Question

Meteorologists in Texas want to increase the amount of rain delivered by thunderheads by seeding the clouds. Without seeding, thunderheads produce, on average, 300 acrefeet. The meteorologists randomly selected 30 clouds which they seeded with silver iodide to test their theory that average acrefeet is more than 300. The sample mean is 370.4 with a sample standard deviation of 300.1.
(a) Estimate the p-value.

Answer 1

Answer:

$t=\frac{370.4-300}{\frac{300.1}{\sqrt{30}}}=1.285$  

$p_v =P(z>1.285)=0.104$

Step-by-step explanation:

Data given and notation  

$\bar X=370.4$ represent the sample mean

$s=300.1$ represent the sample standard deviation for the sample  

$n=30$ sample size  

$\mu_o =300$ represent the value that we want to test  

$\alpha$ represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is more than 300, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 300$  

Alternative hypothesis:$\mu > 300$  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

$t=\frac{370.4-300}{\frac{300.1}{\sqrt{30}}}=1.285$  

P-value  

The degrees of freedom are given by:

$df = n-1=30-1=29$

Since is a right tailed test the p value would be:  

$p_v =P(t_{29}>1.285)=0.104$