Question

The amount of water consumed each day by a healthy adult follows a normal distribution with a mean of 1.22 liters. A sample of 10 adults after the campaign shows the following consumption in liters. A health campaign promotes the consumption of at least 2.0 liters per day:

1.44 1.80 1.20 1.32 1.44 1.32 1.90 1.64 1.32 1.60

At the 0.050 significance level, can we conclude that water consumption has increased? Calculate and interpret the p-value.

a. State the null hypothesis and the alternate hypothesis. (Round your answers to 2 decimal places.)


H0: μ ≤
H1: μ >

b. State the decision rule for 0.050 significance level. (Round your answer to 3 decimal places.)

Reject H0 if t >

c.

Compute the value of the test statistic. (Round your intermediate and final answer to 3 decimal places.)

Value of the test statistic

d. At the 0.050 level, can we conclude that water consumption has increased?



H0 and conclude that water consumption has

.

e. Estimate the p-value.

p-value is

Answer 1

Answer:

a) Null hypothesis:$\mu \leq 1.22$    

Alternative hypothesis:$\mu > 1.22$    

b) $df= n-1= 10-1=9$

Then we need to look at the t distribution with 9 degrees of freedom for a value that accumulates 0.05 of the area on the right tail and on this case this value is:

$t_c = +1.833$

And we can use the following excel code to find it: "=T.INV(1-0.05,9)"

c) $t=\frac{1.498-1.22}{\frac{0.229}{\sqrt{10}}}=3.839$      

d) For this case our calculated value is higher than our critical value so then at 0.05 of significance we can conclude that we can reject the null hypothesis and the mean seems to be higher than 1.22.

e) $p_v =P(t_{9}>3.839)=0.00199$    

Step-by-step explanation:

Previous concepts  and data given  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Data given: 1.44 1.80 1.20 1.32 1.44 1.32 1.90 1.64 1.32 1.60

We can calculate the sample mean and deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s = \frac{\sum_{i=1}^n (x_i -\bar x)^2}{n-1}$

$\bar X=1.498$ represent the sample mean    

$s=0.229$ represent the sample standard deviation  

n=10 represent the sample selected  

$\alpha=0.05$ significance level    

Part a: State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if we have significant difference on the mean of 200, the system of hypothesis would be:    

Null hypothesis:$\mu \leq 1.22$    

Alternative hypothesis:$\mu > 1.22$    

If we analyze the size for the sample is < 30 and we don't know the population deviation so 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".  

Part b: Critical region

For this case we need to find the degrees of freedom first given by:

$df= n-1= 10-1=9$

Then we need to look at the t distribution with 9 degrees of freedom for a value that accumulates 0.05 of the area on the right tail and on this case this value is:

$t_c = +1.833$

And we can use the following excel code to find it: "=T.INV(1-0.05,9)"

Part c: Calculate the statistic  

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

$t=\frac{1.498-1.22}{\frac{0.229}{\sqrt{10}}}=3.839$      

Part d

For this case our calculated value is higher than our critical value so then at 0.05 of significance we can conclude that we can reject the null hypothesis and the mean seems to be higher than 1.22.

Part e: P-value  

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

$df=n-1=10-1 = 9$

Then since is a right tailed test the p value would be:    

$p_v =P(t_{9}>3.839)=0.00199$