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lubasha [3.4K]
3 years ago
12

A grass seed company conducts a study to determine the relationship between the density of seeds planted (in pounds per 500 sq f

t) and the quality of the resulting lawn. Eight similar plots of land are selected and each is planted with a particular density of seed. One month later the quality of each lawn is rated on a scale of 0 to 100. The sample data are given below.
Seed Density Lawn Quality
1 30
1 40
2 40
3 40
3 50
3 65
4 50
5 50

Required:
At the 1% level of significance, is there evidence of an association between seed density and lawn quality?
Mathematics
1 answer:
Setler79 [48]3 years ago
5 0

Answer:

There is no significant evidence to support the claim that an association between seed density and lawn quality exists.

Step-by-step explanation:

Given the data :

Seed Density Lawn Quality

1 30

1 40

2 40

3 40

3 50

3 65

4 50

5 50

Using technology to obtain the correlation Coefficient between seed density and lawn quality, the Coefficient, R = 0.6

H0 : ρ = 0

H1 : ρ ≠ 0

We can then calculate the test statistic :

Test statistic = r / √(1 - r²) / (n - 2)

Test statistic = 0.6 / √(1 - 0.6²) / (8 - 2)

Test statistic = 0.6 / 0.3265986

Test statistic = 1.837

Pvalue from test statistic : = 0.11583

α = 0.01

Pvalue > α ;

We fail to reject the null

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Lunch break: In a recent survey of 655 working Americans ages 25-34, the average weekly amount spent on lunch was $43.5
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Using the Empirical Rule, it is found that:

  • a) Approximately 99.7% of the amounts are between $35.26 and $51.88.
  • b) Approximately 95% of the amounts are between $38.03 and $49.11.
  • c) Approximately 68% of the amounts fall between $40.73 and $46.27.

------------

The Empirical Rule states that, in a <em>bell-shaped </em>distribution:

  • Approximately 68% of the measures are within 1 standard deviation of the mean.
  • Approximately 95% of the measures are within 2 standard deviations of the mean.
  • Approximately 99.7% of the measures are within 3 standard deviations of the mean.

-----------

Item a:

43.5 - 3(2.77) = 35.26

43.5 + 3(2.77) = 51.88

Within <em>3 standard deviations of the mean</em>, thus, approximately 99.7%.

-----------

Item b:

43.5 - 2(2.77) = 38.03

43.5 + 2(2.77) = 49.11

Within 2<em> standard deviations of the mean</em>, thus, approximately 95%.

-----------

Item c:

  • 68% is within 1 standard deviation of the mean, so:

43.5 - 2.77 = 40.73

43.5 + 2.77 = 46.27

Approximately 68% of the amounts fall between $40.73 and $46.27.

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