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StatisticsQ&A LibraryPreliminary data analyses indicate that you can reasonably use a t-test to conduct each of the hypothesis tests required in ExercisesApparel 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.1279145720201682127322232233219216111734268820292166186024441844176522671522201219901751211322021712At 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.)Question

Asked Jan 25, 2020

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Preliminary data analyses indicate that you can reasonably use a t-test to conduct each of the hypothesis tests required in Exercises

**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 1

From the given information, the sample size 25.Using EXCEL function, “AVERAGE(A1:A25)”, the mean is 1922.8 and “STDEV.S(A1:A25)”, the standard deviation is 350.9.

Null hypothesis:

*µ=*1736*.*

Alternative hypothesis:

*µ≠*1736

This is a two tailed test.

Since the population standard deviation is unknown, the appropriate test is one sample *t*-test.

Critical value is obtained below:

Df = n – 1

= 25 – 1

= 24

Step 2

Using table of t-distribution, the critical value is ±2.064.

Thus, the critical value...

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