Respuesta :
Answer:
Yes, there is enough evidence to support the claim that ionizing radiation is beneficial in terms of marketability.
Step-by-step explanation:
This is a hypothesis test for the difference between proportions.
The claim is that ionizing radiation is beneficial in terms of marketability.
Then, the null and alternative hypothesis are:
[tex]H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2> 0[/tex]
Being π1: the true proportion of treated bulbs that can be comercialized after 240 days, and π2: the true proportion of untreated bulbs that can be comercialized after 240 days.
The significance level is 0.05.
The sample 1, of size n1=180 has a proportion of p1=0.85.
[tex]p_1=X_1/n_1=153/180=0.85[/tex]
The sample 2, of size n2=180 has a proportion of p2=0.65.
[tex]p_2=X_2/n_2=117/180=0.65[/tex]
The difference between proportions is (p1-p2)=0.2.
[tex]p_d=p_1-p_2=0.85-0.65=0.2[/tex]
The pooled proportion, needed to calculate the standard error, is:
[tex]p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{153+117}{180+180}=\dfrac{270}{360}=0.75[/tex]
The estimated standard error of the difference between means is computed using the formula:
[tex]s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.75*0.25}{180}+\dfrac{0.75*0.25}{180}}\\\\\\s_{M_d}=\sqrt{0.001+0.001}=\sqrt{0.002}=0.0456[/tex]
Then, we can calculate the z-statistic as:
[tex]z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{0.2-0}{0.0456}=\dfrac{0.2}{0.0456}=4.382[/tex]
This test is a right-tailed test, so the P-value for this test is calculated as (using a z-table):
[tex]P-value=P(t>4.382)=0.000008[/tex]
As the P-value (0.000008) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
There is enough evidence to support the claim that ionizing radiation is beneficial in terms of marketability.
