Respuesta :
Answer:
The null hypothesis will not be rejected at 5% significance level.
Step-by-step explanation:
Consider a Chi-square test for goodness of fit.
The hypothesis can be defined as:
H₀: The observed frequencies are same as the expected frequencies.
Hₐ: The observed frequencies are not same as the expected frequencies.
The test statistic is given as follows:
[tex]\chi^{2}=\sum\limits^{n}_{i=1}\frac{(O_{i}-E_{i})^{2}}{E_{i}}[/tex]
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)
[tex]=\frac{40}{100}\times80=32[/tex]
E (XXX)
[tex]=\frac{40}{100}\times80=32[/tex]
E (Dark night)
[tex]=\frac{20}{100}\times80=16[/tex]
Compute the test statistic as follows:
[tex]\chi^{2}=\sum\limits^{n}_{i=1}\frac{(O_{i}-E_{i})^{2}}{E_{i}}[/tex]
[tex]=\frac{(36-32)^{2}}{32}+\frac{(35-32)^{2}}{32}+\frac{(9-16)^{2}}{16}\\\\=3.844[/tex]
The test statistic value is, 5.382.
The degrees of freedom of the test is:
n - 1 = 3 - 1 = 2
The significance level is, α = 0.05.
Compute the p-value of the test as follows:
p-value = 0.1463
*Use a Chi-square table.
p-value = 0.1463 > α = 0.05.
So, the null hypothesis will not be rejected at 5% significance level.
