Answer to Question #286409 in Microeconomics for mukti

Question

Employment and Social Development Canada (ESDC) reported that the Canada mean

unemployment insurance benefit was $573 per week (Government of Canada, 2020). A

researcher in the state of Manitoba anticipated that sample data would show evidence that mean

weekly unemployment insurance benefit in Manitoba was below the national average.

a) Develop appropriate hypotheses such that rejection of H 0 will support the researcher’s

contention.

b) For a sample of 100 individuals, the sample mean weekly unemployment insurance

benefit was $566 with a sample standard deviation of $80. What is the p-value?

c) At α = 0.05, what is your conclusion?

d) Repeat the preceding hypothesis test using the critical value approach.

Answer 1

$a)$

The hypotheses to be tested are,

$H_0:\mu=573$ $vs$ $H_1:\mu\lt573$

$b)$

$n=100,\space \bar{x}=566,\space s=80$

The test statistic is given as,

$Z={(\bar{x}-\mu)\over{s\over\sqrt{n}}}={566-573\over{80\over10}}={-7\over8}=-0.875$

From the standard normal tables, the p-value is $\phi(-0.875)= 0.1894.$

$c)$

The decision rule is,

Reject the null hypothesis if $p-value\lt \alpha$.

Since $p-value=0.1894\gt\alpha=0.05,$ we fail to reject the null hypothesis and conclude that there is no sufficient evidence to support the researcher's contention that mean weekly unemployment insurance benefit in Manitoba is below the national average at 5% significance level.

$d)$

The critical value is given as,

$Z_{\alpha}=Z_{0.005}=-2.575$

and we reject the null hypothesis if $Z\lt Z_{\alpha}$.

Since $Z=-0.875\gt Z_{0.05}=-2.575$, we fail to reject the null hypothesis and conclude that there is no sufficient evidence to support the researcher's contention that mean weekly unemployment insurance benefit in Manitoba is below the national average at 5% significance level.