Question

The mean consumption of bottled water by a person in the United States is 28.5 gallons per year. You believe that a person consume more than 28.5 gallons in bottled water per year. A random sample of 100 people in the United States has a mean bottled water consumption of 27.8 gallons per year with a standard deviation of 4.1 gallons. At α = 0.10 significance level can you reject the claim?

Answer 1

Answer:

$t=\frac{27.8-28.5}{\frac{4.1}{\sqrt{100}}}=-1.707$    

$p_v =P(t_{99}<-1.707)=0.0455$

If we compare the p value with a significance level for example $\alpha=0.1$ we see that $p_v<\alpha$ so we can conclude that we reject the null hypothesis, so there is not enough evidence to conclude that the mean for the consumption is less than 28.5 gallons at 0.1 of significance, so we can reject the claim that person consume more than 28.5 gallons.

Step-by-step explanation:

Data given and notation    

$\bar X=27.8$ represent the mean for the account balances of a credit company

$s=4.1$ represent the population standard deviation for the sample    

$n=1000$ sample size    

$\mu_o =28.5$ represent the value that we want to test  

$\alpha=0.1$ represent the significance level for the hypothesis test.  

t 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 determine if the mean for the person consume is more than 28.5 gallons, the system of hypothesis would be:    

Null hypothesis:$\mu \geq 28.5$    

Alternative hypothesis:$\mu < 28.5$    

We don't know the population deviation, so for this case we can use the 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:    

$t=\frac{27.8-28.5}{\frac{4.1}{\sqrt{100}}}=-1.707$    

Calculate the P-value    

First we need to calculate the degrees of freedom given by:

$df=n-1=100-1=99$

Since is a one-side lower test the p value would be:    

$p_v =P(t_{99}<-1.707)=0.0455$

In Excel we can use the following formula to find the p value "=T.DIST(-1.707,99)"  

Conclusion    

If we compare the p value with a significance level for example $\alpha=0.1$ we see that $p_v<\alpha$ so we can conclude that we reject the null hypothesis, so there is not enough evidence to conclude that the mean for the consumption is less than 28.5 gallons at 0.1 of significance, so we can reject the claim that person consume more than 28.5 gallons.