Question

The following data lists the ages of a random selection of actresses when they won an award in the category of Best​ Actress, along with the ages of actors when they won in the category of Best Actor. The ages are matched according to the year that the awards were presented. Complete parts​ (a) and​ (b) below.

Actress left parenthesis years right parenthesis

29

25

34

27

38

26

26

45

27

36

Actor left parenthesis years right parenthesis

62

36

39

38

31

33

52

40

41

40

a. Use the sample data with a 0.05 significance level to test the claim that for the population of ages of Best Actresses and Best​ Actors, the differences have a mean less than 0​ (indicating that the Best Actresses are generally younger than Best​ Actors).

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​ actress's age minus the​ actor's age. What are the null and alternative hypotheses for the hypothesis​ test?

Upper H 0: mu Subscript d

▼

greater than

equals

not equals

less than ________year(s)

Upper H 1​: mu Subscript d

▼

equals

greater than

not equals

less than _____year(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.)

What is the conclusion based on the hypothesis​ test?

Since the​ P-value is :

▼

greater than

less than or equal to the significance​ level,

▼

reject

fail to reject

the null hypothesis. There

▼

is not

is

sufficient evidence to support the claim that actresses are generally younger when they won the award than actors.

b. Construct the confidence interval that could be used for the hypothesis test described in part​ (a). What feature of the confidence interval leads to the same conclusion reached in part​ (a)?

The confidence interval is ______year(s)less than mu Subscript d less than_________year(s).

​(Round to one decimal place as​ needed.)

What feature of the confidence interval leads to the same conclusion reached in part​ (a)?

What feature of the confidence interval leads to the same conclusion reached in part​ (a)?

Since the confidence interval contains:

▼only negative numbers,

zero,

only positive numbers,

▼fail to reject

reject

the null hypothesis.

Answer 1

Ho :µd=0

Ha :µd <0

Sample #1	Sample #2	difference , Di =sample1-sample2	(Di - Dbar)²
29	62	-33	533.610
25	36	-11	1.210
34	39	-5	24.010
27	38	-11	1.210
38	31	7	285.610
26	33	-7	8.410
26	52	-26	259.210
45	40	5	222.010
27	41	-14	16.810
36	40	-4	34.81

	sample 1	sample 2	Di	(Di - Dbar)²
sum =	313	412	-99	1386.900
mean=	31.3000	41.2000	-9.90000

Level of Significance , α = 0.05

sample size , n = 10

mean of sample 1,               x̅1=31.3000

mean of sample 2,               x̅2=41.2000

mean of difference , D̅ =-9.9000

std dev of difference , Sd = √ [ (Di-Dbar)²/(n-1) = 12.4137

std error , SE = Sd / √n = 3.9256

t-statistic   = (D̅ - µd)/SE   = -2.52

Degree of freedom,    DF=n - 1 = 9

p-value = 0.016 [excel function:   =t.dist(t-stat,df) ]

Conclusion:                            p-value <α , Reject null hypothesis

Since the​ P-value is

less than or equal to the significance​ level,

reject

the null hypothesis. There

is

sufficient evidence to support the claim that actresses are generally younger when they won the award than actors.

--------------------------------------------

b)

Degree of freedom,    DF=n - 1 = 9

t-critical value = t α/2,df = 2.2622           [excel function:   =t.inv.2t(α/2,df) ]

std dev of difference , Sd =  12.4137

std error , SE = Sd / √n = 3.9256

margin of error, E = t*SE = 8.880228

mean of difference , D̅ =-9.9000

confidence interval is

Interval Lower Limit=D̅ - E =-18.7802

Interval Upper Limit=D̅ + E =-1.0198

so, confidence interval is (-18.8< Dbar < - 1.0)

Since the confidence interval contains:only negative numbers,

reject

the null hypothesis.