The data from the data sample o 10 paired observations are shown: Pair Population 1 Population 2
1 19 24
2 25 27
3 31 36
4 52 53
5 49 55
6 34 34
7 59 66
8 47 51
9 17 20
10 51 55
1. If you wish to test whether these data are sufficient to indicate that the mean for population 2 is larger than that for population 1, what are the appropriate null and alternative hypotheses?
2. Assuming that the within-pair differences are approximately normally distributed, conduct
the test using α = 0.1. What is your decision.
3. Find a 90% confidence interval for µd.

Step-by-step explanation:

Corresponding means for population 1 and population 2 form matched pairs.

The data for the test are the differences between the mean for population 1 and mean for population 2.

μd = the mean for population 1 minus the mean for population 2.

Population 1 population 2 diff

19 24 - 5

25 27 - 2

31 36 - 5

52 53 - 1

49 55 - 6

34 34 0

59 66 - 7

47 51 - 4

17 20 - 3

51 55 - 4

Sample mean, xd

= (- 5 - 2 - 5 - 1 - 6 + 0 - 7 - 4 - 3 - 4)/10 = - 3.7

xd = - 3.7

Standard deviation = √(summation(x - mean)²/n

n = 10

Summation(x - mean)² = (- 5 + 3.7)^2 + (- - 2 + 3.7)^2 + (- 5 + 3.7)^2+ (- 1 + 3.7)^2 + (- 6 + 3.7)^2 + (0 + 3.7)^2 + (- 7 + 3.7)^2 + (- 4 + 3.7)^2 + (- 3 + 3.7)^2 + (- 4 + 3.7)^2 = 73.7

Standard - eviation = √(73.7/10

sd = 2.71

For the null hypothesis

H0: μd ≥ 0

For the alternative hypothesis

H1: μd < 0

The distribution is a students t. Therefore, degree of freedom, df = n - 1 = 10 - 1 = 9

The formula for determining the test statistic is

t = (xd - μd)/(sd/√n)

t = (- 3.7 - 0)/(2.71/√10)

t = - 4.32

We would determine the probability value by using the t test calculator.

p = 0.00097

Since alpha, 0.1 > than the p value, 0.00097, then we would reject the null hypothesis. Therefore, at 0.1 level of significance, we can conclude that these data are sufficient to indicate that the mean for population 2 is larger than that for population 1.

3) for population 1,

Mean = (19 + 25 + 31 + 52 + 55 + 34 + 59 + 47 + 17 + 51)/10 = 38.4

Summation(x - mean)² = (19 - 38.4)^2 + (25 - 38.4)^2 + (31 - 38.4)^2+ (52 - 38.4)^2 + (49 - 38.4)^2 + (34 - 38.4)^2 + (59 - 38.4)^2 + (47 - 38.4)^2 + (17 - 38.4)^2 + (51 - 38.4)^2 = 2042.4

Standard deviation, s1 = √2042.4/10 = 14.3

for population 2,

Mean = (24 + 27 + 36 + 53 + 55 + 34 + 66 + 51 + 20 + 55)/10 = 42.1

Summation(x - mean)² = (24 - 42.1)^2 + (27 - 42.1)^2 + (36 - 42.1)^2 + (53 - 42.1)^2 + (55 - 42.1)^2 + (34 - 42.1)^2 + (66 - 42.1)^2 + (51 - 42.1)^2 + (20 - 42.1)^2 + (55 - 42.1)^2 = 2248.9

Standard deviation, s2 = √2248.9/10 = 15

The formula for determining the confidence interval for the difference of two population means is expressed as

Confidence interval = (x1 - x2) ± z√(s²/n1 + s2²/n2)

For a 90% confidence interval, we would determine the z score from the t distribution table because the number of samples are small

Degree of freedom =

(n1 - 1) + (n2 - 1) = (10 - 1) + (10 - 1) = 18

z = 1.734

x1 - x2 = 38.4 - 42.1 = - 3.7

√(s1²/n1 + s2²/n2) = √(14.3²/10 + 15²/10)

= 6.55

Margin of error = 1.734 × 6.55 = 11.4

The 90% confidence interval is

- 3.7 ± 11.4

javeir earned 56 dollors more than kyle so kyle and javier had 14w but javier made 56 dollors more so 14w plus 56 equals 70w so kyle has 14w

step-by-step explanation: hope i plz give me brainliest with the points so this is not constant cause it says last week so its clearly not every week

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step-by-step explanation: