Question

According to a publication, 14.2% of 18 to 25-year-olds were users of marijuana in 2000. A recent poll of 1201 randomly selected 18 to 25-year-olds revealed that 186 currently use marijuana. At the 10% significance level, do the data provide sufficient evidence to conclude that the percentage of 18 to 25-year-olds who currently use marijuana has changed from the 2000 percentage of 14.2%? Use the one-proportion z-test to perform the appropriate hypothesis test, after checking the conditions for the procedure. What are the hypotheses for the one-proportion z-test? Ho:p= ; Ha:p (Type integers or decimals.) What is the test statistic? Z= (Round to two decimal places as needed.) Identify the critical value(s). (Round to three decimal places as needed.) What is the correct conclusion for the hypothesis test? A. Do not reject Ho; the data do provide sufficient evidence to conclude that the percentage who currently use marijuana has changed from 14.2%. B. Reject Ho; the data do provide sufficient evidence to conclude that the percentage who currently use marijuana has changed from 14.2%. C. Reject Ho; the data do not provide sufficient evidence to conclude that the percentage who currently use marijuana has changed from 14.2%. D. Do not reject Ho; the data do not provide sufficient evidence to conclude that the percentage who currently use marijuana has changed from 14.2%.

Answer 1

We frame the Hypotheis

:

H: P = 14.2%

$H_{1}: P \neq 14.2\%$

To test the we use the Z - Statistic

он

Z |P - P S.E(p)

Therefore the Population Proportion is P = 14.2% = 0.142

Given Sample Size = n = 1201

x = 186

we know that the sample proportion is

2 P n

$p = \frac{186}{1201}$

$\Rightarrow p =0.1549$

we know that

p+ 9 =

$\Rightarrow q= 1-p$

$\Rightarrow q= 1-0.1549$

$\Rightarrow q= 0.8451$

$S.E(p) = \sqrt{\frac{pq}{n}}$

$S.E(p) = \sqrt{\frac{0.1549*0.8451}{1201}}$

$\Rightarrow S.E(p) = \sqrt{0.0001}$

$\Rightarrow S.E(p) = 0.0104$

Therfore

Z |P - P S.E(p)

$\Rightarrow Z = \frac{\left | 0.1549 - 0.142 \right |}{0.0104}$

$\Rightarrow Z = \frac{\left | 0.0129 \right |}{0.0104}$

$\therefore \, \, \, Z_{cal} = 1.2404$

Given Level of Siginificance = 10% = 0.10

$Z_{cri} = Z_{\alpha } = Z_{10\%} = Z_{0.10} = 1.645$

EXCEL COMMAND TO GET CRITICAL VALUE OF Z:

=NORM.S.INV(0.1/2)

Since So, We Accept
он
at 10% Level of Significant.

Therefore we conclude that

Therefore " Do not reject ; the data do provide sufficient evidence to conclude that the percentage who currently use marijuana has changed from 14.2%

H: P = 14.2%