Survey

Survey

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1. You took a random sample of 12 two-slice toasters and found the mean price was $57.79 and the standard deviation was $19.05. Assume the prices are normally distributed. When the sample size is small (e.g. less than 30), and if we can assume the that sample is drawn for a normal population, with independent sampling it is appropriate to construct a 95% confidence interval using the T-distribution. The problem above describes this situation, so we can use the T-distribution. Using the T-distribution, the formula for a 100(1- )% confidence interval for a normal population mean is given by: Where is the upper point of the t distribution (which can be determined by a t distribution table) with degrees of freedom equal to n – 1, is the sample size, is the standard deviation, and is the sample mean. From the problem, we know that , and Since we are constructing . Further, from a t distribution table, we see that a 95% confidence interval, = with n-1=12-1=11 degrees of freedom. Substituting these values into the formula above gives: So the 95% confidence interval is . 2.In a random sample of 70 bolts, the mean length was 1.25 inches and the standard deviation was 0.05 inches. When the sample size is large (e.g. larger than 30), we can use the Z-distribution to estimate the 95% confidence intervals. Since this sample has a large sample of size n=70, we can use the Z-distribution to form the confidence interval using the following formula: Where is the upper distribution table), point of the Z distribution (which can be determined by a normal is the sample size, is the standard deviation, and is the sample mean. From the problem, we know that , and Since we are constructing a 95% confidence interval, . Further, from a normal distribution table, we see that = . Substituting these values into the formula above gives: So a 95% confidence interval is 3.In a random sample of 18 one-person tents, the mean price was $144.19 and the standard deviation was $61.32. Assume the prices are normally distributed. When the sample size is small (e.g. less than 30), and if we can assume the that sample is drawn for a normal population, with independent sampling it is appropriate to construct a 95% confidence interval using the T-distribution. The problem above describes this situation, so we can use the T-distribution. Using the T-distribution, the formula for a 100(1- )% confidence interval for a normal population mean is given by: Where is the upper point of the t distribution (which can be determined by a t distribution table) with degrees of freedom equal to n – 1, is the sample size, is the standard deviation, and is the sample mean. , and Since we are From the problem, we know that constructing a 95% confidence interval, . Further, from a t distribution table, we see = with n-1=18-1=17 degrees of freedom. Substituting these values that into the formula above gives: So the 95% confidence interval is