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# I appreciate you taking the time to answer my question today, please answer as many as...

I appreciate you taking the time to answer my question today, please answer as many as you can, thank you.

Question 1 2 pts

(CO6) From a random sample of 55 businesses, it is found that the mean time that employees spend on personal issues each week is 4.9 hours with a standard deviation of 0.35 hours. What is the 95% confidence interval for the amount of time spent on personal issues?

 (4.84, 4.96)
 (4.83, 4.97)
 (4.81, 4.99)
 (4.82, 4.98)

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Question 2 2 pts

(CO6) If a confidence interval is given from 8.54 to 10.21 and the mean is known to be 9.375, what is the margin of error?

 8.54
 0.835
 1.67
 0.418

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Question 3 2 pts

(CO6) Which of the following is most likely to lead to a small margin of error?

 small standard deviation
 large margin of error
 small mean
 large sample size

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Question 4 2 pts

(CO6) From a random sample of 41 teens, it is found that on average they spend 31.8 hours each week online with a standard deviation of 5.91 hours. What is the 90% confidence interval for the amount of time they spend online each week?

 (30.62, 32.99)
 (25.89, 37.71)
 (29.99, 33.61)
 (30.28, 33.32)

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Question 5 2 pts

(CO6) A company making refrigerators strives for the internal temperature to have a mean of 37.5 degrees with a standard deviation of 0.6 degrees, based on samples of 100. A sample of 100 refrigerators have an average temperature of 37.70 degrees. Are the refrigerators within the 90% confidence interval?

 No, the temperature is outside the confidence interval of (36.90, 38.10)
 No, the temperature is outside the confidence interval of (37.40, 37.60)
 Yes, the temperature is within the confidence interval of (37.40, 37.60)
 Yes, the temperature is within the confidence interval of (36.90, 38.10)

1)

Solution :

Given that,

Point estimate = sample mean = = 4.9

Population standard deviation = = 0.35

Sample size = n = 55

At 95% confidence level the z is , = 1 - 95% = 1 - 0.95 = 0.05 / 2 = 0.05 / 2 = 0.025

Z /2 = Z0.025 = 1.96

Margin of error = E = Z /2* ( / n)

= 1.96 * (0.35 / 55)

= 0.09

At 95% confidence interval estimate of the population mean is, - E < < + E

4.9 - 0.09 < < 4.9 + 0.09

4.81 < < 4.99

(4.81 , 4.99 )

2)

Given that,

Lower confidence interval = 8.54

Upper confidence interval = 10.21 = 9.375

Margin of error = E = Upper confidence interval - = 10.21 - 9.375 = 0.835

Margin of error = 0.835

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