Question

A dietitian wants to know the average time spent on breakfast in a primary school. The dietitian randomly samples 25 students and finds that the average is 12.6 minutes with a standard deviation of 1.53 minutes. Assume that the distribution of the time spent on the breakfast is normally distributed. The dietitian finds a 95% confidence interval for this sample is (11.968, 13.232).

Answer 1

Answer:

The right options are:

a. The margin of error is 0.632

f. There is a 95% chance that the true mean is between 11.968 and 13.232 minutes.

Step-by-step explanation:

The question is incomplete:

Select one or more:

a. The margin of error is 0.632

b. The margin of error is 1264.

C. The margin of error is 0.600.

d. We believe that the true mean time spent on breakfast is between 11.968 and 13.232 minutes.

e. If we take many other samples from this population, 95% of them will have a sample mean that is between 11968 and 13232.

f. There is a 95% chance that the true mean is between 11.968 and 13.232 minutes.

The margin of error can be calculated from the bounds of the confidence interval as:

$E=\dfrac{UL-LL}{2}=\dfrac{13.232-11.968}{2}=\dfrac{1.264}{2}=0.632$

As the confidence interval is calculated at 95% confidence, we are 95% confident that the true mean is within the calculated interval. We are not 100% confident although (that's why Option d is false).

This interval does not give us information about the sampling distribution (option e is false).