The sugar content of the syrup in canned peaches is normally distributed. A random sample of n = 10 cans yields a sample standard deviation of s = 4.8 milligrams. A 95% two-sided confidence interval for σ² is closest to
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- The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.903 g and a standard deviation of 0.308 g. The company that produces these cigarettes claims that it has now reduced the amount of nicotine. The supporting evidence consists of a sample of 48 cigarettes with a mean nicotine amount of 0.832 g. Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly selecting 48 cigarettes with a mean of 0.832 g or less. P(M < 0.832 g) = = Enter your answer as a number accurate to 4 decimal places. NOTE: Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.A sample of 60 obese adults was put on a low carbohydrate diet for a year. The average weight loss was 11 Ibs. and the standard deviation was 9 Ibs. Calculate a 99% lower confidence bound for the true average weight loss. O 8.3 O -8.8 O -10.1 O 8.9 o o oIn a random sample of 20 people, the mean commute time to work was 33.9 minutes and the standard deviation was 7.2 minutes. Assume the population is normally distributed and use a t-distribution to construct a 99% confidence interval for the population mean μ. What is the margin of error of μ? Interpret the results.
- Suppose a randomly selected sample of n = 62 men has a mean foot length of x = 28 cm, and the standard deviation of the sample is 2 cm. Calculate an approximate 95% confidence interval for the mean foot length of men. (Round the answers to one decimal place.) In the study, n = 45 men were put on a diet. The men who dieted lost an average of 5.6 kg, with a standard deviation of 3.6 kg. (a) Compute the standard error of the mean for the men who dieted. (Round the answer to two decimal places.)kg(b) Compute an approximate 95% confidence interval for the mean weight loss for the men who dieted. (Round your answer to the nearest hundredth.)An approximate 95% confidence interval is to kgThe sugar content of the syrup in canned peaches is normally distributed. A random sample of n = 10 cans yields a sample standard deviation of s = 4.841 milligrams. Construct a 95% two-sided confidence interval for o. Round your answers to 2 decimal places.The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.954 g and a standard deviation of 0.301 g. The company that produces these cigarettes claims that it has now reduced the amount of nicotine. The supporting evidence consists of a sample of 40 cigarettes with a mean nicotine amount of 0.902 g. Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly selecting 40 cigarettes with a mean of 0.902 g or less. P(M < 0.902 g) Enter your answer as a number accurate to 4 decimal places. NOTE: Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted. Based on the result above, ¿is it valid to claim that the amount of nicotine is lower? (Let's use a 5% cut-off for our definition of unusual.) O No. The probability of obtaining this data is high enough to have been a chance occurrence. O Yes. The probability of this data is unlikely to have occurred by chance alone.
- In a random sample of 28 people, the mean commute time to work was 31.3 minutes and the standard deviation was 7.3 minutes. Assume the population is normally distributed and use a t-distribution to construct a 90% confidence interval for the population mean μ. What is the margin of error of μ? Interpret the results.In a random sample of 27 people, the mean commute time to work was 32.5 minutes and the standard deviation was 7.2 minutes. Assume the population is normally distributed and use a t-distribution to construct a 80% confidence interval for the population mean μ. What is the margin of error of μ? Interpret the results.A random sample of 100 households was selected as part of a study on electricity usage, and the number of kilowatt-hours (kWh) was recorded for each household .The average usage was found to be 7.5 kWh. In a very large study in the previous year it was found that the standard deviation of the usage was 2.3 kWh.Assuming the standard deviation is unchanged and that the usage is normally distributed, calculate a 95% confidence interval for the mean usage (round to two decimals)
