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 3cm per week. A random sample of 50 shrubs has an average growth of 2.90cm per week with a standard deviation of 0.30cm. Is there overwhelming evidence to support the claim that the growth rate of the new shrub is less than 3cm per week at a 0.100 significance level

Answer 1

Answer:

$t=\frac{2.9-3}{\frac{0.3}{\sqrt{50}}}=-2.357$    

$df=n-1=50-1=49$  

$p_v =P(t_{(49)}$  

Since the p value is less than the significance level of 0.1 we have enough evidence to reject the null hypothesis and we can conclude that the true growth rate is significantly less than 3 cm per week

Step-by-step explanation:

Information given

$\bar X=2.90$ represent the sample mean for the growth

$s=0.30$ represent the sample standard deviation

$n=50$ sample size  

$\mu_o =3$ represent the value to check

$\alpha=0.1$ represent the significance level

t would represent the statistic

$p_v$ represent the p value for the test

System of hypothesis

We want to verify if the true mean for the growth us less than 3cm per week, the system of hypothesis are:  

Null hypothesis:$\mu \geq 3$  

Alternative hypothesis:$\mu < 3$  

We don't know the population deviation so we can use the t statistic:

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

Replacing the info given we got:

$t=\frac{2.9-3}{\frac{0.3}{\sqrt{50}}}=-2.357$    

The degrees of freedom are given by:

$df=n-1=50-1=49$  

The p value for this case would be given by:

$p_v =P(t_{(49)}$  

Since the p value is less than the significance level of 0.1 we have enough evidence to reject the null hypothesis and we can conclude that the true growth rate is significantly less than 3 cm per week