Use the confidence interval to find the margin of error and the sample mean. (0.142,0.300) The margin of error is The sample mean is
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- Use the data below to test the claim that the weight loss was greater with the low fat diet compared to the low carbohydrate diet. Use a 5% level of significance. Low Fat Low Carbohydrate x¯1=4.7lb s1=7.2lb n1=77 x¯1=2.6lb s1=5.9lb n1=79 Identify the tail of the test. What is the P-value? Will the null hypothesis be rejected? Is the initial claim supported?Which values have a z-score less than -0.5Pls solve. I got it wrong on practice quiz and i have no idea what i did wrong. Thanks!!
- Use the normal distribution to find a confidence interval for a proportion p given the relevant sample results. Give the best point estimate for p, the margin of error, and the confidence interval. Assume the results come from a random sample. A 99% confidence interval for p given that p = 0.75 and n = 110. Round your answer for the point estimate to two decimal places, and your answers for the margin of error and the confidence interval to three decimal places. Point estimate = Margin of error = + i The 99% confidence interval is to iR4Use the confidence interval to find the estimated margin of error. Then find the sample mean. A store manager reports a confidence interval of (42.6,78.2)when estimating the mean price (in dollars) for the population of textbooks. The estimated margin of error is (Type an integer or a decimal.) The sample mean is (Type an integer or a decimal.)
- Social Networking Sites A recent survey of social networking sites has a mean of million visitors for a specific month. The standard deviation was million. Find the confidence interval of the true mean. Assume the variable is normally distributed. Round your answers to at least two decimal places.Social Networking Sites A recent survey of 6 social networking sites has a mean of 13 million visitors for a specific month. The standard deviation was 3.2 million. Find the 90% confidence interval of the true mean. Assume the variable is normally distributed. Round your answers to at least two decimal places.Private Four-Year Colleges From the data in CollegeScores4yr we know that about 62% of four-year colleges in the US are private institutions. Suppose that we select samples of n = 75 schools at a time from the population of all four-year colleges and find the proportion of private schools in each sample. (a) Consider the distribution of these sample proportions when n = 75 and p = 0.62. What is its form? Center = i Standard error = (round to three decimal places) (b) What proportion of these samples will have less than 50% private schools? Round to three decimal places. istats
- Q7Q12Use the confidence interval to find the estimated margin of error. Then find the sample mean. A biologist reports a confidence interval of (3.0,4.4) when estimating the mean height (in centimeters) of a sample of seedlings. Question content area bottom Part 1 The estimated margin of error is enter your response here .