A simple random sample of 50 adults was surveyed, and it was found that the mean amount of time that they spend surfing the Internet per day is 54.2 minutes, with a standard deviation of 14.0 minutes. What is the 99% confidence interval (z-score = 2.58) for the number of minutes that an adult spends surfing the Internet per day?
50min
99% = mean ± Z SEm
SEm = SD/√n
To find the confidence interval, we can use the formula:
Confidence Interval = sample mean ± (z-score * standard deviation / √sample size)
Given:
Sample mean = 54.2 minutes
Standard deviation = 14.0 minutes
Sample size = 50
z-score for a 99% confidence level is 2.58.
Plugging in the values into the formula, we get:
Confidence Interval = 54.2 ± (2.58 * 14.0 / √50)
Calculating the square root of the sample size (√50):
√50 ≈ 7.071
Now, we can substitute this value into the formula:
Confidence Interval = 54.2 ± (2.58 * 14.0 / 7.071)
Calculating the value inside the parentheses:
2.58 * 14.0 ≈ 36.12
Dividing 36.12 by 7.071:
36.12 / 7.071 ≈ 5.10
Finally, we can substitute this value into the formula:
Confidence Interval = 54.2 ± 5.10
Calculating the lower bound and upper bound:
Lower bound = 54.2 - 5.10 ≈ 49.10
Upper bound = 54.2 + 5.10 ≈ 59.30
Therefore, the 99% confidence interval for the number of minutes that an adult spends surfing the Internet per day is approximately 49.10 to 59.30 minutes.
To calculate the confidence interval for the mean amount of time adults spend surfing the Internet per day, we can use the formula:
Confidence Interval = Mean ± (Z * (Standard Deviation / √Sample Size))
Given that the sample mean is 54.2 minutes, the standard deviation is 14.0 minutes, and the sample size is 50, and using a Z-score of 2.58 for a 99% confidence interval, we can now calculate the confidence interval.
Confidence Interval = 54.2 ± (2.58 * (14.0 / sqrt(50)))
First, we need to calculate the standard error of the mean (SEM), which is the standard deviation divided by the square root of the sample size:
Standard Error of the Mean (SEM) = Standard Deviation / sqrt(Sample Size)
= 14.0 / sqrt(50)
≈ 1.98
Now, we can calculate the confidence interval:
Confidence Interval = 54.2 ± (2.58 * 1.98)
Using this calculation, we get the confidence interval as follows:
Lower Confidence Limit = 54.2 - (2.58 * 1.98)
≈ 54.2 - 5.10
≈ 49.10
Upper Confidence Limit = 54.2 + (2.58 * 1.98)
≈ 54.2 + 5.10
≈ 59.30
Therefore, the 99% confidence interval for the number of minutes that an adult spends surfing the Internet per day is approximately 49.10 minutes to 59.30 minutes.