To construct a 95% confidence interval, we will use the t-distribution since the population standard deviation is unknown.
First, we need to calculate the sample mean and standard deviation:
Sample mean (x̄) = (2+5+4+6+7+4+2+3+2+3)/10 = 3.8
Sample standard deviation (s) = √[(Σ(x - x̄)²)/(n-1)]
= √[((2-3.8)² + (5-3.8)² + (4-3.8)² + (6-3.8)² + (7-3.8)² + (4-3.8)² + (2-3.8)² + (3-3.8)² + (2-3.8)² + (3-3.8)²)/(10-1))]
= 1.623
Next, we need to determine the t-value for a 95% confidence interval with 9 degrees of freedom (n-1=10-1). From the t-distribution table, the t-value is approximately 2.262.
The confidence interval formula is:
CI = x̄ ± t*(s/√n)
Plugging in the values, we get:
CI = 3.8 ± 2.262*(1.623/√10)
CI = 3.8 ± 1.316
CI = (2.484, 5.116)
Therefore, we can be 95% confident that the true average number of hours per day adults spend in watching television is between 2.484 and 5.116 hours.
The the data show the number of hours per day 12 adults spent infront of screens watching television content and these selected form normal distribution
2 5 4 6 7
4 2 3 2 3
Construct a 95% confidence interval to estimate the average number of hours per day adults spend in watching television.
1 answer