1. YOU WORK FOR A CONSUMER ADVOCATE AGENCY AND WANT TO FIND THE MEAN REPAIR COST OF A WASHING MACHINE. AS PART OF YOUR STUDY, YOU RANDOMLY SELECT 55 REPAIR COSTS AND FIND THE MEAN TO BE $131.00. THE SAMPLE STANDARD DEVIATION IS $17.00. COMPLETE PARTS (a) and (b).
(a) Construct a 905 confidence interval for the population mean repair cost.
The 90% confidence interval is (_______, ______). ROUND TO TWO DECIMAL PLACES AS NEEDED.)
(b) CHANGE THE SAMPLE SIZE TO N=110. CONSTRUCT A 90% CONFIDENCE INTERVAL FOR THE POLULATION MEAN REPAIR COST.
The 90% confidence interval is (________, ________.) ROUND TO TWO DECIMAL PLACES AS NEEDED).
Which confidence interval is wider? Explain. Choose the correct answer below.
A. The n=110 confidence interval is wider because a larger sample is taken, giving more information about the population.
B. The n=55 confidence interval is wider because a smaller sample is taken, giving less information about the population.
C. The two intervals are the same size because the confidence interval is based on the level of confidence and sample standard deviation.
2. You randomly select and measure the contents of 10 bottles of cough syrup. The results (in fluid ounces) are show below.
4.215 4.292 4.258 4.243 4.187
4.238 4.261 4. 245 4.225 4.236
(a) THE CONFIDENCE INTERVAL FOR THE POPULATION VARIANCE IS (______,________.) ROUND TO SIX DECIMAL PLACES AS NEEDED.
(b) THE CONFIDENCE INTERVAL FOR THE POPULATION STNADARD DEVIATION IS (_________,_________.) ROUND TO FOUR DECIMAL PLACES AS NEEDED.
Step by step method for computing confidence interval for mean is given in the answer.