Q. 39 - Elementary Statistics

Question

Beef Consumption. According to Food Consumption, Prices,\and Expenditures, published by the U.S. Department of Agriculture. the mean consumption of beef per person in 2011 was 57.5 lb. A sample of 40 people taken this year yielded the data, in pounds, on last year's beef consumption given on the Weiss Stats site. Use the technology of your choice to do the following.

a. Obtain a normal probability plot, a boxplot, a histogram, and a stem-and-leaf diagram of the data        on beef consumptions. 

b. Decide, at the 5% significance level, whether last year's mean beef consumption is less than the         2011 mean of 57.5 lb. Apply the one mean t-test.

c. The sample data contain four potential outliers: 0, 0, 0, and 13.Remove those four observations,           repeat the hypothesis test in part (b), and compare your result with that obtained in part (b).

d. Assuming that the four potential outliers are not recording errors, comment on the advisability of        removing them from the sample data before performing the hypothesis test.

e. What action would you take regarding this hypothesis test?

Step-by-Step Solution

Answer

Nonparametric test is used to test the population parameters when the variable is not normally distributed. In this situation, the nonparametric test is more suitable when compared to one mean t test.

Step 1. Solution a : probability plot

Construct a normal probability plot by using MINITAB.

MINITAB procedure:

Step 1: Choose Graph > Probability Plot.

Step 2: Choose Single, and then click OK.

Step 3: In Graph variables, enter the column of Consumption and click OK.

Step 2. MINITAB output

Step 3. Boxplot

Construct a boxplot by using MINITAB.

MINITAB procedure:

Step 1: Choose Graph > Boxplot or Stat > EDA Boxplot.

Step 2: Under Multiple Y's, choose Simple. Click OK.

Step 3: In Graph variables, enter the data of Consumption and click OK.

Step 4. MINITAB output

Step 5. Histogram

Construct a histogram by using MINITAB.

MINITAB procedure:

Step 1: Choose Graph> Histogram.

Step 2: Choose Simple, and then click OK. 

Step 3: In Graph variables, enter the corresponding column of Consumption  and click OK.

Step 6. MINITAB output

Step 7. Stem-and-leaf diagram

Construct a stem-and-leaf by using MINITAB.

MINITAB procedure:

Step 1: Select Graph > Stem and leaf.

Step 2: Select the column of variables in Graph variables and click OK.

Step 8. MINITAB output

Step 9. Solution b

Check whether last year's mean beef consumption is less than the 2011 mean of 57.5 lb.

State the null and alternative hypothesis:

Null hypothesis:

$H_0: \mu =57.5 \mathrm{lb}$

That is, the mean beef consumption is not less than the 2011 mean of 57.5 Ib.

Alternative hypothesis:

$H_a: \mu <57.5 \mathrm{lb}$

That is, the mean beef consumption is less than the 2011 mean of 57.5 lb.

Here, the significance level is, $\alpha =0.05.$

Step 10. Test statistic and p-value

MINITAB procedure:

Step 1: Choose Stat > Basic Statistics > 1-Sample Z

Step 2: In Samples in Column, enter the column of Consumption.

Step 3: In Perform hypothesis test, enter the test mean as 57.5.

Step 4: Check Options, enter Confidence level as 95.

Step 5: Choose less than in alternative and click OK in all dialogue boxes.

Step 11. MINITAB output

From MINITAB output, the value of test statistic is $\boxed{-1.87}$and the p-value is $\boxed{0.034}$.

Step 12. P-value approach

Rejection rule:

If $P\le \alpha$, then reject the null hypothesis.

Here, the P-value is 0.034 which is less than the level of significance. That is,

$P(=0.034)<\alpha (=0.05).$

Therefore, the null hypothesis is rejected at 5% level.

Thus, it can be conclude that the test results are statistically significant at 5% level of significance.

Step 13. Interpretation

The data provide sufficient evidence to conclude that the mean beef consumption is less then the 2011 mean of 57.5 lb. at 5% level.

Step 14. Solution c

Remove the outliers and repeat the hypotheses test.

Obtain the test statistic and p-value by using MINITAB.

MINITAB procedure:

Step 1: Choose Stat > Basic Statistics > 1-Sample Z.

Step 2: In Samples in Column, enter the column of Consumption.

Step 3: In Perform hypothesis test, enter the test mean as 57.5.

Step 4: Check Options, enter Confidence level as 95.

Step 5: Choose less than in alternative and click OK in all dialogue boxes.

Step 15. MINITAB output

From MINITAB output, the value of test statistic is $\boxed{-0.21}$ and the p-value is $\boxed{0.417}$.

Step 16. P-value approach

Rejection rule:

If $P\le \alpha$. then reject the null hypothesis.

Here, the P-value is 0.417 which is greater than the level of significance. 

That is, $P(=0.417)>\alpha (=0.05)$. Therefore, the null hypothesis is not rejected at 5% level.

Thus, it can be conclude that the test results are not statistically significant at 5% level of

significance.

Interpretation:

The data do not provide sufficient evidence to conclude that the mean beef consumption is less

than the 2011 mean of 57.5 lb. at 5% level.

Step 17. Comparison

From part b., the mean beef consumption is less than the 2011 mean of 57.5 lb. because the null hypothesis is rejected.

From part c., the mean beef consumption is not less than the 2011 mean of 57.5 lb. because the

null hypothesis is not rejected.

Thus, both results are not similar.

Step 18. Solution d

Explanation:

Here, the outliers are larger effect on the hypothesis test result.

If the outliers are not removed from the data, then the result of the hypothesis test produce valid conclusions regarding the population. Moreover, the distribution of the data is skewed to left.

If the outliers are removed from the data, then the result of the hypothesis test does not produce valid conclusions regarding the population. Moreover, the distribution of the data is roughly symmetric

Step 19. Solution e

Nonparametric test:

Nonparametric test is used to test the population parameters when the variable is not normally

distributed.

In this situation, the nonparametric test is more suitable when compared to one mean t-test.