Answer :
Answer:
The pvalue of the test is 0.0058 < 0.01, which means that there is significant evidence to conclude the proportion of infants who receive Prevnar and experience of a loss of appetite is different from 0.135.
Step-by-step explanation:
Test if the proportion of infants who receive Prevnar and experience of a loss of appetite is different from 0.135
This means that the null hypothesis is:
[tex]H_{0}: p = 0.135[/tex]
And the alternate hypothesis is:
[tex]H_{a}: p \neq 0.135[/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.
0.135 is tested at the null hypothesis:
This means that [tex]\mu = 0.135, \sigma = \sqrt{0.135*0.865}[/tex]
Of the 710 infants,121 experienced a loss of appetite.
This means that [tex]n = 710, X = \frac{121}{710} = 0.1704[/tex]
Value of the test-statistic:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{0.1704 - 0.135}{\frac{\sqrt{0.135*0.865}}{\sqrt{710}}}[/tex]
[tex]z = 2.76[/tex]
Pvalue of the test and decision:
The pvalue of the test is the pobability that the population proportion differs from the tested proportion by at least 0.1704 - 0.135 = 0.0354, which is P(|z| > 2.76). This probability is 2 multiplied by the pvalue of z = -2.76.
Looking at the z-table, z = -2.76 has a pvalue of 0.0029
2*0.0029 = 0.0058
The pvalue of the test is 0.0058 < 0.01, which means that there is significant evidence to conclude the proportion of infants who receive Prevnar and experience of a loss of appetite is different from 0.135.