Answer:
(a) The test statistic value is, 5.382.
(b) Retain the null hypothesis.
Step-by-step explanation:
A Chi-square test for goodness of fit will be used in this case.
The hypothesis can be defined as:
<em>H₀</em>: The observed frequencies are same as the expected frequencies.
<em>Hₐ</em>: The observed frequencies are not same as the expected frequencies.
The test statistic is given as follows:
![\chi^{2}=\sum\limits^{n}_{i=1}{\frac{(O_{i}-E_{i})^{2}}{E_{i}}}](https://tex.z-dn.net/?f=%5Cchi%5E%7B2%7D%3D%5Csum%5Climits%5E%7Bn%7D_%7Bi%3D1%7D%7B%5Cfrac%7B%28O_%7Bi%7D-E_%7Bi%7D%29%5E%7B2%7D%7D%7BE_%7Bi%7D%7D%7D)
The information provided is:
Observed values:
Half Pint: 36
XXX: 35
Dark Night: 9
TOTAL: 80
The expected proportions are:
Half Pint: 40%
XXX: 40%
Dark Night: 20%
Compute the expected values as follows:
E (Half Pint) ![=\frac{40}{100}\times 80=32](https://tex.z-dn.net/?f=%3D%5Cfrac%7B40%7D%7B100%7D%5Ctimes%2080%3D32)
E (XXX) ![=\frac{40}{100}\times 80=32](https://tex.z-dn.net/?f=%3D%5Cfrac%7B40%7D%7B100%7D%5Ctimes%2080%3D32)
E (Dark night) ![=\frac{20}{100}\times 80=16](https://tex.z-dn.net/?f=%3D%5Cfrac%7B20%7D%7B100%7D%5Ctimes%2080%3D16)
Compute the test statistic as follows:
![\chi^{2}=\sum\limits^{n}_{i=1}{\frac{(O_{i}-E_{i})^{2}}{E_{i}}}](https://tex.z-dn.net/?f=%5Cchi%5E%7B2%7D%3D%5Csum%5Climits%5E%7Bn%7D_%7Bi%3D1%7D%7B%5Cfrac%7B%28O_%7Bi%7D-E_%7Bi%7D%29%5E%7B2%7D%7D%7BE_%7Bi%7D%7D%7D)
![=[\frac{(36-32)^{2}}{32}]+[\frac{(35-32)^{2}}{32}]+[\frac{(9-16)^{2}}{16}]](https://tex.z-dn.net/?f=%3D%5B%5Cfrac%7B%2836-32%29%5E%7B2%7D%7D%7B32%7D%5D%2B%5B%5Cfrac%7B%2835-32%29%5E%7B2%7D%7D%7B32%7D%5D%2B%5B%5Cfrac%7B%289-16%29%5E%7B2%7D%7D%7B16%7D%5D)
![=3.844](https://tex.z-dn.net/?f=%3D3.844)
The test statistic value is, 5.382.
The degrees of freedom of the test is:
<em>n</em> - 1 = 3 - 1 = 2
The significance level is, <em>α</em> = 0.05.
Compute the <em>p</em>-value of the test as follows:
<em>p</em>-value = 0.1463
*Use a Ch-square table.
<em>p</em>-value = 0.1463 > <em>α</em> = 0.05.
So, the null hypothesis will not be rejected at 5% significance level.
Thus, concluding that the production of the premium lagers matches these consumer preferences.