Question

The amount of water consumed each day by a healthy adult follows a normal distribution with a mean of 1.44 liters. A health campaign promotes the consumption of at least 2.0 liters per day. A sample of 10 adults after the campaign shows the following consumption in liters:

1.80	1.94	1.70	1.50	1.64	1.40	1.64	1.70	1.66	1.54

At the 0.100 significance level, can we conclude that water consumption has increased? Calculate and interpret the p-value.

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

2. State the decision rule for 0.100 significance level. (Round your answer to 3 decimal places.)

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

4. At the 0.100 level, can we conclude that water consumption has increased?

5. Estimate the p-value.

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 1.44
Alternative Hypothesis, Ha: μ > 1.44

Rejection Region
This is right tailed test, for α = 0.1 and df = 9
Critical value of t is 1.383.
Hence reject H0 if t > 1.383

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (1.652 - 1.44)/(0.1524/sqrt(10))
t = 4.399

It can be concluded that the consumption has increased.

P-value Approach
P-value = 0.0009
As P-value < 0.1, reject the null hypothesis.