discussion 1

Hours of TV Per Week

0

4

8

14

20

25

0

5

8

15

20

28

0.5

5

10

15

20

30

1

5

10

15

20

30

1

5

10

15

21

30

2

6

12

15

25

30

2

6

14

18

25

35

2

6

14

20

25

40

Sample Size (n)

48

Mean ( x̅ )

14.32

Population Standard Deviation

uknown

Sample Standard Deviation (s)

10.468

T Critical Value

-1.679

State the Null & Alternative Hypotheses

in Words & Symbols

Which Test Is Appropriate For This and WHY?

I hypothesize that the population watches, on average, less than 16 hours of TV a week.

Null Hypothesis:

: µ = 16

Alternative Hypothesis:

: µ < 16

I will use a significance level of for this test.

Because the Population Standard Deviation is NOT known and it is not based on proportions/percentages,

The t-test is appropriate for this situation.

Calculate the Value of the Test Statistic

Find the Critical Value and/or P-Value

= = ≈ -1.11

To find the t-critical value, you go to the t-table in the back of the textbook.

· This is a 1-tailed test (since the alternative hypothesis is <).

· The alpha-level is 0.05.

· There are 47 degrees of freedom (48-1). I will use the closest number (45) on the t-table since 47 is not shown.

· The critical value is -1.679. The value is negative because it is a left-tailed test.

What is the Statistical Conclusion and WHY?

What is the Simple Conclusion?

Since my data is located outside of the rejection region, my statistical decision is:

Fail to Reject

The data does not support the claim that the average person watches less than 16 hours a week.