using a 5% level of significance, test the claim that the subject distribution of books in the library fits…

Explanation:

Step1: Identify null and alternative hypotheses

The null hypothesis $H_0$ is that the distributions are the same, and the alternative hypothesis $H_1$ is that the distributions are different.

Step2: Recall significance level

The significance level $\alpha$ is given as 5% or 0.05.

Step3: Calculate expected frequencies

Let the total number of books checked out be $n=299 + 203+211 + 109+66=888$. For Business: $E_1 = 0.32\times888=284.160$ For Humanities: $E_2=0.25\times888 = 222.000$ For Natural Science: $E_3=0.20\times888=177.600$ For Social Science: $E_4=0.15\times888 = 133.200$ For All other subjects: $E_5=0.08\times888=71.040$

Step4: Calculate chi - square statistic

The chi - square statistic $\chi^{2}=\sum\frac{(O - E)^{2}}{E}$, where $O$ is the observed frequency and $E$ is the expected frequency. For Business: $\frac{(299 - 284.160)^{2}}{284.160}=\frac{14.84^{2}}{284.160}=\frac{220.2256}{284.160}\approx0.775$ For Humanities: $\frac{(203 - 222)^{2}}{222}=\frac{(- 19)^{2}}{222}=\frac{361}{222}\approx1.626$ For Natural Science: $\frac{(211 - 177.6)^{2}}{177.6}=\frac{33.4^{2}}{177.6}=\frac{1115.56}{177.6}\approx6.281$ For Social Science: $\frac{(109 - 133.2)^{2}}{133.2}=\frac{(-24.2)^{2}}{133.2}=\frac{585.64}{133.2}\approx4.397$ For All other subjects: $\frac{(66 - 71.040)^{2}}{71.040}=\frac{(-5.04)^{2}}{71.040}=\frac{25.4016}{71.040}\approx0.358$ $\chi^{2}=0.775 + 1.626+6.281+4.397+0.358 = 13.437$

Answer:

(a) 0.05; $H_0$: The distributions are the same, $H_1$: The distributions are different (b) 13.437