Answer: We reject the null hypothesis, and we use Normal distribution for the test.
Explanation:
Since we have given that
We claim that
Null hypothesis :
![H_0:\mu\geq 50000](https://img.qammunity.org/2020/formulas/mathematics/college/mfz8jtfx6och2qiwfxzbnfe0ilf3pimzgs.png)
Alternate hypothesis :
![H_1:\mu<50000](https://img.qammunity.org/2020/formulas/mathematics/college/ye2hbf4rcy680l226wk2ht35eslnt4pnha.png)
There is 5% level of significance.
![\bar{X}=46800\\\\\sigma=9800\\\\n=29](https://img.qammunity.org/2020/formulas/mathematics/college/c88s7aa689s0s1x1evyke1ck24z9baphdx.png)
So, the test statistic would be
![z=\frac{\bar{X}-\mu}{(\sigma)/(√(n))}\\\\z=(46800-50000)/((9800)/(√(29)))\\\\z=-1.75](https://img.qammunity.org/2020/formulas/mathematics/college/l4lkmjipjbhzbw9qi5908a0ac2ecubvnsx.png)
Since alternate hypothesis is left tailed test.
So, p-value = P(z≤-2.31)=0.0401
And the P-value =0.0401 is less than the given level of significance i.e. 5% 0.05.
So, we reject the null hypothesis, and we use Normal distribution for the test.