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3 years ago

65 points! PLEASE HELP!!!

Mathematics

1 answer:

kow [346]3 years ago

8 0

Answer:

Holly's score is an outlier because the mean of the data is 83, and 30 is very far then 83.

The mean is 83

The median is 87

Removing the outlier change the mean because the mean became higher.

The mean was most affected by the outlier because the mean changed to 86. 31

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3 years ago

The number of goals scored by a hockey team in each of its first 10 games is 2,4,0,3,4,1,1 and 5. Find the mode of each of the n

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4 years ago

The bacterial strain Acinetobacter has been tested for its adhesion properties, which is believed to follow a normal distributio

pantera1 [17]

Answer:

a) Alternative hypothesis should be one sided. Because Null and Alternative hypotheses are:

$H_{0}$ : μ=2.66 dyne-cm.

$H_{a}$ : μ<2.66 dyne-cm.

b) the hypothesis that mean adhesion is at least 2.66 dyne-cm is true

Step-by-step explanation:

Let μ be the mean adhesion in dyne-cm.

Null and alternative hypotheses are:

$H_{0}$ : μ=2.66 dyne-cm.

$H_{a}$ : μ<2.66 dyne-cm.

First we need to calculate test statistic and then the p-value of it.

test statistic of sample mean can be calculated as follows:

t= $\frac{X-M}{\frac{s}{\sqrt{N} } }$ where

X is the sample mean

M is the mean adhesion assumed under null hypothesis (2.66 dyne-cm)

s is the standard deviation known (0.7 dyne-cm_2)

N is the sample size(5)

Sample mean is the average of 2.69, 5.76, 2.67, 1.62 and 4.12 dyne-cm, that is $\frac{2.69+5.76+2.67+1.62+4.12}{5}$ ≈ 3.37

using the numbers we get

t= $\frac{3.37-2.66}{\frac{0.7}{\sqrt{5} } }$ ≈ 2.27

The p-value is ≈ 0.043. Taking significance level as 0.05, we can conlude that sample proportion is significantly higher than 2.66 dyne-cm.

Thus, according to the sample the hypothesis that mean adhesion is at least 2.66 dyne-cm is true

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4 years ago