Question

An environmental scientist wants to use a 95% confidence interval to estimate the mean number of hours per day her solar panel receives direct sunlight. She observed the panel for 48 randomly selected days and found that the solar panel received an average of 7.1 hours of sunlight per day and the standard deviation was 2.3 hours. a) Compute the 95% confidence interval for the mean. b) Explain your confidence interval from part (a) in context.

Answer 1

Answer:

a) The 95% confidence interval for the mean number of hours per day her solar panel receives direct sunlight is between 2.5 and 11.7.

b) The confidence interval is built from a sample of 48 days. However, we can be 95% sure that for all days, the mean number of hours per day her solar panel receives direct sunlight is between 2.5 and 11.7.

Step-by-step explanation:

We have the standard deviation for the sample. So we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 48 - 1 = 47

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 47 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$. So we have T = 2.0117

The margin of error is:

M = T*s = 2.0117*2.3 = 4.6

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 7.1 - 4.6 = 2.5 hours of sunlight per day

The upper end of the interval is the sample mean added to M. So it is 7.1 + 4.6 = 11.7 hours of sunlight per day

a) Compute the 95% confidence interval for the mean.

The 95% confidence interval for the mean number of hours per day her solar panel receives direct sunlight is between 2.5 and 11.7.

b) Explain your confidence interval from part (a) in context.

The confidence interval is built from a sample of 48 days. However, we can be 95% sure that for all days, the mean number of hours per day her solar panel receives direct sunlight is between 2.5 and 11.7.