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yarga [219]
3 years ago
11

A biologist was interested in determining whether sunflower seedlings treated with and an extract from Vinca minor roots resulte

d in a lower average height of sunflower seedlings that the standard height of 15.7 cm. The biologist treated a random sample of 33 seedlings with the extract and subsequently measured the height of those seedlings. At the 0.01 significance level, is there evidence that the true average height of the seedlings treated with an extract from Vinca minor roots is less than 15.7 cm?
Height
15.5
15.8
15.7
15.1
15.1
15.5
15.2
15.7
15.8
15.4
16.2
15.5
16.2
15.5
15.4
16.3
14.9
15.3
15.1
16.1
15.3
15.4
15.1
15.3
14.6
15.1
15.0
15.3
15.8
15.5
14.8
15.2
14.8

a. State the null and alternative hypotheses.
b. Report the value of the test statistic. Round answer to 2 decimal places. (Either calculate or use software such as minitab)
c. Using the p-value, do you reject the null hypothesis or fail to reject the null hypothesis? Explain your decision.
d. Based on your decision in part (c), write a conclusion within the context of the problem.
Mathematics
1 answer:
Aleksandr-060686 [28]3 years ago
4 0

Answer:

Kindly check explanation

Step-by-step explanation:

H0 : μ = 15.7

H1 : μ < 15.7

This is a one sample t test :

Test statistic = (xbar - μ) ÷ (s/√(n))

n = sample size = 33

Using calculator :

The sample mean, xbar = 15.41

The sample standard deviation, s = 0.419

Test statistic = (15.41 - 15.70) ÷ (0.419/√(33))

Test statistic = - 3.976

Using the Pvalue calculator :

Degree of freedom, df = n - 1 ; 33 - 1 = 32

Pvalue(-3.976, 32) = 0.000187

Decison region :

Reject H0 if Pvalue < α

Since Pvalue < α ; we reject H0

There is significant evidence to conclude that the true average height of the seedlings treated with an extract from Vinca minor roots is less than 15.7 cm.

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Then, we have to subtract this number from x^2, so we have

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x = \dfrac{3-51}{2} = \dfrac{-48}{2} = -24

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