Question

5) The amount of water consumed each day by a healthy adult follows a normal distribution with a mean of 1.56 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:

1.88 1.74 1.98 1.70 1.86 1.72 1.74 1.98 1.68 1.50

At the 0.025 significance level, can we conclude that water consumption has increased? Interpret the p-value.

a) State the null hypothesis and the alternate hypothesis.

b) State the decision rule for 0.025 significance level. (Round your answer 3 decimal places)

Reject HO 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) State the decision rule for 0.025 significance level. (Round your answer to 3 decimal places) reject HO if t > _____

Answer 1

= (1.88 + 1.74 + 1.98 + 1.70 + 1.86 + 1.72 + 1.74 + 1.98 + 1.68 + 1.5)/10 = 1.778

s = sqrt(((1.88 - 1.778)^2 + (1.74 - 1.778)^2 + (1.98 - 1.778)^2 + (1.70 - 1.778)^2 + (1.86 - 1.778)^2 + (1.72 - 1.778)^2 + (1.74 - 1.778)^2 + (1.98 - 1.778)^2 + (1.68 - 1.778)^2 + (1.5 - 1.778)^2)/9) = 0.1483

H₀:   = 1.56

H₁:   > 1.56

B) At alpha = 0.025, the critical value is t^* = 2.262

Reject H₀ if t > 2.262

C) The test statistic t = ()/(s/)

= (1.778 - 1.56)/(0.1483/)

= 4.649

D) Since the test statistic value is greater than the critical value (4.649 > 2.262), so we should reject the null hypothesis.

So at 0.025 significance level we can conclude that the water consumption has increased.