Question

(CO6) From a random sample of 68 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?
(4.55, 5.25)
(4.82, 4.98)
(4.71, 5.09)
(4.83, 4.97)

2.

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

small mean
large sample size
small standard deviation
large maximum error

3.

(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 37.53 degrees. Are the refrigerators within the 90% confidence interval?

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

4.

(CO6) What is the 97% confidence interval for a sample of 104 soda cans that have a mean amount of 15.10 ounces and a standard deviation of 0.08 ounces?

(15.083, 15.117)
(15.020, 15.180)
(12.033, 12.067)
(15.940, 15.260)

Answer 1

Answer:

(1) (4.82, 4.98)

(2) Large sample size

(3) Yes, the temperature is within the confidence interval of (37.40, 37.60)

(4) (15.083, 15.117)

Step-by-step explanation:

Confidence Interval (CI) = mean + or - (t×sd)/√n

(1) mean = 4.9, sd = 0.35, n = 68, degree of freedom = n-1 = 68 - 1 = 67

t-value corresponding to 67 degrees of freedom and 95% confidence level is 1.9958

CI = 4.9 + (1.9958×0.35)/√68 = 4.98

CI = 4.9 - (1.9958×0.35)/√68 = 4.82

CI is (4.82, 4.98)

(2) Error margin = (t-value × standard deviation)/√sample size

From the formula above, error margin varies inversely as the square root of the sample size. Since the relationship between the error margin and sample size is inverse, increase in one (sample size) will conversely lead to a decrease in the other (error margin)

(3) mean = 37.5, sd = 0.6, n= 100, degree of freedom = n-1 = 100-1 = 99

t-value corresponding to 99 degrees of freedom and 90% confidence level is 1.6602

CI = 37.5 + (1.6602×0.6)/√100 = 37.5 + 0.10 = 37.60

CI = 37.5 - (1.6602×0.6)/√100 = 37.5 - 0.10 = 37.40

37.53 is within the confidence interval (37.40, 37.60)

(4) mean = 15.10, sd =0.08, n = 104, degree of freedom = n-1 = 104-1 = 103

t-value corresponding to 103 degrees of freedom and 97% confidence interval is 2.2006

CI = 15.10 + (2.2006×0.08)/√104 = 15.10 + 0.017 = 15.117

CI = 15.10 - (2.2006×0.08)/√104 = 15.10 - 0.017 = 15.083

CI is (15.083, 15.117)