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A horticulturist working for a large plant nursery is conducting experiments on the growth rate of a new shrub. Based on previous research, the horticulturist feels the average weekly growth rate of the new shrub is 4cm per week. A random sample of 40 shrubs has an average growth of 1.50cm per week with a standard deviation of 5cm. Is there overwhelming evidence to support the claim that the growth rate of the new shrub is less than 4cm per week at a 0.200 significance level?
Find the value of the test statistic. Round your answer to three decimal places, if necessary.

Sagot :

Answer:

Following are the answer to the given question:

Step-by-step explanation:

Following are the null and alternative Hypothesis:

[tex]Null \ Hypothesis: \mu = 4\\\\Alternative \ Hypothesis: \mu < 4[/tex]

Rejection Zone

This is left tailed test, for α = 0.2 and df = 39

[tex]critical\ value(t) = -0.851\\\\reject\ H_0 \ when = t < -0.851[/tex]

Testing statistic:

[tex]t = \frac{(\bar{x} - \mu)}{( \frac{s}{\sqrt{n}})}[/tex]

  [tex]= \frac{1.5 - 4}{\frac{5}{\sqrt{40}}}\\\\= \frac{-2.5}{\frac{5}{6.32}}\\\\= \frac{-2.5}{0.79}\\\\ = -3.162[/tex]

[tex]P-value = 0.0015[/tex]

P-value < 0.2, reject the null hypothesis.

Approach to the Rejection Zone:

The null hypothesis is rejected since the test statistic t is outside of the critical range value. Only at the 0.200 significance mark, there's significant evidence that its new shrub's growth rate is less than 4 cm per week.

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