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Sagot :
Answer:
The p-value of the test is 0.0018, which means that for a level of significance higher than this, there is sufficient evidence to claim that less than 50% of Internet users get their health questions answered online.
Step-by-step explanation:
Test the claim that less than 50% of Internet users get their health questions answered online.
At the null hypothesis, we test that 50% of Internet users get their health questions answered online, that is:
[tex]H_0: p = 0.5[/tex]
At the alternate hypothesis, we test that this proportion is less than 50%, that is:
[tex]H_a: p < 0.5[/tex]
The test statistic is:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.
0.5 is tested at the null hypothesis:
This means that [tex]\mu = 0.5, \sigma = \sqrt{0.5*0.5} = 0.5[/tex]
A random sample of 1318 Internet users was asked where they will go for information the next time they need information about health or medicine; 606 said that they would use the Internet.
This means that [tex]n = 1318, X = \frac{606}{1318} = 0.4598[/tex]
Test statistic:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{0.4598 - 0.5}{\frac{0.5}{\sqrt{1318}}}[/tex]
[tex]z = -2.92[/tex]
P-value of the test and decision:
The p-value of the test is the probability of finding a sample proportion below 0.4598, which is the p-value of z = -2.92.
Looking at the z-table, z = -2.92 has a p-value of 0.0018.
The p-value of the test is 0.0018, which means that for a level of significance higher than this, there is sufficient evidence to claim that less than 50% of Internet users get their health questions answered online.
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