Q. 9.116 - Elementary Statistics

Question

Apparel and Services. According to the document Consumer Expenditures, a publication of the Bureau of Labor Statistics, the average consumer unit spent \( 1736 on apparel and services in 2012  That same year, 25 consumer units in the Northeast had the following annual expenditures, in dollars, on apparel and services.

1279	1457	2020	1682	1273
2223	2233	2192	1611	1734
2688	2029	2166	1860	2444
1844	1765	2267	1522	2012
1990	1751	2113	2202	1712

At the 5 % significance level, do the data provide sufficient evidence to conclude that the 2012 mean annual expenditure on apparel and services for consumer units in the Northeast differed from the national mean of \)1736? (Note: The sample mean and sample standard deviation of the data are \(1922.76 and \)350.90, respectively.)

Step-by-Step Solution

Answer

At a $5\%$ level of significance, the test findings can be concluded to be statistically significant.

Step 1: Given Information

Examine if the data is sufficient to show that the Northeast's mean annual clothing and service spending differed from the national average of $1,736 in 2012.

The following should be given as the null and alternative hypotheses:

Nullity hypothesis:

$H_o:\mu =\$1,736$

In 2012, the average annual expenditure on apparel and services for consumer units in the Northeast was $1,736 vs $1,736 overall.

Alternative hypothesis:

$H_o:\mu \ne \$1736$

In 2012, the Northeast's mean annual consumer unit spend on clothing and services was $1,736 compared to the national average of $1,736.

Determine the level of significance.

The significance level in this case is$\alpha =0.05$

Step 2: Expression

Using the Excel add-in, calculate the value of the test statistic (PHStat).

Excel add-in (PHStat) procedure:

Step 1: In EXCEL, Select Add-Ins > PHStat > One-Sample Tests >t Test for the mean, Sigma unknown.

Step 2: Enter 1,736 under Null hypothesis.

Step 3: Enter 0.05 under Level of significance.

Step 4: In Sample Statistic Options, choose Sample Statistic Unknown.

Step 5: Enter $\mathit{A}\mathbf{1}\mathbf{:}\mathit{A}\mathbf{26}$ under Sample Cell Range.

Step 6: In Test Options, choose Two Tail Test.

Step 7: In Output Options, enter a Title and click OK.

Step 3: Expression

Excel add-in (PHStat) output:

t Test for Hypothesis of the Mean

Data
Null Hypothesis $\mu =$	1736
Level of significance	0.05
Sample size	25
Sample Mean	1922.76
Sample Standard Deviation	350.8991166

Intermediate Calculations
Standard Error of the Mean	70.1798
Degrees of Freedom	24
t Test Statistic	2.6612

Two-Tail Test
Lower Critical Value	-2.0639
Upper Critical Value	2.0639
p- Value	0.0137
Reject the null hypothesis

From the output, the value of test statistic is $2.6612$, the critical value is $\pm 2.0639$ and the P-value is $0.0137$

Step 4: Calculation

Critical value:

The crucial value is derived from the Excel add-in (PHStat) output is $\pm 2.0639$

P-value:

The P-Value is calculated using the Excel add-in (PHStat) output. is $0.0137$

Step 5: Definition

Methodology for determining critical value:

The following is the rule for rejection:

Reject the null hypothesis if the result of the test statistic is in the rejection range.$H_o$.

If the values of test statistic do not falls in the rejection region, then do not reject the null hypothesis $H_o$.

Here, the value of test statistic falls in the rejection region. That is, $t(=2.6612)>t_{crit}(+2.0639)$

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.

P - value approach:

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

Here, the P-value is 0.0137 which is less than the level of significance. That is, $P(=0.0137)<\alpha (=0.05)$.

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 6: Definition

Interpretation:

The data is sufficient to show that the Northeast's average annual clothing and service spend in 2012 differed from the national average of $1,736.