The amount of water consumed each day by a healthy adult follows a normal distribution with a mean of 1.50 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:
1.52 | 1.64 | 1.66 | 1.40 | 1.82 | 1.70 | 1.90 | 1.45 | 1.78 | 1.92 |
At the 0.010 significance level, can we conclude that water consumption has increased? Calculate and interpret the p-value.
Reject H0 if t > _____
Value of the test statistic_____
______ H0 and conclude that water consumption has _______
Solution:
Here, we have to use one sample t test for the population mean.
State the null hypothesis and the alternate hypothesis.
The null and alternative hypotheses are given as below:
Null hypothesis: H0: The average amount of water consumed each day by a healthy adult is 1.50 liters.
Alternative hypothesis: Ha: The average amount of water consumed each day by a healthy adult is more than 1.50 liters.
H0: µ = 1.5 versus Ha: µ > 1.5
This is an upper tailed test.
State the decision rule for 0.010 significance level.
n = 10
df = n – 1 = 9
α = 0.01
Critical value = 2.8214
(by using t-table or excel)
Reject H0 if t > 2.8214
Compute the value of the test statistic.
The test statistic formula is given as below:
t = (Xbar - µ)/[S/sqrt(n)]
From given data, we have
µ = 1.5
Xbar = 1.679
S = 0.181012584
t = (Xbar - µ)/[S/sqrt(n)]
t = (1.679 – 1.5)/[ 0.181012584/sqrt(10)]
t = 3.1271
At the 0.010 level, can we conclude that water consumption has increased?
Test statistic value is greater than critical value, so we reject the null hypothesis.
Reject H0 and conclude that water consumption has increased.
There is sufficient evidence to conclude that the average amount of water consumed each day by a healthy adult is more than 1.50 liters.
Estimate the p-value.
P-value = 0.0061
(by using t-table)
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