# Week 6 Statistics Qnt 561

**Topics:**Statistical hypothesis testing, Null hypothesis, Normal distribution

**Pages:**7 (1112 words)

**Published:**September 27, 2011

H0: game length is >= 3.5 hours

Ha: game length is < 3.5 hours

mean = 2.9553

stdev = 0.5596

Get the t test statistic:

t = (x-mu)/(stdev/sqrt(N))

t = (2.9553-3.5)/(0.5596/sqrt(17))

t = -4.0133

Get the critical value for df = N-1 = 16, one tail, alpha is 0.05: -1.7459

Since our test statistic is much lower than the critical value, we reject the null hypothesis. There is enough evidence to conclude that games are shorter than 3.50 hours.

Chapt 11 # 58

The amount of income spent on housing is an important component of the cost of living. The total costs of housing for homeowners might include mortgage payments, property taxes, and utility costs (water, heat, electricity). An economist selected a sample of 20 homeowners in New England and then calculated these total housing costs as a percent of monthly income, five years ago and now. The information is reported below. Is it reasonable to conclude the percent is less now than five years ago?

Home OwnerFive years agoNow

117%10%

22039

32937

44327

53612

64341

74524

81926

94928

104926

1135%32%

121632

132321

143312

154440

164442

172822

182919

193935

202212

SOLUTION

Before After

1 17 10

2 20 39

3 29 37

4 43 27

5 36 12

6 43 41

7 45 24

8 19 26

9 49 28

10 49 26

11 35 32

12 16 32

13 23 21

14 33 12

15 44 40

16 44 42

17 28 22

18 29 19

19 39 35

20 22 12

Since these are before and after measurements, we need to treat this as a paired t-test.

Here's how the test statistic is calculated.

First we find the different of the samples. I'll subtract the afters from the befores. Here is the column of differences:

7

-19

-8

16

24

2

21

-7

21

23

3

-16

2

21

4

2

6

10

4

10

Now we need to calculate the standard deviation of this column. We find it to be: 12.4778

And the average of this column is 6.3.

So the test-statistic is (6.3-0)/(12.4778/sqrt(20)) = 2.258.

The null hypothesis is that the difference is 0.

The alternative hypothesis is that the difference is greater than 0.

We choose alpha = 0.05.

The critical value for this test is found from the t-table to be 1.645.

Since the test-statistic exceeds the critical value, we conclude that the percent is less now than it was 5 years ago.

Chapt 12 # 42

Martin Motors has in stock three cars of the same make and model. The president would like to compare the gas consumption of the three cars (labeled car A, car B, and car C) using four different types of gasoline. For each trial, a gallon of gasoline was added to an empty tank, and the car was driven until it ran out of gas. The following table shows the number of miles driven in each trial.

|of Gasoline |Distance (miles) | | |Car A |Car B |Car C | |Regular |22.4 |20.8 |21.5 | |Super regular |17.0 |19.4 |20.7 | |Unleaded |19.2 |20.2 |21.2 | |Premium unleaded |20.3 |18.6 |20.4 |

Using the .05 level of significance:

A. Is there a difference among types of gasoline?

B. Is there a difference in the cars?

We use two-way ANOVA to analyze the data. The output is as follows:

Anova: Two-Factor Without Replication

SUMMARY Count Sum Average Variance

Regular 3 64.7 21.56666667 0.643333333

Super Regular 3 57.1 19.03333333 3.523333333

Unleaded 3 60.6 20.2 1

Premium Unleaded 3 59.3 19.76666667 1.023333333

Car A 4 78.9 19.725 5.0625

Car B 4 79 19.75 0.916666667

Car C 4 83.8 20.95 0.243333333

ANOVA

Source of Variation SS df MS F P-value F...

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