According to the National lnstitute on Alcohol Abuse and Alcoholism, 41% of college students nationwide engage in "binge drinking" behavior, having 5 or more drinks in one occasion during the past two weeks. A college president wonders if the proportion of students enrolled at her college that binge drink is actually lower than the national proportion.

ln a commissioned study, 348 students are selected randomly from a list of all students enrolled at the college. Of these, 132 admitted to having engaged in binge drinking.

At the 5% level of significance, is there enough evidence to conclude students enrolled at this college that binge drink is lower than the national proportion of students that binge drink?
2) Step 2: Collect the Data. This step means to find the summary data for the sample, and to assess normality, and to find the test statistic.
a) What is the sample proportion? Use 4 decimal places.

b) What is the standard deviation for this problem?

c) Are the normality assumptions met? Show or Explain fully.

d) What is the test statistic? If computing by hand do not used rounded values in your calculation, use all 8 decimals. Then round at the end of the calculation. Alternatively, obtain the test statistic from running the hypothesis test in the calculator. Use 4 decimal places.

Step-by-step explanation:

Null hypothesis: National proportion P ≥ sample proportion

Alternative hypothesis: National proportion P < sample proportion

Collect the Data.

Sample size n = 348, population proportion P = 0.41,

a. What is the sample proportion? Use 4 decimal places.

Sample proportion p = 132/348 = 0.3793

b. What is the standard deviation for this problem?

The standard deviation can be calculated using this formula:

√[(p x (1-p)) / n] = √[(0.3793 x (1 - 0.3793) / 348

= √[(0.3793 x 0.6203) / 348]

 = √(0.2354 / 348)

= √0.0006765

SD = 0.026

c. The conditions are met:

The sampling method is simple random sampling: Yes

Each sample point can result in just two possible outcomes: a success; have engaged in binge drinking and a failure: have not engaged in binge drinking

The sample includes at least 216 failures and 132 successes

The population size is at least 20 times as big as the sample size: college students nationwide are even 20 times more than 348.

d) What is the test statistic?

z = (p - P) / σ

z = (0.3793 - 0.41) / 0.026

z = (-0.0307 / 0.026)

z = -1.18

At a 5% level of significance, where α= 0.05, p value for -1.18 for a one tailed test using the p value calculator is  0.119, the result is not significant at p < .05 thus, concluding that we fail to reject the null as there is not enough evidence to conclude students enrolled at this college that binge drink is lower than the national proportion of students that binge drink

neither i nor ii 56 / 4.9 = 11.91489362 and 82 / (11+5) =5.125 so i is wrong

ii. (24/8+2) times 3 = 15 and (42+9)/2 =25.5

step-by-step explanation:

56

step-by-step explanation: you take each of the numbers and divide by the amount then you take the range