In: Statistics and Probability
Solution :
The null and alternative hypotheses are as follows :
H0 : p = 0.75 i.e. The proportion of all college student who support cracking down on underage drinking is 0.75 .
H1 : p < 0.75 i.e. The proportion of all college student who support cracking down on underage drinking is less than 0.75.
To test the hypothesis we shall use z-test for single proportion. The test statistic is given as follows :
Where, p̂ is sample proportion, p is hypothesized value of population proportion under H0, q = 1 - p and n is sample size.
Sample proportion of college student who supported cracking down on underage drinking is,
p = 0.75, q = 1 - 0.75 = 0.25 and n = 2100
The value of the test statistic is -2.8221.
Since, our test is left-tailed test, therefore we shall obtain left-tailed p-value for the test statistic. The left-tailed p-value is given as follows :
p-value = P(Z < value of the test statistic)
p-value = P(Z < -2.8221)
p-value = 0.0024
The p-value is 0.0024.
We make decision rule as follows :
If p-value is less than the significance level, then we shall reject the null hypothesis (H0) at given significance level.
If p-value is greater than the significance level, then we shall be fail to reject the null hypothesis (H0) at given significance level.
In our question, significance level is not given. Generally significance level of 0.01 or 0.05 is used. We shall use significance level of 0.01.
Significance level (α) = 0.01
(0.0024 < 0.01)
Since, p-value is less than the significance level of 0.01, therefore we shall reject the null hypothesis (H0) at 0.01 significance level.
Conclusion : At 0.01 significance level, there is sufficient evidence to conclude that the less than 75% of all college student support cracking down on underage drinking.
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