The below table gives us the number of books issued from a certain library on the various days of the week.
Test at $5\%$ level whether issuing book is a day dependent (Given $\chi_{5:0.05}=11.07$)
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We have to use the chi-square test of independence, right?
H0: Issuing book is a day independent.
H1: Issuing book is a day dependent.
The sum of the row is equal to $120+130+110+115+135+110=720$
We have $\chi_{5:0.05}=11.07$.
Which degrees of freedom do we have here?
Then we have to calculated the corresponding p-value, right?
I'm no expert, so take my answer with a grain of salt, although I did study Chi-squared tests recently, however, I'm not certain it is the "best test", or I have "applied it in the best way", or whatever. But I feel like in statistics these things are arguable depending on your level. Anyway, this is how I would do it based on the OCR A-Level Further Maths Statistics syllabus.
Quoting from Further Maths OCR A Level Statistics:
Our "model" is a table of expected frequencies where the numbers in each box are $120$, because that's what you would expect if the days are all independent.
Degrees of freedom = number of bins - number of constraints = $6-1$ (the total number of books) $=5.$
To be clear, there is $1$ constraint: the total number of books.
So from the table of critical values, the critical value is $\chi_{5:0.05}=11.07.$
Comparing the data with the observed frequencies, we get:
$\chi^2_{\text{calc}}=\frac{(120-120)^2}{120} + \frac{(130-120)^2}{120} + \frac{(110-120)^2}{120} + \frac{(115-120)^2}{120} + \frac{(135-120)^2}{120} + \frac{(110-120)^2}{120} = \ldots$
If $\chi^2_{\text{calc}} > 11.07$ then reject $H_0$ as there is significant evidence of dependence. If $\chi^2_{\text{calc}} < 11.07$ then accept $H_0$: there is no significant evidence of dependence.
You then should write a conclusion, putting your result into the context of the question, as you usually do with hypothesis tests.
I'm not sure what $p-$values have to do with this?