Question

A development economist is studying income growth in a rural area of a developing country. The last census of the population of this area, several years earlier, showed that mean household annual income was 425 dollars, and the variance of household income was 2500 (dollars-squared). A current random sample of 100 households yields a sample mean income of $433.75. Assume that household annual incomes are approximately normally distributed, and that the population variance is known still to be 2500. Test the null hypothesis that population mean income has not increased against the alternative hypothesis that it has increased, at a 1% level of significance.

What is the form of the rejection region that should be used to conduct this hypothesis test?

Answer 1

If we compare the p value and the significance level given
`\alpha=0.01` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL reject the null hypothesis, so we can say that the population mean is not singificantly higher than 425.

Explanation:

Data given and notation

`\bar X=433.75` represent the mean height for the sample

`\sigma=√(2500)=50` represent the population standard deviation for the sample

`n=100` sample size

`\mu_o =425` represent the value that we want to test

`\alpha=0.01` represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

`p_v` represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean height actually is higher than the mean height for men in 1960, the system of hypothesis would be:

Null hypothesis:
`\mu \leq 425`

Alternative hypothesis:
`\mu > 425`

Since we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

`z=(\bar X-\mu_o)/((\sigma)/(√(n)))` (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

`z=(433.75-425)/((50)/(√(100)))=1.75`

P-value

Since is a one right tail test the p value would be:

`p_v =P(z>1.75)=1-P(z<1.75)=0.040`

Conclusion

If we compare the p value and the significance level given
`\alpha=0.01` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL reject the null hypothesis, so we can say that the population mean is not singificantly higher than 425.