<span>5.378 E9
=</span><span>5.378 x 10^9
= 5,378,000,000</span>
Answer:
The value of the test statistic is 6.642.
Step-by-step explanation:
A Chi-square test for Goodness of fit will be used to determine whether the viewing audience proportions changed.
The hypothesis is defined as:
<em>H₀</em>: The viewing audience proportions has not changed.
<em>Hₐ</em>: The viewing audience proportions has changed.
The test statistic is:
![\chi^{2}=\sum \frac{(O-E)^{2}}{E}](https://tex.z-dn.net/?f=%5Cchi%5E%7B2%7D%3D%5Csum%20%5Cfrac%7B%28O-E%29%5E%7B2%7D%7D%7BE%7D)
Here,
O = observed frequencies (O - E)²
E = expected frequencies = n × p
The table below shows the observed and expected frequencies.
Compute the test statistic value as follows:
![\chi^{2}=\sum \frac{(O-E)^{2}}{E}\\=0.011+1.282+4.379+0.970\\=6.642](https://tex.z-dn.net/?f=%5Cchi%5E%7B2%7D%3D%5Csum%20%5Cfrac%7B%28O-E%29%5E%7B2%7D%7D%7BE%7D%5C%5C%3D0.011%2B1.282%2B4.379%2B0.970%5C%5C%3D6.642)
Thus, the value of the test statistic is 6.642.
Browse means to glance at, for example when looking at a magazine you most likely skim or glance at the small paragraphs next to the big pictures.
Answer:
Second Option
Step-by-step explanation:
(pythagorean theorem)
![PQ^2 =74](https://tex.z-dn.net/?f=PQ%5E2%20%3D74)
![PQ = \sqrt{74}](https://tex.z-dn.net/?f=PQ%20%3D%20%5Csqrt%7B74%7D)