The amount of water consumed each day by a healthy adult follows a normal distribution with a mean of 1.28 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: |
1.90 1.62 1.78 1.30 1.68 1.46 1.46 1.66 1.32 1.52 |
At the 0.100 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 0.100 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 0.100 level, can we conclude that water consumption has increased? |
Fail to reject or Reject H0 Conclude that water consumption has Increased or not increased. |
e. | Estimate the p-value. |
p-value is: greater than 0.10 between 0.025 and 0.05 less than 0.0005 between 0.01 and 0.025 between 0.005 and 0.01 between 0.05 and 0.10 between 0.0005 and 0.005 |
a)Ho: mu = muo= 1.28 V/s H1: mu <muo= 1.28
Ho : The water consumption has not increased.
H1: The water consumption has increased.
b) The 10% level of significance.
c) To .the test statistics is given by
tn-1 = (Xbar- - muo)/S/√n
X bar = 15.7/10 = 1.57
S^2 = ( sum (X)^2 - n * (Xbar)^2 / n-1)
= (24.9868). - 10*(1.57)^2 / 10-1
= 24.9868- 24.649 / 9
= 0.3378/9
= 0.03753
S= 0.03753
t9 = ( 1.57- 1.28) / 0.1937/√10
= 4.7344. ( Calculated value)
Tabulated value t9,0.10 = 1.833
Calculated value > Tabulated value
Therefore we reject null hypothesis Ho.
d) Conclusion: The water consumption has may be increased.
e) p value = P( t9 < 4.7344) = 0.0000011226
alpha = 0.10
pvalue < alpha
ie we reject Ho . The water consumption may be increased.
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