In a recent study of 35 freshman, the mean number of hours per week that they played video games was 16.6. Assume the population standard deviation is 2.8. Determine the 99% confidence interval. Round your values to the nearest thousandth.

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In a recent study of 35 freshman, the mean number of hours per week that they played video games was 16.6. Assume the population standard deviation is 2.8. Determine the  99% confidence interval. Round your values to the nearest thousandth.

Bytelearn · Accepted Answer

Calculate z-score: To calculate the 99% confidence interval for the mean, we need to use the formula for the confidence interval of the mean when the population standard deviation is known: CI = x̄ ± (z * (σ/√n)) where x̄ is the sample mean, z is the z-score corresponding to the confidence level, σ is the population standard deviation, and n is the sample size. First, we need to find the z-score for a 99% confidence level.Find z-score for 99% confidence: We can look up the z-score for a 99% confidence level in a standard normal distribution table or use a calculator that provides this functionality. The z-score that corresponds to a 99% confidence level is approximately 2.576.Calculate margin of error: Now we have all the values needed to calculate the confidence interval: x̄ = 16.6 (sample mean) z = 2.576 (z-score for 99% confidence) σ = 2.8 (population standard deviation) n = 35 (sample size) Let's calculate the margin of error (ME) using the formula: ME = z * (σ/√n)Calculate standard error: First, calculate the standard error (σ/√n): Standard error = σ/√n = 2.8/√35 Standard error ≈ 2.8/5.9161 Standard error ≈ 0.4732 Now, round this to the nearest thousandth. Standard error ≈ 0.473Calculate margin of error: Next, calculate the margin of error (ME): ME = z * Standard error ME = 2.576 * 0.473 ME ≈ 1.2179 Now, round this to the nearest thousandth. ME ≈ 1.218Calculate confidence interval: Finally, calculate the confidence interval using the margin of error: Lower limit = x̄ - ME = 16.6 - 1.218 Lower limit ≈ 15.382 Upper limit = x̄ + ME = 16.6 + 1.218 Upper limit ≈ 17.818 Now, round both values to the nearest thousandth. Lower limit ≈ 15.382 Upper limit ≈ 17.818

In a recent study of $35$ freshman, the mean number of hours per week that they played video games was $16$ . $6$ . Assume the population standard deviation is $2$ . $8$ . Determine the $99 \%$ confidence interval. Round your values to the nearest thousandth.

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In a recent study of $35$ freshman, the mean number of hours per week that they played video games was $16$ . $6$ . Assume the population standard deviation is $2$ . $8$ . Determine the $99 \%$ confidence interval. Round your values to the nearest thousandth.