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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Q. 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.

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: $\newline$ CI = $\bar{x} \pm (z \times (\sigma/\sqrt{n}))$ $\newline$ where $\bar{x}$ is the sample mean, $z$ is the z-score corresponding to the confidence level, $\sigma$ is the population standard deviation, and $n$ is the sample size. $\newline$ 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: $\newline$ $\bar{x} = 16.6$ (sample mean) $\newline$ $z = 2.576$ (z-score for $99$ % confidence) $\newline$ $\sigma = 2.8$ (population standard deviation) $\newline$ $n = 35$ (sample size) $\newline$ Let's calculate the margin of error (ME) using the formula: $\newline$ $ME = z \times (\sigma/\sqrt{n})$
Calculate standard error: First, calculate the standard error $\sigma/\sqrt{n}$ : $\newline$ Standard error = $\sigma/\sqrt{n} = 2.8/\sqrt{35}$ $\newline$ Standard error $\approx 2.8/5.9161$ $\newline$ Standard error $\approx 0.4732$ $\newline$ Now, round this to the nearest thousandth. $\newline$ Standard error $\approx 0.473$
Calculate margin of error: Next, calculate the margin of error (ME): $\newline$ $ME = z \times \text{Standard error}$ $\newline$ $ME = 2.576 \times 0.473$ $\newline$ $ME \approx 1.2179$ $\newline$ Now, round this to the nearest thousandth. $\newline$ $ME \approx 1.218$
Calculate confidence interval: Finally, calculate the confidence interval using the margin of error: $\newline$ Lower limit = $\bar{x} - \text{ME} = 16.6 - 1.218$ $\newline$ Lower limit $\approx 15.382$ $\newline$ Upper limit = $\bar{x} + \text{ME} = 16.6 + 1.218$ $\newline$ Upper limit $\approx 17.818$ $\newline$ Now, round both values to the nearest thousandth. $\newline$ Lower limit $\approx 15.382$ $\newline$ Upper limit $\approx 17.818$