the 95% confidence interval for the mean score, μ, of all students taking the test
Q: A study of 36 golfers showed that their mean score at The Links at Arundel was 83. The standard…
A: n=sample size= 36 X¯=sample mean=83σ=sample standard deviation=5 So the point estimate=X=83 Since…
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A: Given, Standard Deviation =σ=4.7Sample size =n = 47Mean = x=28
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A: sample size(n)=75Mean()=50standard deviation(s)=13confidence interval=95%
Q: Fran is training for her first marathon, and she wants to know if there is a significant difference…
A: Sample mean (x̅1) = 42.8Sample mean (x̅2) = 40.9Sample size (n1) = 34Sample size (n2) = 41Standard…
Q: You randomly select 20 cars of the same model that were sold at a car dealership and determine the…
A: Answer:- Given, Sample size, n = 20 Sample mean, X¯= 9.75 Sample standard deviation, S = 2.39…
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A: What is the standard error of mean?
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A: 5. Given that, n=300x¯=67.7s=23.6 Confidence level=90% Significance level=1-0.90-0.10…
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Q: university in England and each student in a random sample of 38 female students from the same…
A: The following information has been provided: For Males : Sample Mean 1 (Xˉ1)(\bar X_1)= 282282…
Q: The test statistic is The p-value is Should the null hypothesis be rejected?
A: here from given information for gamer in age- group 16-25 x1¯ = 42.5 σ1 = 9.5 n1 = 351 for gamer…
Q: The local Health Department conducted a study in 2009 of 900 adults asking each adult the number of…
A: Given Information: Sample size (n) = 900 Sample mean (x¯) = 7.946 Standard deviation = 6.567…
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Q: The combined SAT scores for students at a local high school are normally distributed with a mean of…
A: X follows a Normal distribution with mean = 1497 and S.D = 293 Then z=(x-mean)/SD = (x-1497)/293…
Q: Dr. Vegapunk thinks that watching anime (Japanese animated shows) decreases social skills in college…
A: Dr. Vegapunk thinks that watching anime (Japanese animated shows) decreases social skills in college…
Q: From the scores in subject statistics for 8 students, it was found that the mean score is 73 and the…
A: We have given that, Sample mean (x̄) = 73, standard deviation (s) = 5.15 and sample size (n) = 8…
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A: According to the given information in this question We need to find the 95% confidence interval
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A: Solution: Hypothesis testing: From the above hypothesis test we can conclude that the low fat…
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Q: A person must score in the upper 0.15% of admission test scores in order to qualify for a very…
A: A person must score in the upper 0.15% of admission test scores in order to qualify for a very…
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A: We know that the If μ is the mean score and σ is the standard deviation for this set of test scores,…
Q: Find a 90% interval estimate of the mean daily time spent on rigorous exercise by all employees.
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A: Given data,n1=40x1=46.6σ1=4.6n2=45x2=48.8σ2=2.4Compute value of test statistic?
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A: The sample size is 45, sample mean is 3.8 and standard deviation is 1.5. The degrees of freedom is,…
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A: It is given that Sample size n = 35 Sample mean M = 4.50 Sample SD s = 1.12 Confidence level = 99%…
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A: Given that n=13 , Mean=X̄=790 , Standard deviation=S=8.3 95% confidence interval
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A: Let be the group runners training for marathons be population 1 and be the individual…
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A: Let X be the score of an IQ test.Given that, X is known to be normally distributed with a mean of…
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Q: When 40 people used Weight Watchers for one year, their mean weight loss was 3.0 lbs and their…
A: The test hypotheses are given below: H0:μ=0 lbsHa:μ>0 lbs From the given information, the…
Q: 20% of drivers don't use turn signals, observed of 140 cars. What is the standard deviation of…
A: Given that 20% of drivers don't turn signals and there are 140 cars. That is, p=0.20 and n=140
Q: Sam computed a 90% confidence interval for mean from a specific random sample size n. He claims that…
A: Given Information: Sam computed a 90% confidence interval for mean from a specific random sample of…
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A: From the provided information, The claim is that the mean number of hours of sleep that seniors in…
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A: Given, n=62 data points standard deviation SD= 2.1, calculated a 92% two-sided confidence interval…
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- Suppose the scores of students on an exam are normally distributed with a mean of 558 and a standard deviation of 61. According to the empirical rule, what percentage of students scored between 375 and 741 on the exam?The combined SAT scores for the students at a local high school are normally distributed with a mean of 1521 and a standard deviation of 301. The local college includes a minimum score of 888.9 in its admission requirements. What percentage of students from this school earn scores that fail to satisfy the admission requirement?Suppose the scores of students on an exam are normally distributed with a mean of 253 and a standard deviation of 80. According to the empirical rule, what percentage of students scored between 13 and 493 on the exam?
- for a sample of 121 that has a mean of 6540 and a standard deviation of 780, calculate the 90% confidence interval20 TV remote controls were followed over the years to find the mean lifespan of this model. The sample gave a mean of 7.2 years, and the standard deviation of 0.6 years. The distribution of the population is unknown. Construct a 90% confidence interval for the mean lifespan of these remote controls.Are low-fat diets or low-carb diets more effective for weight loss? A sample of 61 subjects went on a low-carbohydrate diet for six months. At the end of that time, the sample mean weight loss was 2.1 kilograms with a sample standard deviation of 4.71 kilograms. A second sample of 63 subjects went on a low-fat diet. Their sample mean weight loss was 3.7 kilograms with a standard deviation of 5.67 kilograms. Can you conclude that the mean weight loss of subjects having low-carb diets is less than the mean weight loss of subjects having low-fat diets? Let μ1 denote the mean weight loss on the low-carb diet and μ2 denote the mean weight loss on the low-fat diet. Use the =α0.05 level and the critical value method. 1) State the appropriate null and alternate hypotheses. 2)Choose a significance level and find the critical value. 3)Compute the test statistic. To compute the test statistic, we use the formula 4)Determine whether to reject H0.
- Participants were enrolled in a health-coaching program for eight weeks to try to improve physical health and overall wellness. At the end of the trial period, the participants take a personal health improvement survey. The mean health improvement survey score for all participants in the program was 74.4 with a standard deviation of 10.3. We have reason to think that one of the health coaches (Coach A) who participated in an additional training program will have subjects who have higher health improvement survey scores than the overall program average. We know that Coach A trains 30 subjects with the average health improvement survey score of 80.1. a) We want to test the hypothesis that subjects who work with Coach A have higher health improvement survey scores compared with the rest of the program participants. Use alpha of 0.05 b) Suppose a continuous random variable can only take on values between −1 and +1 on a standard normal curve, what is the area under the curve? Use…In a simple random sample of 144 households in a county in Virginia, the average number of children in these households was 3.62 children. The standard deviation from this sample was 2.40 children. A 90% confidence interval for the mean number of children in all households in this county is?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 large number of bags of cement have been manufactured. You sample 10 of the bags and find that they have a mean weight of 91.5 pounds with a standard deviation of 8.8 pounds. Find the 90 % confidence limit for the mean of the weight of all of the cement bags.A researcher is attempting to calculate a 95% confidence interval for the mean average wingspan of Statsian Swallows in a particular rainforest. They take a sample and find that of the 16 swallows surveyed, the mean average was 3.14 inches and the standard deviation was 0.2 inches. What is the 95% confidence interval for this sample?An agriculture researcher plants twenty five plots with a new variety of corn. The average yielded for these plots is x-bar = 150 bushels per acre. Assume that the yield per acre for the new variety of corn follows a normal distribution with unknown mean and standard deviation = 10 bushels per acre. A 90% Confidence interval for mean is?
