Answer:
Step-by-step explanation:
Hello!
The objective is to test the claim that less than 41% of the publisher's readers own a laptop.
To do so, a sample of 320 readers was taken and the proportion of readers that own a laptop resulted in 36%
Be X: number of readers that own a laptop.
X~Bi(n;p)
n=320
Sample proportion p'= 0.36
The hypotheses are:
H₀: p ≥ 0.41
H₁: p < 0.41
α: 0.05
[tex]Z=\frac{p'-p}{\sqrt{\frac{p(1-p)}{n} } }[/tex]≈N(0;1)
[tex]Z_{H_0}=\frac{0.36-0.41}{\sqrt{\frac{0.41(1-0.41)}{320} } }= -1.82[/tex]
This test is one-tailed to the left, meaning, that you'll reject the null hypothesis to small values of Z, there is only one critical value that defined the rejection region:
[tex]Z_{\alpha }= Z_{0.05}= -1.645[/tex]
The decision rule is:
If [tex]Z_{H_0}[/tex] ≤ -1.645, reject the null hypothesis.
If [tex]Z_{H_0}[/tex] > -1.645, do not reject the null hypothesis.
The calculated value is less than the critical value, then the decision is to reject the null hypothesis.
So at a 5% significance level, the test is significant. You can conclude that the population proportion of the publisher's readers that own a laptop is less than 41%.
I hope this helps!
