Answer: We reject the null hypothesis, and we use Normal distribution for the test.
Step-by-step explanation:
Since we have given that
We claim that
Null hypothesis : [tex]H_0:\mu\geq 50000[/tex]
Alternate hypothesis : [tex]H_1:\mu<50000[/tex]
There is 5% level of significance.
[tex]\bar{X}=46800\\\\\sigma=9800\\\\n=29[/tex]
So, the test statistic would be
[tex]z=\dfrac{\bar{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}\\\\z=\dfrac{46800-50000}{\dfrac{9800}{\sqrt{29}}}\\\\z=-1.75[/tex]
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.