Question

13. Refer to the data set in the accompanying table. Assume that the paired sample data is a simple random sample and the differences have a distribution that is approximately normal. Use a significance level of 0.10 to test for a difference between the number of words spoken in a day by each member of 30 different couples.

Couple Male      Female

1              12320    11172

2              2410       1134

3              16390    9702

4              9550       9198

5              11360    10248

6              9450       3024

7              13310    19698

8              6670       6090

9              15090    9114

10           17650    11256

11           12730    14826

12           13050    12306

13           6980       2646

14           5860       3906

15           9410       3360

16           6960       7602

17           3270       630

18           6830       3570

19           7570       5922

20           12430    8568

21           6050       2730

22           8820       11886

23           13610    8946

24           11420    12810

25           6160       5880

26           7720       3864

27           2800       5922

28           14330    6426

29           6440       8400

30           8720       9198

In this example mu Subscript d is the mean value of the differences d for the population of all pairs of​ data, where each individual difference d is defined as the number of words spoken by a male minus the number of words spoken by a female in a couple. What are the null and alternative hypotheses for the hypothesis​ test?

Identify the test statistic.

T=

(Round to two decimal places as​ needed.)

Identify the​ P-value.

​P-value=____

​(Round to three decimal places as​ needed.)

Since the​ P-value is (less/greater) than the significance​ level, (fail to reject/reject) the null hypothesis. There (is/is not) sufficient evidence to support the claim that there is a difference between the number of words spoken in a day by each member of 30 different couples.

Answer 1

Hypotheses:

Ho: μd = 0 and Ha: μd ≠ 0

Test statistic:

n = n1 = n2 = 30

d-bar = 1844.2

s = 3429.95

SE = s/√n = 3429.95/√30 = 626.2203287

t = d-bar/SE = 1844.2/626.220328704948 = 2.94

p- value:

0.006

Conclusion:

Since the​ P-value is lesser than the significance​ level, reject the null hypothesis. There is sufficient evidence to support the claim that there is a difference between the number of words spoken in a day by each member of 30 different couples.

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