# 13. Refer to the data set in t

13. Refer to the data set in the accompanying table. Assume thatthe paired sample data is a simple random sample and thedifferences have a distribution that is approximately normal. Use asignificance level of 0.10 to test for a difference between thenumber of words spoken in a day by each member of 30 differentcouples.

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 thedifferences d for the population of all pairs of​ data, where eachindividual difference d is defined as the number of words spoken bya 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/isnot) sufficient evidence to support the claim that there is adifference between the number of words spoken in a day by eachmember of 30 different couples.

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 yt = 2.945, df = 29, p-value = 0.006306alternative hypothesis: true difference in means is not equal to090 percent confidence interval:780.172 2908.228sample estimates:mean of the differences1844.2

Since the​ P-value(0.006306) is less than the significance​level (0.10), reject) the null hypothesis. There is sufficientevidence to support the claim that there is a difference betweenthe number of words spoken in a day by each member of 30 differentcouples.

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