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grin007 [14]
3 years ago
8

An apple farm claims that the average weight of all their apples is 8oz. A sample of 20 apples were picked that had a mean of 7.

77oz and standard deviation of 0.95oz. We want to test whether the mean apple weight is different from 8oz. Assume that the apple weights are normally distributed.
Mathematics
1 answer:
Nimfa-mama [501]3 years ago
5 0

Answer:

The mean apple weight is different from 8oz

Step-by-step explanation:

Claim : An apple farm claims that the average weight of all their apples is 8oz.

H_0:\mu \neq 8\\H_a:\mu = 8

n = 20

Standard deviation =s = 0.95

Since n < 30

So we will use t-test

x = 7.77

t =\frac{x-\mu}{\frac{s}{\sqrt{n}}}

Substitute the values :

t =\frac{7.77-8}{\frac{0.95}{\sqrt{20}}}

t =−1.0827

Since we are not given the significance level

So, we will take 5%

So,α = 0.05

Degree of freedom = df = n-1 = 20-1 = 19

So, using t table

t_{\frac{\alpha}{2} , df}=2.093

Since t critical > t statistic

So, we accept the null hypothesis.

So, The mean apple weight is different from 8oz

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KIM [24]

The z-score of a person who scored 128 on the exam is equal to 2.9

<h3>What is a z-score?</h3>

A z-score is also referred to as a z-value or standard score and it can be defined as a measure of the distance between a raw score and the mean, when standard deviation units are used.

Mathematically, the z-score of a given sample score in a normal distribution can be calculated by using this formula:

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Substituting the given parameters into the formula, we have;

Z-score, z = (128 - 70)/(20)

Z-score, z = 58/20

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Read more on z-scores here: brainly.com/question/26714379

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Please help me with the last question
Greeley [361]
You'd solve this by creating the equation
20 + 6x - 1 + x + 14 = 180
33 + 7x = 180
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÷ 7
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And then you'd substitute in the value of x, so
6 × 21 = 126
126 - 1 = 125° = Angle A
21 + 14 = 35° = Angle C
I hope this helps!
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