72 116.2 The Hypotheses for this problem are: H0: μ1 = μ2 H1: μ1μ2 Find the p-value. Round answer to 4 decimal places. Make sure you put the 0 in front of the decimal. p-value =A researcher was interested in knowing if mean BMI was different between males and females. He obtained the following test results. Choose the correct interpretation of the results: F test p value = 0.30 Two sample t test, assume equal variance = 0.02 Two sample t test, assume unequal variance = 0.61 Mean BMI is not significantly different between males and females. None Mean BMI is higher in males than females. Mean BMI is significantly different between males and females.SEE MORE QUESTIONS
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