Answer to Question #315923 in Statistics and Probability for celestia

Question

A certain group of welfare recipients receives relief goods with a mean amount of Php 500.00 per week. A random sample of 75 recipients is survived and found that the mean amount of relief goods they received in a week is Php 600 and a standard decision of Php. 50.00. Test the claim at 1% level of significance is not Php 500.00 per week and assume that the population is approximately normally distributed.

Determine the following

Null and alternative hypothesis:

Critical value and rejection region:

Computed test statistic:

Conclusion:

Answer 1

$H_0:\mu =500\\H_1:\mu \ne 500\\T=\sqrt{n}\frac{\bar{x}-\mu}{s}=\sqrt{75}\frac{600-500}{50}=17.3205\sim t_{n-1}=t_{74}\\Critical\,\,value: \\t_{n-1,\frac{1+\gamma}{2}}=t_{74,0.995}=2.6439\\\mathrm{Re}jection\,\,region: \left| T \right|>2.6439\\17.3205>2.6439\Rightarrow \mu \ne 500$