Following are observed frequencies. The null hypothesis is $H _ { 0 } : p _ { 1 } = 0.35 , p _ { 2 } = 0.25$
$p _ { 3 } = 0.2 , p _ { 4 } = 0.1 , p _ { 5 } = 0.1$
$\begin{array} { l | l l l l l }
\text { Category } & 1 & 2 & 3 & 4 & 5 \
\hline \text { Observed } & 49 & 34 & 14 & 9 & 10
\end{array}$
Test the hypothesis that the distribution of the observed frequencies is as given by the nul hypothesis. Use the $\alpha = 0.05$ level of significance.
A) Do not reject $H _ { 0 }$
B) Reject $H _ { 0 }$

Question

Following are observed frequencies. The null hypothesis is $H _ { 0 } : p _ { 1 } = 0.35 , p _ { 2 } = 0.25$
$p _ { 3 } = 0.2 , p _ { 4 } = 0.1 , p _ { 5 } = 0.1$
$\begin{array} { l | l l l l l } 
\text { Category } & 1 & 2 & 3 & 4 & 5 \
\hline \text { Observed } & 49 & 34 & 14 & 9 & 10
\end{array}$
Test the hypothesis that the distribution of the observed frequencies is as given by the nul hypothesis. Use the $\alpha = 0.05$ level of significance.
A) Do not reject $H _ { 0 }$
B) Reject $H _ { 0 }$

Quizplus · Accepted Answer

To test the hypothesis, we calculate the expected frequencies for each category based on the null hypothesis and then use the chi-square test. The expected frequencies are calculated by multiplying the total number of observations by the probability of each category under $H_0$. The total number of observations is $49 + 34 + 14 + 9 + 10 = 116$. Thus, the expected frequencies are:- For category 1: $116 \times 0.35 = 40.6$- For category 2: $116 \times 0.25 = 29$- For category 3: $116 \times 0.2 = 23.2$- For category 4: $116 \times 0.1 = 11.6$- For category 5: $116 \times 0.1 = 11.6$Next, we calculate the chi-square statistic:$$\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}$$where $O_i$ is the observed frequency and $E_i$ is the expected frequency for category $i$. Plugging in the values:$$\chi^2 = \frac{(49-40.6)^2}{40.6} + \frac{(34-29)^2}{29} + \frac{(14-23.2)^2}{23.2} + \frac{(9-11.6)^2}{11.6} + \frac{(10-11.6)^2}{11.6}$$$$\chi^2 \approx 8.91$$The degrees of freedom (df) for the chi-square test is $n - 1$, where $n$ is the number of categories. Here, $df = 5 - 1 = 4$. At the $\alpha = 0.05$ level of significance and with 4 degrees of freedom, the critical value from the chi-square distribution table is approximately 9.49.Since the calculated $\chi^2$ value (8.91) is less than the critical value (9.49), we do not have sufficient evidence to reject the null hypothesis. Therefore, we conclude that the observed frequencies do not significantly differ from the expected frequencies given by the null hypothesis.

Following Are Observed Frequencies H0:p1=0.35,p2=0.25H _ { 0 } : p _ { 1 } = 0.35 , p _ { 2 } = 0.25H0​:p1​=0.35,p2​=0.25

Following Are Observed Frequencies $H _ { 0 } : p _ { 1 } = 0.35 , p _ { 2 } = 0.25$