Question

In a recent survey of drinking laws, a random sample of 1000 women showed that 65% were in favor of increasing the legal drinking age. In a random sample of 1000 men, 60% favored increasing the legal drinking age. Test the claim that the percentage of men and women favoring a higher legal drinking age is different at (alpha 0.05).

Answer 1

Explanation:

Given that:

Let sample size of women be
`n_1` = 1000

Let the proportion of the women be
`p_1` = 0.65

Let the sample size of the men be
`n_2` = 1000

Let the proportion of the mem be
`p_2` = 0.60

The null and the alternative hypothesis can be computed as follows:

`H_0: p_1 = p_2`

`H_0a: p_1 \\eq p_2`

Thus from the alternative hypothesis we can realize that this is a two tailed test.

However, the pooled sample proportion p =
`\frac{p_1n_1+p_2n_2 } {n_1 +n_2}`

p =
`\frac{0.65 * 1000+0.60*1000 } {1000 +1000}`

p =
`\frac{650+600 } {2000}`

p = 0.625

The standard error of the test can be computed as follows:

`SE = \sqrt{p(1-p) ( \frac{1} {n_1}+ (1)/(n_2) )}`

`SE = \sqrt{0.625(1-0.625) ( \frac{1} {1000}+ (1)/(1000) )}`

`SE = √(0.625(0.375) ( 0.001+0.001 ))`

`SE = √(0.234375 (0.002))`

`SE = \sqrt{4.6875 * 10^(-4)}`

`SE = 0.02165`

The test statistics is :

`z =(p_1-p_2)/(S.E)`

`z =(0.65-0.60)/(0.02165)`

`z =(0.05)/(0.02165)`

`z =2.31`

At level of significance of 0.05 the critical value for the z test will be in the region between - 1.96 and 1.96

Rejection region: To reject the null hypothesis if z < -1.96 or z > 1.96

Conclusion: Since the value of z is greater than 1.96, it lies in the region region. Therefore we reject the null hypothesis and we conclude that the percentage of men and women favoring a higher legal drinking age is different.