Determine the t critical value for a two-sided confidence interval in each of the following situations. (Round your answers to three decimal places.) USE SALT (a) Confidence level = 95%, df = 5 (b) Confidence level = 95%, df = 10 (c) Confidence level = 99%, df = 10
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- A 95% confidence interval for a proportion is (0.62, 0.88). From this we can infer: The population proportion is 0.75 The population proportion is equal to the sample proportion O The population proportion is between 0.62 and 0.88 O The population proportion is likely to be between 0.62 and 0.88Virginia wants to estimate the percentage of students who live more than three miles from the school. She wants to create a 95% confidence interval which has a margin of error of at most 5%. How many students should be polled to create the confidence interval? z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576 Use the table of values above.IMR in Singapore. The infant mortality rate (IMR) is the number of infant deaths per 1000 live births. Suppose that you have been commissioned to estimate the IMR in Singapore. From a random sample of 1109 live births in Singapore, you find that 0.361% of them resulted in infant deaths. You next find a 90% confidence interval: or 0.000647 to 0.00657. You then conclude, “I can be 90% confident that the IMR in Singapore is somewhere between 0.647 and 6.57.” How did you do?
- If n = 20, the sample mean = 250, SD = 15, SEM = 3.354, what is the confidence interval (using 95% confidence level). What percent confidence do you have that the population mean is between what two valuesIn a survey, 23 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $47 and standard deviation of $9. Construct a confidence interval at a 80% confidence level.Give your answers to one decimal place.________ ± ________Interpret your confidence interval in the context of this problem.Tina catches a 14-pound bass. She does not know the population mean or standard deviation. So she takes a sample of five friends and they say the last bass they caught was 9, 12, 13, 10, and 10 pounds. Find the t and calculate a 95% (α = .05) confidence interval.
- During the 2009 holiday season, 436 randomly sampled American adults were surveyed regarding their average spending on Black Friday. A 95% confidence interval based on this sample is ($80.31,$89.11). (c) Determine the sample mean x and the sample standard deviation s of the sample. (d) Which would be the confidence level corresponding to an interval ($81.27,$88.15) for the data found in (c)? (e) If the values of the sample mean and sample standard deviation found in (c) do not change, find the sample size N for which a 99% confidence interval would have a margin of error smaller than 5.How to calculate a 90% confidence interval?Beth wants to determine a 99% confidence interval for the true proportion of high school students in the area who attend their home basketball games. How large of a sample must she have to get a margin of error less than 0.04? Assume we have no prior estimate of the proportion and want a conservative choice for the sample size. n =
- #34a researcher wishes to develop an interval estimate for response scores on user surveys. She's not sure about the standard deviation, but in previous years the scores have ranged from 25 to 86. How large a sample should she draw if she wants to construct a 95% confidence interval level that will be within 3 points of the population mean score?