Statistics 1 Final Examination ECON2140 L01

ECON2140 L01代写 A consumer testing service wants to study the noise level of a new vacuum cleaner. Measuring the noise level of a random sample of

1. ECON2140 L01代写

A consumer testing service wants to study the noise level of a new vacuum cleaner. Measuring the noise level of a random sample of n=12 machines, it gets the following data in decibels: 74, 79, 77, 76, 74, 76, 77, 76, 74, 71, 80, 75. Construct a 95% confidence interval for the average noise level of such vacuum cleaners.  

2.  ECON2140 L01代写

The mean combined score (Verbal + Math) for the SAT is 1000 and the standard deviation is 200. If 1000 high school seniors take the SAT, how many should have a score between 1100 and 1300? Assume a normal distribution among the applicants.

3.

It is desired to estimate the average number miles that a radial automobile tire will be safe to use. If σ= 500 miles for these tires, how large a sample will be needed to assert with a probability of .95 that the sample mean will be off by not more than 100 miles? Round answer up to the nearest whole number.

4. ECON2140 L01代写

A random sample of one hundred couples married in Reno during 1985 indicated that 26 were still married. Others were either single or remarried to someone else. A similar sample was taken in Las Vegas of 60 couples married during 1985 with the result that 20 are still together. Use the α =.05 significance level to test whether the proportion of couples still married is the same in both locations.

5.  ECON2140 L01代写

In a study of the status 225 mortgage applications, some applications have been accepted, some rejected and some pending.  The bank is interested in knowing whether the status is related to the difference between residential and commercial applications. Use a chi-squared contingency test at the α =.05 significance level to test this dependence. Specify the null hypothesis and its alternative and perform the test based on the data below.

Ans.       H_o :

               H_a :

Critical value of  x²=

 x²=

Conclusion: