Answer:
According to the result of the hypothesis test, there is no significant difference between the true population proportion of students who binge drink according to the results of the Professor's survey and the national standard of 0.44.
So, the Professor shouldn't be surprised by this result.
Step-by-step explanation:
- The known standard proportion of students for binge drinking is 44% = 0.44.
- In the Professor's random sample, 96 out of 244 students admitted to binge drinking in the past week.
- So, we need to check if the result of the random sample is in keeping with the known national standard.
For hypothesis testing, the first thing to define is the null and alternative hypothesis.
The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.
While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.
For this question, the null hypothesis is that there is no significant difference between the true population proportion of students who binge drink according to the results of the Professor's survey and the national standard of 0.44.
The alternative hypothesis will now be that there is significant difference between the true population proportion of students who binge drink according to the results of the Professor's survey and the national standard of 0.44.
Mathematically,
The null hypothesis is represented as
H₀: p = 0.44
The alternative hypothesis is given as
Hₐ: p ≠ 0.44
To do this test, we will use the t-distribution because no information on the population standard deviation is known
So, we compute the t-test statistic
t = (x - μ)/σₓ
x = sample proportion of the Professor's poll = (96/244) = 0.393
μ = p₀ = the national standard = 0.44
σₓ = standard error = √[p(1-p)/n]
where n = Sample size = 244
σₓ = √[0.393×0.607/244] = 0.0313
t = (0.393 - 0.44) ÷ 0.0313
t = -1.50
checking the tables for the p-value of this t-statistic
Degree of freedom = df = n - 1 = 244 - 1 = 243
Significance level = 0.05 (most tests are performed at 5% significance level)
The hypothesis test uses a two-tailed condition because we're testing in the two directions.
p-value (for t = -1.50, at 0.05 significance level, df = 243, with a two tailed condition) = 0.134913
The interpretation of p-values is that
When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.
So, for this question, significance level = 0.05
p-value = 0.134913
0.134913 > 0.05
Hence,
p-value > significance level
This means that we fail to reject the null hypothesis & say that there is not enough evidence to conclude that there is a significant difference between the true population proportion of students who binge drink according to the Professor's survey and the national standard.
That is, there is no significant difference between the true population proportion of students who binge drink according to the results of the Professor's survey and the national standard of 0.44.
Hope this Helps!!!
