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

Hypothesis :

Clearly it is paired t-test .

Couple	Male	Female	Difference (di)	di^2
1	12320	11172	1148	1317904
2	2410	1134	1276	1628176
3	16390	9702	6688	44729344
4	9550	9198	352	123904
5	11360	10248	1112	1236544
6	9450	3024	6426	41293476
7	13310	19698	-6388	40806544
8	6670	6090	580	336400
9	15090	9114	5976	35712576
10	17650	11256	6394	40883236
11	12730	14826	-2096	4393216
12	13050	12306	744	553536
13	6980	2646	4334	18783556
14	5860	3906	1954	3818116
15	9410	3360	6050	36602500
16	6960	7602	-642	412164
17	3270	630	2640	6969600
18	6830	3570	3260	10627600
19	7570	5922	1648	2715904
20	12430	8568	3862	14915044
21	6050	2730	3320	11022400
22	8820	11886	-3066	9400356
23	13610	8946	4664	21752896
24	11420	12810	-1390	1932100
25	6160	5880	280	78400
26	7720	3864	3856	14868736
27	2800	5922	-3122	9746884
28	14330	6426	7904	62473216
29	6440	8400	-1960	3841600
30	8720	9198	-478	228484
Total			55326	443204412

Therefore ,

Now we have to find & for find test statistic.

Test Statistic

For p-value

> x=c(12320,2410,16390,9550,11360,9450,13310,6670,15090,17650,12730,13050,6980,5860,9410,6960,3270,6830,7570,12430,6050,8820,13610,11420,6160,7720,2800,14330,6440,8720)
> y=c(11172,1134,9702,9198,10248,3024,19698,6090,9114,11256,14826,12306,2646,3906,3360,7602,630,3570,5922,8568,2730,11886,8946,12810,5880,3864,5922,6426,8400,9198)
> t.test(x,y,paired=T,conf.level=0.90)

Paired t-test

data: x and y
t = 2.945, df = 29, p-value = 0.006306
alternative hypothesis: true difference in means is not equal to 0
90 percent confidence interval:
780.172 2908.228
sample estimates:
mean of the differences
1844.2

Since the​ P-value(0.006306) is less than the significance​ level (0.10), 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.