The average life of a certain type of motor is 10 years with a standard deviatio

Question

The average life of a certain type of motor is 10 years with a standard deviation of 2 years. You are interested in measuring the life time of this motor by conducting a survey asking customers experiences. Use this information and answer Question 1a to 1h. Question 1e: After a month-long survey, you collect 49 responses and find that the average life of the motor is 10.28 years, what is the upper limit of a two-sided 95% confidence interval of the true average life of the motor? (Use four decimal places) Question 1f You would like to narrow the half width of the confidence interval that you have calculated for Question 1e. Which of the following could you do? (Select all that apply) Reduce the significance level Reduce the confidence level Increase the significance level Increase the confidence level Collect more survey responses Question 1g: After a month-long survey, you collected 367 responses. You find the average of the life of the motor is 10.3 years with a standard deviation of 1.9. How many more samples would you need to collect so that a 99.012% confidence interval of the average life of the motor is +/-of 0.1 year? Question 1h After a month-lo the life of the motor is greater than 10 years. You would like to understand what proportion of the motors last longer than 10 years. How many surveys would you need to collect in total in order to obtain a margin of error of +/-3% with 98.02% confidence? ng survey, you collect 49 responses. You found that 94% of the customers indicated

Explanation / Answer

E) At 95% cinfidence interval the critical value = 1.96

Confidence interval = mean +/- Z*(SD/sqrt(n ))

= 10.28 +/- 1.96 * (2/sqrt (49))

= 10.28 +/- 0.56

= 9.72, 10.84

SO the upper limit is 10.84

F) Option-B is the correct answer.

G) Margin of error = 0.1

Or, Z * SD/sqrt (n) = 0.1

Or, 2.575 * 1.9/sqrt(n) = 0.1

Or, sqrt(n) = 48.925

Or, n = 2393.66 = 2394

The more samples are required = 2394 - 367 = 2027

H) n = 49 * (94/100) = 46.06 = 46

P(X > 10) = P((X - mean)/(SD/sqrt (n) > (10 - 10)/(2/sqrt(46))

= P(Z > 0)

= 1 - P(Z < 0)

= 1 - 0.5 = 0.5

At 98.02% confidence interval the critical value Z* = 2.33

Margin of error = 0.03

Or, Z* * SD/sqrt (n ) = 0.03

Or, 2.33 * 2/sqrt(n) = 0.03

Or, sqrt(n) = 155.33

Or, n = 24127.41 = 24127

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