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Since the confidence interval contains (only positive numbers/zero/only negative

ID: 3326147 • Letter: S

Question


Since the confidence interval contains (only positive numbers/zero/only negative numbers) (fail to reject/reject) the null hypothesis. Listed below are the numbers of words spoken in a day by each member of eight different randomly selected couples Complete parts (a) and (b) below Male 16.305 27,155 1370 7910 19,175 15.30914071 25,357 Female 25,085 13,395 19,026 18,376 13,134 16,336 15,685 19,023 a. Use a 0.05 significance level to test the claim that among couples, males speak fewer words in a day than females. In this example, 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 words spoken by the male minus words spoken by the female. What are the null and alternative hypotheses for the hypothesis test? Ho:Ha! word(s) Type integers or decimals. Do not round) 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 )

Explanation / Answer

Solution:

Part a

Here, we have to use paired t test. Null and alternative hypothesis for this test are given as below:

H0: µd = 0

H1: µd < 0

We are given

= 0.05

Test statistic formula is given as below:

t = Dbar / [Sd/sqrt(n)]

From given data, we have

Sample size = 8

Dbar = -1301.0000

Sd = 10401.0140

df = n – 1 = 8 – 1 = 7

Calculation table is given as below:

Male

Female

Di

(Di - DBar)^2

16305

25085

-8780

55935441

27155

13395

13760

226833721

1370

19026

-17656

267486025

7910

18376

-10466

83997225

19175

13134

6041

53904964

15309

13336

1973

10719076

14071

15685

-1614

97969

25357

19023

6334

58293225

t = -1301.0000/[10401.0140/sqrt(8)]

t = -0.35

P-value = 0.3670

(By using t-table or excel)

Since the P-value is greater than the significance level, we do not reject the null hypothesis. There is not sufficient evidence to support the claim that males speak fewer words in a day than females.

Part b

From given data, we have

Sample size = 8

Dbar = -1301.0000

Sd = 10401.0140

df = n – 1 = 8 – 1 = 7

= 0.05

Critical value = t =

Confidence interval = Dbar -/+ t*Sd/sqrt(n)

Confidence interval = -1301.0000 -/+ 2.3646*10401.0140/sqrt(8)

Confidence interval = -1301.0000 -/+ 2.3646* 3677.313765

Confidence interval = -1301.0000 -/+ 8695.4653

Lower limit = -1301.0000 - 8695.4653 = -9996.5

Upper limit = -1301.0000 + 8695.4653 = 7394.5

Since the confidence interval contains zero, so we fail to reject the null hypothesis.

Male

Female

Di

(Di - DBar)^2

16305

25085

-8780

55935441

27155

13395

13760

226833721

1370

19026

-17656

267486025

7910

18376

-10466

83997225

19175

13134

6041

53904964

15309

13336

1973

10719076

14071

15685

-1614

97969

25357

19023

6334

58293225