A sports news station wanted to know whether people who live in the North or the South are bigger sports fans. For its study, 124 randomly selected Southerners were surveyed and found to watch a mean of 3.9 hours of sports per week. In the North, 122 randomly selected people were surveyed and found to watch a mean of 4.5 hours of sports per week. Find a $90 \%$ confidence interval for the true difference between the mean numbers of hours of sports watched per week for the two regions if the South has a population standard deviation of 1.7 hours per week and the North has a population standard deviation of 1.4 hours per week. Let Population 1 be people who live in the South and Population 2 be people who live in the North. Round the endpoints of the interval to one decimal place, if necessary.
Answer (2)
The 90% confidence interval for the difference in mean hours of sports watched per week between Southerners and Northerners is (0.3, 0.9). This indicates that we are 90% confident the true difference in mean hours lies within this range. The calculation involved using the means, standard deviations, sample sizes, and a critical value from the z-distribution.
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Calculate the difference in sample means: 4.5 − 3.9 = 0.6 .
Find the critical value for a 90% confidence level: z 0.05 = 1.645 .
Calculate the standard error: SE ≈ 0.1985 .
Determine the confidence interval: ( 0.6 − 1.645 × 0.1985 , 0.6 + 1.645 × 0.1985 ) ≈ ( 0.3 , 0.9 ) .
L o w er E n d p o in t = 0.3 U pp er E n d p o in t = 0.9
Explanation
Understand the problem and provided data We are asked to find a 90% confidence interval for the difference in mean hours of sports watched per week between Southerners and Northerners. We are given the sample means, population standard deviations, and sample sizes for both groups.
Define the variables Let's define the following:
Population 1: South Population 2: North Sample size of Southerners: n 1 = 124 Sample size of Northerners: n 2 = 122 Sample mean of Southerners: x ˉ 1 = 3.9 hours Sample mean of Northerners: x ˉ 2 = 4.5 hours Population standard deviation of Southerners: σ 1 = 1.7 hours Population standard deviation of Northerners: σ 2 = 1.4 hours Confidence level: 90% = 0.90
Calculate the difference of means First, calculate the point estimate of the difference between the two population means: x ˉ 2 − x ˉ 1 = 4.5 − 3.9 = 0.6
Find the critical value Next, we need to find the critical value z α /2 for a 90% confidence level. Since it is a two-tailed test, α = 1 − 0.90 = 0.10 , so α /2 = 0.05 . The z-score such that the area to the right is 0.05 is z 0.05 = 1.645 .
Calculate the standard error Now, calculate the standard error of the difference between the two population means: SE = n 1 σ 1 2 + n 2 σ 2 2 = 124 1. 7 2 + 122 1. 4 2 = 124 2.89 + 122 1.96 ≈ 0.0233 + 0.0161 = 0.0394 ≈ 0.1985
Calculate the margin of error Calculate the margin of error: ME = z α /2 × SE = 1.645 × 0.1985 ≈ 0.3265
Calculate the lower endpoint Calculate the lower endpoint of the confidence interval: ( x ˉ 2 − x ˉ 1 ) − ME = 0.6 − 0.3265 ≈ 0.2735 Round to one decimal place: 0.3
Calculate the upper endpoint Calculate the upper endpoint of the confidence interval: ( x ˉ 2 − x ˉ 1 ) + ME = 0.6 + 0.3265 ≈ 0.9265 Round to one decimal place: 0.9
State the final answer Therefore, the 90% confidence interval for the true difference between the mean numbers of hours of sports watched per week for the two regions is ( 0.3 , 0.9 ) .
Examples
Confidence intervals are used in sports analysis to estimate the range within which the true difference in performance metrics (like average game scores or player statistics) between two teams or groups of athletes likely falls. For example, a sports analyst might want to determine if there's a significant difference in the average points scored per game between two basketball teams. By calculating a confidence interval, they can state with a certain level of confidence (e.g., 95%) that the true difference in average scores lies within a specific range. This helps in making informed decisions about team strategies, player evaluations, and predictions.
