95 confidence level sample 25 randomly selected mean test score of 81.5 standard deviation 10.2
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A: Given information significance level α = 0.05 S1 = 1.63 S2 = 1.49 n1 = 50 n2 = 50
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A: Hello, thanks for posting your question here. To solve this problem, population standard deviation…
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- What are the hypotheses that must be established in a statistical test? Toggle and null sample mean and variance Average and proportions Interval Estimation and Point EstimationHow many computers in a random sample of 155 household the simple mean number of personnel computers was 1.15 assume the population standard deviation is 0.78 construct a 99.5% confidence interval for the mean number of personal computers run the answer to the least two decimal places a 99.5% conference interval for the main number of personal computers is blank <mean<blankAssum a sample is used to estimate a populatation mean. Find the 95% confidence interval for a sample size of 41 with a mean of 64.2 and a standard deviation of 13.4. Enter answer as an open interval (i.e. parentheses) accurate to 3 decimal places.
- Verdadero=true falso=falseFor 30 randomly selected rolling stones concerts, the mean gross is 2.3 million dollars. Assuming a population standard deviation gross earnings of 0.51 million dollars, obtain a 99% confidence interval for the mean gross earnings of a rolling stones concert (in millions). How can we're interpret the answer?We want to estimate the true mean number of ounces of water a high school athlete drinks each day at a 95% confidence level. A random sample of 15 athletes from a school is selected and their typical daily water consumption is reported below: 64, 72, 80, 88, 128, 128, 128, 132, 145, 155, 160, 175, 180, 200, 225 What is the value of the standard error of the mean? (When calculating, use the standard deviation of the sample rounded to 3 decimal places and then round your standard error answer to 3 decimal places.) =
- best point estimate for 0.678>p>0.788 with a 95% confidence intervalIf a sample of 25 fish from my pond has mean weight 157 grams with a standard deviation of 8.3 grams and we know the population standard deviation for the fish in my pond is actually 9.4 grams, then WHAT IS the MARGIN OF ERROR in the 99 percent confidence interval for true mean weight of fish in my pond? Group of answer choicesSample 2 10, 15, 22, 1, 12, 1, 5, 12, 5, 6, 6, 1, 1, 25, 8, 10 what is the 95% Confidence interval to 11.69?
- Average hours per week on the internet. Students monitor surveys 1200 undergraduates from 100 colleges semiannually to understand trends among college students. Recently the student monitor reported that the average amount of time spent per week on the internet was 19.0 hours. You suspect that this amount is far too small for your campus and plan a survey. a) You feel that a reasonable estimate of the standard deviation is 10.0 hours. What sample size is needed so the expected margin of error of your estimate is not larger than one hour for 95% confidence? b) The distribution of times is likely to be heavily skewed to the right. Do you think that this skewness will invalidate the use of the t confidence interval in this case? Explain your answerWomen N Mean Standard Deviation New Drug 40 38.88 3.97 Placebo 41 39.24 4.21 Men N Mean Standard Deviation New Drug 10 45.25 1.89 Placebo 9 39.06 2.22 There is no statistical difference between the new drug and the placebo in women. Is there a significant difference between the placebo and the new drug among men in this study? The pooled standard deviation is equal to 2.05. Please complete the 5 steps of hypothesis testing. Remember that the n of both groups is less than 30.In a sample of 100 individuals, the average height is 170 cm and the standard deviation is 15 cm. Using a 95% confidence level, what is the margin of error for the mean height of the population?
