The amount of water consumed each day by a healthy adult follows a normal distri

Question

The amount of water consumed each day by a healthy adult follows a normal distribution with a mean of 1.34 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:

1.35 1.40 1.72 1.46 1.58 1.50 1.40 1.35 1.36 1.50

At the 0.010 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:

H1: >

b. State the decision rule for 0.010 significance level. (Round your answer to 3 decimal places.)

Reject H0 if t > ______

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

Value of the test statistic_________

Explanation / Answer

there are n=10 observations

The amount of water consumed each day by a healthy adult follows a normal distribution with a mean of 1.34 liters with the variance unknown.

at 0.01 significance level we want to test whether water consumption has increased or not.

let X denotes the amount of water consumed each day by a healthy adult

so X~N(1.34,sigma2) where sigma is unknown.

we want to test whether consumption has increases or not.

a) so Null hypothesis is H0: u<=1.34

and alternative hypothesis is H1: u>1.34 [answer]

to test this we have a sample of n=10 observation with sample mean=Xbar=1.462 and sample standard deviation=s=0.1186

since the population standard deviation is unknown hence it is estimated by sample standard deviation s

the test statistic is given by

T=(Xbar-134)*sqrt(n)/s which under H0 follows a t distribution with df n-1

since the alternative hypothesis is right tailed hence H0 is rejected if t>talpha;n-1 where t is the value of test statistic and talpha;n-1 is the upper alpha point of a t distribution with df n-1

here alpha=level of significance=0.01 and n-1=9

so t0.01;9=2.821

b) hence reject H0 if t>2.821

c) now Xbar=1.462 s=0.1186 n=10

hence value of the test statistic is t=(1.462-1.34)*sqrt(10)/0.1186=3.253 [answer]

hence t>2.821. hence at 0.01 level of significance H0 is rejected and the conclusion is that the water consumption has increased.

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