Answer to Question #286167 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) Appropriate hypotheses:

$H_0: \mu=573$

$H_1: \mu\not =573$

b) p-value:

Sample size is greater than 30, so z-test is appropriate to compute test statistic.

$Z=\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}$

$Z=\frac{573-566}{\frac{80}{\sqrt{100}}}=0.875$

From normal distribution table, it is found that p--value is 0.809 in respect to above Z-score.

c) Conclusion:

Hence, p-value is greater than alpha (α = 0.05 < p-0.809). So, we fail to reject the null hypothesis. Based on this, it can conclude that the Canada mean unemployment insurance benefit was $573 per week.

d) Hypothesis test using the critical value approach:

Critical value at α = 0.05 is 1.645.

Z-test score is less than critical value (0.875<1.645), so  the null hypothesis can't be rejected.