Question

In: Statistics and Probability

13. Refer to the data set in the accompanying table. Assume that the paired sample data...

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.

Solutions

Expert Solution

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.


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