Question

A United Nations report shows the mean family income for Mexican migrants to the United States is $27,300 per year. A FLOC (Farm Labor Organizing Committee) evaluation of 20 Mexican family units reveals a mean to be $28,500 with a sample standard deviation of $9,651. Does this information disagree with the United Nations report? Apply the 0.01 significance level.

a. State the null hypothesis and the alternate hypothesis.



b. State the decision rule for 0.01 significance level. (Negative amounts should be indicated by a minus sign. Round your answers to 3 decimal places.)



c. Compute the value of the test statistic. (Round your answer to 2 decimal places.)



d. Does this information disagree with the United Nations report?

Answer 1

The null hypothesis is that the mean family income for Mexican migrants to the United States is $27,300 per year. The decision rule for a 0.01 significance level is to reject the null hypothesis if the test statistic falls outside the critical region. The test statistic can be calculated using the formula z = (sample mean - population mean) / (sample standard deviation/sqrt (sample size)). If the test statistic falls outside the critical region, there is disagreement with the United Nations report.

Step-by-step explanation:

a. Null hypothesis: The mean family income for Mexican migrants to the United States is $27,300 per year.

Alternate hypothesis: The mean family income for Mexican migrants to the United States is not $27,300 per year.

b. Decision rule for 0.01 significance level: The decision rule for a two-tailed test at the 0.01 significance level is to reject the null hypothesis if the test statistic falls outside the critical region, which is determined by the z-score corresponding to a 0.005 probability in each tail.

c. Test statistic: To compute the test statistic, we can use the formula: z = (sample mean - population mean) / (sample standard deviation/sqrt (sample size)). Using the given information, the test statistic is z = (28500 - 27300) / (9651 / sqrt(20)).

d. Disagreement with the United Nations report: To determine if the information disagrees with the United Nations report, we compare the test statistic with the critical values for the 0.01 significance level. If the test statistic falls outside the critical region, we can reject the null hypothesis and conclude that there is disagreement with the United Nations report.

Answer 2

A United Nations report shows the mean family income for Mexican migrants to the United States is $27,300 per year. A FLOC (Farm Labor Organizing Committee) evaluation of 20 Mexican family units reveals a mean to be $28,500 with a sample standard deviation of $9,651.

stated mean is 27,300 per year.

sample of 20 family units reveals a mean of 28,500 with a sample standard deviation of 9,651.

sample standard error = standard deviation of sample divided by square root of sample size.

sample standard error = 9,651 / sqrt(20) = 2158.029205.

since standard deviation is from sample rather than from population, use t-score rather than z-score.

degrees of freedom is equal to sample size minus 1 = 20 - 1 = 19.

t-score = (sample mean minus assumed population mean) / standard error.

t-score = (28,500 - 27,300) / 2158.029205 = .5560629102

alpha for .01 significance level for two sided distribution is the same as .005 significance level for one sided distribution.

if using t-score tables, you would look for critical t-score of .01 for two sided distribution or critical t-score of .005 for one sided distribution.

either one is for 19 degrees of freedom.

using the ti-84 plus scientific calculator, i looked for critical t-score of.005, since the calculator assumed one sided distribution table.

critical t-score is between -2.86 and 2.86 according to TI-84 Plus.

critical t-score table confirms this value is true at .01 alpha for two tailed distribution and at .005 alpha for 1 tailed distribution.

i'm reasonably certain this is accurate.

since the t-score is .5560629102, it is well within the confidence interval limits.

you would therefore reject the alternate hypothesis becauwse the results of the tewt are not significant.

they are significant if outside the limits and not significant if inside the limits.

your alternate hypothesis is that the scores are not equal to the assumed mean.

your null hypothesis is that they are equal to the assumed mean.

since the t-score is well within the confidence limits, this mean that the sample mean is most likely different from the assumed population mean due to random variation in sample means of size 20.

in order for you to reject the null hypothesis, the t-score would have had to be either less than -2.86 or greater than 2.86.

then you could have considered that the sample means difference from the assumed population mean would not have been likely to chance variations at the .01 significance level.

you can assume that the data does not contradict the report.

that means you aqcept H0.