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:
![H_{0}: p = 0.135](https://tex.z-dn.net/?f=H_%7B0%7D%3A%20p%20%3D%200.135)
And the alternate hypothesis is:
![H_{a}: p \neq 0.135](https://tex.z-dn.net/?f=H_%7Ba%7D%3A%20p%20%5Cneq%200.135)
The test statistic is:
![z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D)
In which X is the sample mean,
is the value tested at the null hypothesis,
is the standard deviation and n is the size of the sample.
0.135 is tested at the null hypothesis:
This means that ![\mu = 0.135, \sigma = \sqrt{0.135*0.865}](https://tex.z-dn.net/?f=%5Cmu%20%3D%200.135%2C%20%5Csigma%20%3D%20%5Csqrt%7B0.135%2A0.865%7D)
Of the 710 infants,121 experienced a loss of appetite.
This means that ![n = 710, X = \frac{121}{710} = 0.1704](https://tex.z-dn.net/?f=n%20%3D%20710%2C%20X%20%3D%20%5Cfrac%7B121%7D%7B710%7D%20%3D%200.1704)
Value of the test-statistic:
![z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D)
![z = \frac{0.1704 - 0.135}{\frac{\sqrt{0.135*0.865}}{\sqrt{710}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7B0.1704%20-%200.135%7D%7B%5Cfrac%7B%5Csqrt%7B0.135%2A0.865%7D%7D%7B%5Csqrt%7B710%7D%7D%7D)
![z = 2.76](https://tex.z-dn.net/?f=z%20%3D%202.76)
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.