Use this information to answer the following 3 questions: A high school achievement test has been...
Use this information to answer the following 3 questions:
A high school achievement test has been designed for American students such that the average score of US seniors on this test is 80, and the standard deviation is 10 points. You want to know whether French students score differently than Americans, on average. To study this question, you administer the test (in French) to 300 French high school seniors and observe a mean of 82.5.
1.State the null and alternative hypotheses in symbols (note: feel free to spell out your symbols if it's hard to insert symbols into the answer field).
2.Compute a z-statistic to test this null hypothesis. Assume that the SD of the American population is the population SD.
3.To determine whether the value of z you computed is “significant”, you will compare it to a standard normal (z) distribution. What does this distribution represent, in other words, what is this a distribution of?
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A high school achievement test has been...
An achievement test given to high school seniors is used to predict college GPA. The following...
An achievement test given to high school seniors is used to predict college GPA. The following relationship between the test scores and college GPA exists within the entire population of students: 1. Y 0.10+3.94 X+ E In the above, Y (response) is the college GPA of a given student, X is the student's score on the achievement test (predictor, scored as proportion correct, between 0 and 1), and e is random error, with E~N(0,.005), where the value.005 is the variance...
Achievement test scores of all high school seniors in a state have mean 60 and variance...
Achievement test scores of all high school seniors in a state have mean 60 and variance 64. A random sample of n = 100 students from one large high school had a mean score of 58. Is there evidence to suggest that this high school is inferior? (Calculate the probability that the sample mean is at most 58 when n = 100.) Show the appropriate graph of the sample mean.
1. Suppose the scores for high school seniors on the verbal portion of the SAT test...
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The mean exam score for 44 male high school students is 20.1 and the population standard...
The mean exam score for 44 male high school students is 20.1 and the population standard deviation is 4.7. The mean exam score for 58 female high school students is 19.6 and the population standard deviation is 4.1. At a = 0.01, can you reject the claim that male and female high school students have equal exam scores? Complete parts (a) through (e). Click here to view page 1 of the standard normal distribution table. Click here to view page...
You administered a survey to a sample of 401 high school sophomores and found a mean narcissism score of 91.3 with a st...
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4. You want to estimate the mean SATM score for 250,000 high school seniors in California. Only a...
4. You want to estimate the mean SATM score for 250,000 high school seniors in California. Only about 45% of California students take the SAT. These self-selected students are planning to attend college and are not representative of all California seniors A simple random sample (SRS) of 500 California high school seniors is tested. The mean score of the sample is Y 461. What could you say about the mean score, n-508 in the population of all 250,000 seniors? Assume...
An SRS of 380 high school seniors gained an average of.... An SRS of 380 high...
An SRS of 380 high school seniors gained an average of....
An SRS of 380 high school seniors gained an average of -23.11 points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation σ = 48.85. We want to estimate the mean change in score μ in the population of all high school seniors. (a) Using the 68-95-99.7 Rule or the z-table (Table A), give a 99.7%...
An SRS of 380 high school seniors gained an average of i -23.51 points in their...
An SRS of 380 high school seniors gained an average of i -23.51 points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation σ 48.51, we want to estimate the mean change in score u in the population of all high school seniors. (a) Using the 68-95-99.7 Rule or the z-table (Table A), give a 95% confidence interval (a, b) for based on this sample. Enter your...
The mean score on a science assessment test for 49 randomly selected male high school students...
The mean score on a science assessment test for 49 randomly selected male high school students was 153. Assume the population standard deviation is 36. The mean score on the same test for 50 randomly selected female high school students was 147. Assume the population standard deviation is 34. At a= 0.05, can you support the claim that the mean score on the science assessment test for male high school students is greater than the mean score for female high...
At a local high school, the juniors and seniors compete to see who has a better...
At a local high school, the juniors and seniors compete to see who has a better average on a statistics test. Ten randomly selected junior had an average of 38.0 points with a standard deviation of 2.9 and 14 randomly selected seniors had an average of 31.2 points with a standard deviation of 6.2. Assume both populations follow a normal distribution We want to perform a test to see if on average juniors perform better on the test. a. Define...
An achievement test given to high school seniors is used to predict college GPA. The following relationship between the test scores and college GPA exists within the entire population of students: 1. Y 0.10+3.94 X+ E In the above, Y (response) is the college GPA of a given student, X is the student's score on the achievement test (predictor, scored as proportion correct, between 0 and 1), and e is random error, with E~N(0,.005), where the value.005 is the variance...
The mean exam score for 44 male high school students is 20.1 and the population standard deviation is 4.7. The mean exam score for 58 female high school students is 19.6 and the population standard deviation is 4.1. At a = 0.01, can you reject the claim that male and female high school students have equal exam scores? Complete parts (a) through (e). Click here to view page 1 of the standard normal distribution table. Click here to view page...
4. You want to estimate the mean SATM score for 250,000 high school seniors in California. Only about 45% of California students take the SAT. These self-selected students are planning to attend college and are not representative of all California seniors A simple random sample (SRS) of 500 California high school seniors is tested. The mean score of the sample is Y 461. What could you say about the mean score, n-508 in the population of all 250,000 seniors? Assume...
An SRS of 380 high school seniors gained an average of....
An SRS of 380 high school seniors gained an average of -23.11 points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation σ = 48.85. We want to estimate the mean change in score μ in the population of all high school seniors. (a) Using the 68-95-99.7 Rule or the z-table (Table A), give a 99.7%...
An SRS of 380 high school seniors gained an average of i -23.51 points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation σ 48.51, we want to estimate the mean change in score u in the population of all high school seniors. (a) Using the 68-95-99.7 Rule or the z-table (Table A), give a 95% confidence interval (a, b) for based on this sample. Enter your...
At a local high school, the juniors and seniors compete to see who has a better average on a statistics test. Ten randomly selected junior had an average of 38.0 points with a standard deviation of 2.9 and 14 randomly selected seniors had an average of 31.2 points with a standard deviation of 6.2. Assume both populations follow a normal distribution We want to perform a test to see if on average juniors perform better on the test. a. Define...
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