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)
