Answer:
Check the explanation
Explanation:
Going by the first attached image below we reject H_o against H_1 if obs.
![T > t_(\alpha /2;n-1)](https://img.qammunity.org/2021/formulas/mathematics/college/bso5jz0d3dbkbnqfn5frgo9okmtisf09bn.png)
here obs.T=1.879
![\therefore obs.T \\gtr 2.447=t_(0.025;6)](https://img.qammunity.org/2021/formulas/mathematics/college/rh7jwysjz64qwfh37e6x8yxa561l740zdq.png)
we accept
at 5% level of significance.
i.e there is no sufficient evidence to indicate that the special study program is more effective at 5% level of significance.
1.
this problem is simillar to the previous one except the alternative hypothesis.
Let X_i's denote the bonuses given by female managers and Y_i's denote the bonuses given by male managers.
we assume that
independently
We want to test
![H_0:\mu_(1)=\mu_(2) vs H_1:\mu_(1)\\eq \mu_(2)](https://img.qammunity.org/2021/formulas/mathematics/college/7j1wo6m10bc0t2gyuuyfj7tyi1k6447u55.png)
define
![D_i=X_i-Y_i , i=1(1)8](https://img.qammunity.org/2021/formulas/mathematics/college/h4trfs4nbdmldayavay1vpnazma6kv63rq.png)
now
![D_i\sim N(\mu _(1)-\mu _(2)=\mu _(D),\sigma _(1)^(2)+\sigma _(2)^(2)=\sigma _(D)^(2)) , i=1(1)8](https://img.qammunity.org/2021/formulas/mathematics/college/1u6c659lh28lggulkihzzza32s3ddk2rgh.png)
the hypothesis becomes
![H_0:\mu_(D)=0 vs H_1:\mu_(D)\\eq 0](https://img.qammunity.org/2021/formulas/mathematics/college/muu5e39wry70311zd4lyd5nh3buuw4tow9.png)
in the third attached image, we use the same test statistic as before
i.e at 5% level of significance there is not enough evidence to indicate a difference in average bonuses .