Answer: the correct option is A
Step-by-step explanation:
We want to find 95% confidence interval for the mean of the weight of pretzels.
Number of samples. n = 15 bags
Mean, u = 10.2
Standard deviation, s = 0.25
We will use the t- test
Degree of freedom = n - 1 = 15 - 1= 14
Alpha, a = (1-confidence interval )/2
a = (1-0.95)/2 = 0.025
Looking at the t-distribution table, the corresponding z value is 2.131
Confidence interval = z × standard deviation/√n
Confidence interval = 2.131 × 0.25/√15
Confidence interval = 0.13755545851
Approximately 0.138
At 95% confidence interval,
The lower end is 10.2 - 0.138 = 10.062 Approximately 10.06
The upper end is 10.2 - 0.138 = 10.338. Approximately 10.34
