# 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 hypothesistest?

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 significancelevel, (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.

Answer:

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 significancelevel (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.