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Dimas [21]
2 years ago
13

A men's softball league is experimenting with a yellow baseball that is easier to see during night games. One way to judge the e

ffectiveness is to count the number of errors. In a preliminary experiment, the yellow baseball was used in 10 games and the traditional white baseball was used in another 10 games. The number of errors in each game was recorded and is listed here. Can we infer that there are fewer errors on average when the yellow ball is used?
Yellow 5 2 6 7 2 5 3 8 4 9

White 7 6 8 5 9 11 8 3 6 10
Mathematics
1 answer:
sergiy2304 [10]2 years ago
5 0

Answer:

There cannot be equal errors in both and yellow has fewer errors.

Step-by-step explanation:

we can do paired t test for these two colours

H_0: \bar = \bar y\\H_a: \bar <  \bar y\\

(one tailed test)

df = 9

The data can be tabulated as follows:

Yellow white

 

5 7

2 6

6 8

7 5

2 9

5 11

3 8

8 3

4 6

9 10

t-Test: Paired Two Sample for Means  

 

Yellow white

Mean 5.1 7.3

Variance 5.877777778 5.788888889

Observations 10 10

Pearson Correlation -0.139051655  

Hypothesized Mean Difference 0  

df 9  

t Stat -1.908439275  

P(T<=t) one-tail 0.044341411  

t Critical one-tail 1.833112923  

P(T<=t) two-tail 0.088682822  

t Critical two-tail 2.262157158  

Since p value one tailed = 0.0443 and it is <0.05 our significance level, we reject null hypothesis.

There cannot be equal errors in both and yellow has fewer errors.

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kicyunya [14]

Answer: 2%, second option is correct.

Step-by-step explanation:

To state 1/50 in percent, divide 1 by 50, then multiply by 100

=( 1 ÷ 50) x 100

= 0.02 x 100

= 2%

I hope this helps, please mark as brainliest.

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2 years ago
1/2(6+3p)-3/4(8-10p)<br>Expand and simplify <br>show all steps
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2 years ago
Jenny earned a 77 on her most recent test. Jenny’s score is no less than 5 points greater than 4/5 of Terrance’s score. If t rep
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Read 2 more answers
A population consists of the following five values: 1, 3, 4, 4, and 6. List all samples of size 2 from left to right without rep
aniked [119]

Answer:

The sample of sizes 2 and their mean are given below.

Step-by-step explanation:

The population consist of  5 values, S = {1, 3, 4, 4, 6}.

The number of samples of size 2 (without replacement) that can be formed from these 5 values is:

{5\choose 2}=\frac{5!}{2!(5-2)!} =10

Th formula to compute the mean is:

\bar x=\frac{1}{n}\sum x_{i}

List the 10 samples and their mean as follows:

<u>Sample</u>                       <u>Mean</u>

(1, 3)                    \bar x=\frac{1}{2}[1+3]=\frac{4}{2}=2.0

(1, 4)                    \bar x=\frac{1}{2}[1+4]=\frac{5}{2}=2.5

(1, 4)                    \bar x=\frac{1}{2}[1+4]=\frac{5}{2}=2.5

(1, 6)                    \bar x=\frac{1}{2}[1+6]=\frac{7}{2}=3.5

(3, 4)                   \bar x=\frac{1}{2}[3+4]=\frac{7}{2}=3.5

(3, 4)                   \bar x=\frac{1}{2}[3+4]=\frac{7}{2}=3.5

(3, 6)                   \bar x=\frac{1}{2}[3+6]=\frac{9}{2}=4.5

(4, 4)                   \bar x=\frac{1}{2}[4+4]=\frac{8}{2}=4.0

(4, 6)                   \bar x=\frac{1}{2}[4+6]=\frac{10}{2}=5.0

(4, 6)                   \bar x=\frac{1}{2}[4+6]=\frac{10}{2}=5.0

8 0
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First you would do $325.49 divided by 20 and you would get $16.27. so you would need $16.27 a week.
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