Do the data support the claim that the two machines produce material with the same mean wear?
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A:
Q: A study was performed to determine whether men and women differ in repeatability in assembling…
A: Given: n1=28n2=25s1=1.98s2=2.31α=0.02
Q: a Can we conclude that there is a difference in the mean measurements between the two methods?
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A: The provided sample means are shown below: Xˉ1=31\bar X_1 = 31 Xˉ2=38\bar X_2 = 38 X Also, the…
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Q: (a) What is the level of significance? State the null and alternate hypotheses. H0: ?2 = 0.18; H1:…
A: According to the answering guidelines, we can answer only 3 subparts of a question and the rest can…
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Q: A cigarette manufacturer sent each of two laboratories presumably identical samples of tobacco. Each…
A: Solution-: Given: n1=5,n2=5,α=0.05 (A) We identify the hypothesis and interpret the result. (B)…
Q: a) Find t. (Give your answer correct to two decimal places.)
A: n = 44, Σd = 224, Σd2 = 6268 The mean is given by The standard deviation is given by
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A: Two companies manufacture a rubber material in tended for use in an automotive application. The part…
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A: Given Information: Sample size(n) = 43 Mean HCT level (x) = 33.8 pg/ml Standard Deviation (σ) =…
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A: From the provided information, Mean (µ) = 95 Variance (σ2) = 6 Sample size (n) = 50 Sample average…
Q: The fan blades on commercial jet engines must be replaced when wear on these parts indicates too…
A: a) The level of significance, α=0.01. As the null hypothesis, assume that the variance of…
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A: The sample size n is 30.
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Q: A pilot study showed a mean of 11,8 kg per pack and a variance of 0,49 kg. How many packs should the…
A: Given : A pilot study showed a mean of 11.8 kg per pack and a variance of 0.49 kg i.e Mean (μ) =…
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A: Given data: 0.50, 0.49, 0.45, 0.54, 0.44, 0.49, 0.51, 0.48, 0.51, 0.53, 0.51, 0.63, 0.69, 0.76,…
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A: From the provided information, Sample size (n) = 15 Sample mean (x̄) = 110 Standard variance (s2) =…
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A: The sample of size n1=20 has a mean x¯1=66.0 and standard deviation s1=10.0. The sample of size…
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A: Assume that μ defines the true porosity factor of coke.
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A: Group who checks it once a day : sample mean , x1 = 43, standard deviation , s1 = 4.77 n1 = 15…
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A: From the provided information, The degree of freedom = n1 + n2 – 2 = 70 + 50 – 2 = 118 As, the…
Q: C. For the test in part (i), how many degrees of freedom does the distribution of the test statistic…
A: (c) Since p-value associated with Levine's test for equality of variance is 0.048, Which indicates…
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A: We have given that Treatment with sample size of 42 and mean of 0.69 and standard deviation of 0.25…
Q: population sta ke use of the C periment might
A: Sample sizes are: nA=26nB=26 Sample means are: x¯A=20.1x¯B=24.1 So, x¯A-x¯B=4 Population standard…
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A: Given that, Sample size n=29 Sample mean x¯=18 Sample standard deviation s=3
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A: The null and alternative hypotheses are: H0:σ2≤0.18 mm2 H1:σ2>0.18 mm2
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Q: Medium Lead Level High Lead Level 72 92 n2 = 11 92 X2 = 89.451 85 88 S2 = 9.686 97 83 92 99 111 91
A: GivenFor High lead level:n2=11x¯2=89.451s2=9.686Let μ2 is population mean of IQ scores for subjects…
Q: Patients with rheumatoid arthritis are at a greater risk of developing osteoporosis. The reasons are…
A: Solution: Given information: n= 43 Sample size x= 33.8 pg/ml Sample mean σ=23.6 pg/ml…
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A: Let X be the weights of a bag Population mean μ=9, sample standard deviations σ=0.64…
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A:
Q: An educator has developed a program to improve math scores on the Texas STAAR test. The average…
A: To test:H0:μ=1472H1:μ≠1472
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A: sample mean, x = 1.21 sample variance = 160.7824 sample standard deviation, s = 160.7824 = 12.68…
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A: Given information: Sample size, n = 43 men The sample mean, x¯=33.8 pg/mL The population standard…
In one company, two machines producing tire material intended for commercial airplane tires are subjected to resistance
abrasion test in the field application. The QC engineer conducted their regular inspection of the two machines’ outputs. Upon
test, the following data was recovered from the 29 selected samples. For Machine 1, the sample average and standard deviation
was recorded to be 20mg per 1000 cycles and 2mg per 1000 cycles. For Machine 2, the sample average and standard deviation
is 15mg and 8mg per 1000 cycles. Do the data support the claim that the two machines produce material with the same mean
wear? Use α =0.10, and assume that each population is
