Questions: According to a survey of 1,200 adult college students, the mean number of alcoholic beverages consumed per week is 14 , with a standard deviation of 5 beverages. What test statistic is calcuated for this scenario? p -value mean test t-value null test z-score Submit Question

According to a survey of 1,200 adult college students, the mean number of alcoholic beverages consumed per week is 14 , with a standard deviation of 5 beverages. What test statistic is calcuated for this scenario?
p -value
mean test
t-value
null test
z-score
Submit Question

Transcript text: According to a survey of 1,200 adult college students, the mean number of alcoholic beverages consumed per week is 14 , with a standard deviation of 5 beverages. What test statistic is calcuated for this scenario? p -value mean test t-value null test z-score Submit Question

Solution

Solution Steps

Step 1: Calculate the Standard Error

The standard error \( SE \) is calculated using the formula:

\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{5}{\sqrt{1200}} \approx 0.1443 \]

Step 2: Calculate the Test Statistic

The test statistic \( Z_{test} \) is calculated using the formula:

\[ Z_{test} = \frac{\bar{x} - \mu_0}{SE} = \frac{14 - 14}{0.1443} = 0.0 \]

Step 3: Calculate the P-value

For a two-tailed test, the P-value is calculated as:

\[ P = 2 \times (1 - T(|z|)) = 1.0 \]

Final Answer

The test statistic is \( Z_{test} = 0.0 \) and the P-value is \( 1.0 \).

Thus, the final answer is:

\[ \boxed{Z_{test} = 0.0, \, P = 1.0} \]

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