91影视

The data forthe success of treatment and type of treatment is provided.

The level of significance is 0.01.

Theexpected frequency is computed as,

\(E = \frac{{\left( {row\;total} \right)\left( {column\;total} \right)}}{{\left( {grand\;total} \right)}}\)

The table for the observed values and the totals of row and column is represented as,

Theexpected frequency tableis represented as,

Here, all expected values are greater than 5.

Assuming the subjects are gathered from random selection and assigned randomly to the groups, the requirements of chi-square test is satisfied.

The claim for independence of type of treatment and the success of treatment is tested using the following hypotheses:

\({H_0}:\)Success is independent of the type of treatment

\({H_1}:\)Success is dependent of the type of treatment.

The value of the test statisticis computed as,

\[\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \\ = \frac{{{{\left( {60 - 67.571} \right)}^2}}}{{67.571}} + \frac{{{{\left( {23 - 15.429} \right)}^2}}}{{15.429}} + ... + \frac{{{{\left( {6 - 13.571} \right)}^2}}}{{13.571}}\\ = 9.750\end{aligned}\]

Therefore, the value of the test statistic is 9.750.

The degrees of freedomare computed as,

\(\begin{aligned}{c}\left( {r - 1} \right)\left( {c - 1} \right) = \left( {2 - 1} \right)\left( {2 - 1} \right)\\ = 1\end{aligned}\)

Therefore, the degrees of freedom are 1.

From chi-square distribution table, the critical value for row corresponding to 1 degrees of freedom and at 0.01 level of significance is 6.635.

Also, the p-value is obtained as 0.002.

Since the critical (6.635) is less than the value of test statistic (9.750). In this case, the null hypothesis is rejected.

Therefore, the decision is to reject the null hypothesis.

There is not enough evidencein favour of the claim that success is independent of the type of treatment.

It suggests that the choice of treatment would be essential to determine the success of the treatment in the carpal tunnel syndrome.