Question

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 2cm per week. A random sample of 48 shrubs has an average growth of 1.80cm per week with a standard deviation of 0.50cm. Is there overwhelming evidence to support the claim that the growth rate of the new shrub is less than 2cm per week at a 0.010 significance level

Answer 1

Answer:

$t=\frac{1.8-2}{\frac{0.5}{\sqrt{48}}}=-2.771$    

The degrees of freedom are given by:

$df=n-1=48-1=47$  

And the p value would be:

$p_v =P(t_{(47)}$    

Since the p value is lower than the significance level we have enough evidence to conclude that the true mean for this case for the growth rate is less than 2cm per week

Step-by-step explanation:

Information given

$\bar X=1.8$ represent the sample mean for the growth

$s=0.5$ represent the sample standard deviation    

$n=48$ sample size    

$\mu_o =2$ represent the value that we want to compare

$\alpha=0.01$ represent the significance level

t would represent the statistic    

$p_v$ represent the p value

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean is less than 2cm per week, the system of hypothesis are :    

Null hypothesis:$\mu \geq 2$    

Alternative hypothesis:$\mu < 2$    

Since we don't know the population deviation the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)    

Replacing the info given we got:

$t=\frac{1.8-2}{\frac{0.5}{\sqrt{48}}}=-2.771$    

The degrees of freedom are given by:

$df=n-1=48-1=47$  

And the p value would be:

$p_v =P(t_{(47)}$    

Since the p value is lower than the significance level we have enough evidence to conclude that the true mean for this case for the growth rate is less than 2cm per week