A simple random sample of 10 households, the number of TV's that each household had is as follows:
1, 3, 1, 2, 0, 2, 1, 1, 1, 5
Assume that it is reasonable to believe that the population is approximately normal and the population standard deviation is 0.56. What is the lower bound of the 95% confidence interval for the mean number of TV's?
Mean = 17/10 = 1.7
(1.7 -1.96 * .56 /sqrt(10)), 1.7 +1.96 * .56 /sqrt(10))
To find the lower bound of the 95% confidence interval for the mean number of TVs, we can use the formula:
Lower bound = Sample mean - Margin of error
1. Calculate the sample mean:
- Add up all the observations: 1 + 3 + 1 + 2 + 0 + 2 + 1 + 1 + 1 + 5 = 17
- Divide the sum by the number of observations: 17 / 10 = 1.7
2. Calculate the margin of error:
- Since the population standard deviation is known (0.56), we can use the formula for the margin of error in the case of a known standard deviation:
Margin of error = (Z * population standard deviation) / square root of sample size
- In this case, we want a 95% confidence interval, so the Z-value for a 95% confidence level is 1.96.
- The sample size is 10, so the square root of the sample size is √10 ≈ 3.16
- Plug in the values: (1.96 * 0.56) / 3.16 ≈ 0.347
3. Calculate the lower bound:
- Lower bound = Sample mean - Margin of error
- Substitute the values: 1.7 - 0.347 ≈ 1.353
Therefore, the lower bound of the 95% confidence interval for the mean number of TVs is approximately 1.353.