5-A random sample of 144 observations has a mean of 20, a median of 21, and a mo

Question

5-A random sample of 144 observations has a mean of 20, a median of 21, and a mode of 22. The population standard deviation is known to equal 3.6. The 80% confidence interval for the population mean is 2-A random sample of 144 observations has a mean of 20, a median of 21, and a mode of 22, and a standard deviation of 3.6. The 95% confidence interval for the population mean is 3-A random sample of 64 students at a university showed an average age of 20 years and a sample standard deviation of 4 years. The 95% confidence interval for the true average age of all students in the university is

Explanation / Answer

5- A random sample of 144 observations has a mean of 20, a median of 21, and a mode of 22. The population standard deviation is known to equal 3.6. The 80% confidence interval for the population mean is

Confidence Intervals for a population mean (n > 30):

For large random samples a confidence interval for a population mean is given by

sample mean ±z * s /n

where sample mean = 20

s = standard deviation of population = 3.6

n = sample size = 144

where z = 1.28 is a multiplier at 80% level of confidence

confidence interval for a population mean = 20 + 1.28 * 3.6 /144 to 20 - 1.28 * 3.6 /144

                                                                             = 20 +0.384 to 20 – 0.384 = 20.384 to 19.616

2- A random sample of 144 observations has a mean of 20, a median of 21, and a mode of 22. The population standard deviation is known to equal 3.6. The 95% confidence interval for the population mean is

Confidence Intervals for a population mean (n > 30):

For large random samples a confidence interval for a population mean is given by

sample mean ±z * s /n

where sample mean = 20

s = standard deviation of population = 3.6

n = sample size = 144

where z = 1.96 is a multiplier at 95% level of confidence

confidence interval for a population mean = 20 + 1.96 * 3.6 /144 to 20 - 1.96 * 3.6 /144

                                                                             = 20 +0.588 to 20 – 0.588 = 20.588 to 19.412

3-A random sample of 64 students at a university showed an average age of 20 years and a sample standard deviation of 4 years. The 95% confidence interval for the true average age of all students in the university is

Confidence Intervals for a population mean (n > 30):

For large random samples a confidence interval for a population mean is given by

sample mean ±z * s /n

where sample mean = 20

s = standard deviation of population = 4

n = sample size = 64

where z = 1.96 is a multiplier at 95% level of confidence

confidence interval for the true average age of all students in the university is

= 20 + 1.96 * 4 /64 to 20 - 1.96 * 4 /64

= 20 +0.98 to 20 – 0.98 = 20.98 to 19.02

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