Question

9. A group of statistics students decided to conduct a survey at their university to find the average (mean) amount of time students spent studying per week. Assuming a population standard deviation of three hours, what is the required sample size if the error should be less than a half hour with a 99% level of confidence?

10.University officials say that at least 70% of the voting student population supports a fee increase. If the 95% confidence interval estimating the proportion of students supporting the fee increase is [0.75, 0.85], what conclusion can be drawn?

11.A survey of an urban university (population of 25,450) showed that 870 of 1,100 students sampled supported a fee increase to fund improvements to the student recreation center. Using the 95% level of confidence, what is the confidence interval for the proportion of students supporting the fee increase?

12.A university surveyed recent graduates of the English Department for their starting salaries. Four hundred graduates returned the survey. The average salary was $25,000. The population standard deviation was $2,500. What is the 95% confidence interval for the mean salary of all graduates from the English Department?

13.A local retail company wants to estimate the mean amount spent by customers. Their store's budget limits the number of surveys to 225. What is their maximum error of the estimated mean amount spent for a 99% level of confidence and an estimated standard deviation of $10.00?

14.A bank wishes to estimate the mean credit card balance owed by its customers. The population standard deviation is estimated to be $300. If a 98% confidence interval is used and an interval of $75 is desired, how many customers should be sampled?

Answer 1

Solution:

Question 9)

Given:

a population standard deviation = three hours

that is: hours

c = confidence level = 99%

Margin of Error = E= 0.5 hour

We have to find sample size n:

Z_c is z critical value for c = 0.99 confidence level.

Find Area = ( 1+c)/2 = ( 1 + 0.99 ) / 2 = 1.99 /2 = 0.9950

Thus look in z table for Area = 0.9950 or its closest area and find corresponding z critical value.

From above table we can see area 0.9950 is in between 0.9949 and 0.9951 and both are at same distance from 0.9950, Hence corresponding z values are 2.57 and 2.58

Thus average of both z values is 2.575

Thus Z_c = 2.575

Thus

The required sample size if the error should be less than a half hour with a 99% level of confidence is n = 239