Question

how often do u.s. adults attend live music or theater events in a given year? gallup asked a random sample of 1025 u.s. adults about this and other leisure activities. in the sample, the mean number of live music and theater events attended was 3.8 with a standard deviation of 6.95. calculate a 95% confidence interval for the mean number of live music and theater events u.s. adults attend in a given year. assume the conditions are met. (round each bound to 4 decimal places.) incorrect answer 3.3745 to incorrect answer 4.2255
sorry, thats incorrect. remember... if the correct df isnt listed, use the greatest df available that is less than the correct df.
ounding up\ to a larger df will result in confidence intervals that are too narrow. the intervals wont be wide enough to capture the population value as often as suggested by the confidence level.

Explanation:

Step1: Identify given values

Sample size $n=1025$, sample mean $\bar{x}=3.8$, sample standard deviation $s=6.95$, confidence level $95\%$

Step2: Find t-critical value

Degrees of freedom $df=n-1=1024$. For 95% confidence, $t^* \approx 1.9623$ (using large df approximation or t-table for df=1000, closest smaller df)

Step3: Calculate standard error

$\text{SE} = \frac{s}{\sqrt{n}} = \frac{6.95}{\sqrt{1025}}$
$\text{SE} \approx \frac{6.95}{32.0156} \approx 0.2171$

Step4: Compute margin of error

$ME = t^* \times \text{SE} = 1.9623 \times 0.2171$
$ME \approx 0.4260$

Step5: Calculate confidence interval

Lower bound: $\bar{x} - ME = 3.8 - 0.4260 = 3.3740$
Upper bound: $\bar{x} + ME = 3.8 + 0.4260 = 4.2260$

Answer:

3.3740 to 4.2260