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8090 [49]
3 years ago
5

11.1.21 A survey​ asked, "How many tattoos do you currently have on your​ body?" Of the males​ surveyed, responded that they had

at least one tattoo. Of the females​ surveyed, responded that they had at least one tattoo. Construct a ​% confidence interval to judge whether the proportion of males that have at least one tattoo differs significantly from the proportion of females that have at least one tattoo. Interpret the interval. Let represent the proportion of males with tattoos and represent the proportion of females with tattoos. Find the ​% confidence interval for . The lower bound is nothing. The upper bound is nothing. ​(Round to three decimal places as​ needed.)
Mathematics
1 answer:
hram777 [196]3 years ago
7 0

Complete question :

The Harris Poll conducted a survey in which they asked, "How many tattoos do you currently have on your body?" Of the 1205 males surveyed, 181 responded that they had at least one tattoo. Of the 1097 females surveyed, 143 responded that they had at least one tattoo. Construct a 95% confidence interval to judge whether the proportion of males that have at least one tattoo differs significantly from the proportion of females that have at least one tattoo. Interpret the interval.

Answer:

(−0.0085 ; 0.0481)

Step-by-step explanation:

Given that :

n1 = 1205 ; x1 = 181 ; n2 = 1097 ; x2 = 143 ; α = 95%

Zα/2 = 1.96 ( Z table)

Confidence interval : (p1 - p2) ± E

E =Zα/2 * √[(p1q1/n1) + (p2q2/n2)]

p1 = x1 /n1 =181/1205 = 0.1502

q1 = 1 - p1 = 1 -0.1502 = 0.8498

p2 = x2/n2 = 143/1097 = 0.1304

q2 = 1 - p2 = 1 -0.1304 = 0.8696

E = 1.96 * √(0.0001059 + 0.0001033)

E = 0.0283

p1 - p2 = 0.1502 - 0.1304 = 0.0198

Lower boundary = 0.0198 - 0.0283 = −0.0085

Upper boundary = 0.0198 + 0.0283 = 0.0481

(−0.0085 ; 0.0481)

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MariettaO [177]

Answer:

Step-by-step explanation:

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area of composite figure: (25pi ft^2) + 200ft^2

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3 years ago
Please help me on 15ab!! will mark brainliest
forsale [732]

15a) 1.2 x 10^6

15b) 1.2 x 10^-4

6 0
3 years ago
Avanety of two types of snack packs are delivered to a store. The box plots compare the number of calories in each
AveGali [126]

Answer:

The number of calories in the packs of trail mix have a greater variation than the number of calories in the packs

of crackers

Step-by-step explanation:

IQR of trail mix data = 105 - 90 = 15

The range of cracker data = 100 - 70 = 30.

Therefore, the first option is NOT TRUE.

To check if option 2 is correct, calculate the lower limit to see if 70 is below the lower limit. If 70 is below the lower limit, then it is an outlier in the trail mix data.

Thus, Lower Limit = Q_1 - 1.5(IQR)

Q1 = 90,

IQR = 105 - 90 = 15

Lower Limit = 90 - 1.5(15)

Lower Limit = 90 - 22.5 = 67.5

70 is not less than the lower limit, therefore, 70 is not an outlier for the trail mix data. The second option is NOT TRUE.

The upper quartile of the trail mix data = 105.

The maximum value of the cracker data = 100.

Therefore, the third option is NOT TRUE.

Range can be used to determine how much variable there is in a data represented on a box plot. The greater the range value, the greater the variation.

Range of trail mix data = 115 - 70 = 45

Range of cracker data = 100 - 70 = 30.

<em>The range value for the number of calories in trail mix is greater than that for cracker, therefore, the number of calories in the packs of trail mix have a greater variation than the number of calories in the packs</em>

<em>of crackers.</em>

The fourth option is TRUE.

7 0
3 years ago
Read 2 more answers
A snack-size bag of M&amp;Ms candies is opened. Inside, there are 12 red candies, 12 blue, 7 green, 13 brown, 3 orange, and 10 y
Anestetic [448]

Answer:

\frac{1}{4180}

Step-by-step explanation:

GIVEN: A snack-size bag of M&Ms candies is opened. Inside, there are 12 red candies, 12 blue, 7 green, 13 brown, 3 orange, and 10 yellow. Three candies are pulled from the bag in succession, without replacement.

TO FIND: What is the probability that the first two candies drawn are orange and the third is green.

SOLUTION:

Total  candies in the bag =57

Probability that first ball is orange, P(A)=\frac{\text{total orange candies in bag}}{\text{total candies in bag}}

                                                       =\frac{3}{57}=\frac{1}{19}

Probability that second ball is orange, P(B)=\frac{\text{total orange candies in bag}}{\text{total candies in bag}}

                                                        =\frac{2}{56}=\frac{1}{28}

Probability that third ball is green, P(C)=\frac{\text{total green candies in bag}}{\text{total candies in bag}}

                                                                 =\frac{7}{55}

Now, probability that first two balls are orange and third is green is

=P(A)\times P(B)\times P(C)

=\frac{1}{19}\times\frac{1}{28}\times\frac{7}{55}

=\frac{1}{19}\times\frac{1}{4}\times\frac{1}{55}

=\frac{1}{4180}

Hence,  probability that first two balls are orange and third is green is \frac{1}{4180}

3 0
3 years ago
Work out the product of 2 2/3 and 1 5/7
joja [24]

Answer:

= 3 5/9

Step-by-step explanation:

2 2/3 × 1 5/7

= 8/3 × 12/7

= (8×12) / (3×7)

= 96/27

= 3 15/27

= 3 5/9

2 2/3 × 1 5/7 = 3 5/9

8 0
2 years ago
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