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- A physical therapist wanted to know whether the mean step pulse of men was less than the mean step pulse of women. She randomly selected 54 men and 70 women to participate in the study. Each subject was required to step up and down a 6-inch platform. The pulse of each subject was then recorded. The following results were obtained. Two sample T for Men vs Women N Mean StDev SE Mean Men Women 98% CI for mu Men - mu Women (- 12.20, - 1.00) T-Test mu Men = mu Women (vs H2 O C. Ho: H1 = H2; Ha: H1 #H2 (b) Identify the P-value and state the researcher's conclusion if the level of significance was a = 0.01. What is the P-value? P-value =A study was conducted by a group of neurosurgeons. They compared a dynamic system (Z-plate) and a static system (ALPS plate) in terms of the number of acute postoperative days in the hospital spent by the patients. The descriptive statistics for these data are as follows: for 14 patients with dynamic system, the sample mean number of acute postoperative days was 7.36 with standard deviation of 1.22; for 6 patients with static system the sample mean number of acute postoperative days was 10.5 with sample standard deviation of 4.59. Assume that the numbers of acute postoperative days in both populations are normally distributed. We wish to estimate µ1 − µ2 with a 99 percent confidence interval. Can you assume unknown population variance are equal. a. Degrees of freedom and t-value are b. Margin of error and confidence interval are c. Based on your interval in part (b), we can state with 99 percent confidence that the average numbers of acute postoperative days in two populations i.…The amount of chlorine in 100 cm3 water taken from a water tank was measured, the average was 2.15 mg was found. The expert claims that the amount of chlorine in the water is greater than that measured. For this reason, 100 households were randomly selected, with an average chlorine content of 2.55mg and a variance of 0.49 mg. According to this data, find the account value z. 13 - O A) 8,15 O B) 5,71 O C) -5,71 O D) -8, 15 O E) 0.54
- Snow avalanches can be a real problem for travelers in the western United States and Canada. A very common type of avalanche is called the slab avalanche. These have been studied extensively by David McClung, a professor of civil engineering at the University of British Columbia. Suppose slab avalanches studied in a region of Canada had an average thickness of u = 67 cm. The ski patrol at Vail, Colorado, is studying slab avalanches in its region. A random sample of avalanches in spring gave the following thicknesses (in cm). 59 51 76 38 65 54 49 62 68 55 64 67 63 74 65 79 (i) Use a calculator with sample mean and standard deviation keys to find x and s. (Round your answers to two decimal places.) X= cm S= cm (ii) Assume the slab thickness has an approximately normal distribution. Use a 1% level of significance to test the claim that the mean slab thickness in the Vail region is different from that in the region of Canada. (a) What is the level of significance? State the null and…200 people were randomly sampled and asked what they regularly eat for breakfast or lunch. Each person was identified as either a consumer or a non consumer of high-fiber cereals, and the number of calories consumed at lunch was measured and recorded. These data are summarized below; Consumer of high fiber cereals Non consumer of high fiber cereals η1 =41 η2 = 159 Mean 1 =603 Mean 2 =639 Stanadard deviation 1 = 110 Standard deviation 2 = 141 If the scientist claims that people who eat high fiber cereals for breakfast do consume on average fewer calories for lunch than people who don’t eat high fiber cereals for breakfast, and if it is true, high fiber cereal manufacturer will be able to claim another advantage of eating their products-potential weight reduction for dieter. REQUIRED Are there sufficient evidence at 5% significance level to support the above claim?I need the p value, please.
- The fan blades on commercial jet engines must be replaced when wear on these parts indicates too much variability to pass inspection. If a single fan blade broke during operation, it could severely endanger a flight. A large engine contains thousands of fan blades, and safety regulations require that variability measurements on the population of all blades not exceed ?2 = 0.18 mm2. An engine inspector took a random sample of 81 fan blades from an engine. She measured each blade and found a sample variance of 0.28 mm2. Using a 0.01 level of significance, is the inspector justified in claiming that all the engine fan blades must be replaced? (b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.) Find or estimate the P-value of the sample test statistic. P-value > 0.1000.050 < P-value < 0.100 0.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.010P-value < 0.00 Find a 90% confidence interval…A researcher wants to determine if there is a significant difference in the average math scores of two different teaching methods (Method A and Method B). He collects a sample of 50 students from each teaching method and calculates their scores. The sample mean score for Method A is 85 and the sample mean score for Method B is 80. The sample standard deviation for Method A is 4 and the sample standard deviation for Method B is 3. Conduct an independent samples t-test to determine if there is a significant difference in the average math scores of the two teaching methods at a significance level of 0.01.In a class of twenty-five 15-year old girls the average height is M = 62.60 inches. The national average for girls at this age is reported as follows: µ = 63.80 inches with a standard deviation, s = 2.66 inches. 1. What is the effect size Cohen’s d and what does it tell us about the effect? 2. Compute the z-statistic.
- The void volume within a textile fabric affects comfort, flammability, and insulation properties. Permeability of a fabric refers to the accessibility of void space to the flow of a gas or liquid. An article gave summary information on air permeability (cm3/cm2/sec) for a number of different fabric types. Consider the following data on two different types of plain-weave fabric: Fabric Type Sample Size Sample Mean Sample Standard Deviation Cotton 10 51.41 0.75 Triacetate 10 132.15 3.59 Assuming that the porosity distributions for both types of fabric are normal, let's calculate a confidence interval for the difference between true average porosity for the cotton fabric and that for the acetate fabric, using a 95% confidence level. Before the appropriate t critical value can be selected, df must be determined: df = 0.5625 10 + 12.8881 10 2 (0.5625/10)2 9 + (12.8881/10)2 9 = 1.8092 0.1849…Find Test Statistic, P value, and Z score.Low-density lipoprotein, or LDL, is the main source of cholesterol buildup and blockage in the arteries. This is why LDL is known as "bad cholesterol." LDL is measured in milligrams per deciliter of blood, or mg/dL. In a population of adults at risk for cardiovascular problems, the distribution of LDL levels is normal, with a mean of 123 mg/dL and a standard deviation of 41 mg/dL. If an individual's LDL is at least 1 standard deviation or more above the mean, he or she will be monitored carefully by a doctor. What percentage of individuals from this population will have LDL levels 1 or more standard deviations above the mean? Use the 68–95–99.7 rule. (Enter your exact answer as a whole number.) percentage %