Question

The amount of water consumed each day by a healthy adult follows a normal distribution with a mean of 1.46 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:

1.78   1.90   1.38   1.60   1.80   1.30    1.56    1.90    1.90    1.68

At the 0.050 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.

H0: μ ≤ _____

H1: μ >_____

B.) State the decision rule for 0.050 significance level. (round to 3 decimal places)

C.) Compute the value of the test statistic. (round to 3 decimal places)

D.) At the 0.050 level, can we conclude that water consumption has increased?

E.) Estimate the p-value.

Answer 1

Solution:

Given:

The amount of water consumed each day by a healthy adult follows a normal distribution with a mean of 1.46 liters.

That is: Population Mean =

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:

1.78   1.90   1.38   1.60   1.80   1.30    1.56    1.90    1.90    1.68

We have to test if water consumption has increased at the 0.050 significance level.

Part A.) State the null hypothesis and the alternate hypothesis.

Since we have to test water consumption has increased , this is directional and right tailed test.

Thus H0 and H1 are:

Part B.) State the decision rule for 0.050 significance level.

Since population standard deviation is unknown and sample size n is small, we use t distribution.

Thus find t critical value.

df = n - 1 = 10 - 1 = 9

significance level = 0.050

Look in t table for df = 9 and one tail area = 0.050 and find t critical value.

thus t critical value = 1.833

Thus decision rule is: Reject H0, when t test statistic value > 1.833 , otherwise we fail to reject H0.

Part C.) Compute the value of the test statistic.

Formula:

where

Thus we need to make following table:

x	x^2
1.78	3.1684
1.90	3.6100
1.38	1.9044
1.60	2.5600
1.80	3.2400
1.30	1.6900
1.56	2.4336
1.90	3.6100
1.90	3.6100
1.68	2.8224

and

Thus t test statistic value is:

Part D.) At the 0.050 level, can we conclude that water consumption has increased?

Since t test statistic value = > t critical value = 1.833 , we reject H0.

Thus At the 0.050 level of significance, we conclude that, water consumption has increased.

Part E) Estimate the p-value.

df = 9

Look in t table for df = 9 row and find the interval in which t = 3.202 fall and then find corresponding one tail area.

That would be the range of p-value.

t = 3.202 fall between 2.821 and 3.250

thus corresponding one tail area is between 0.005 and 0.01

thus p-value is:

0.005 < p-value < 0.01

To get exact p-value, we use following excel command:

=T.DIST.RT( x , df)

=T.DIST.RT( 3.202 , 9 )

=0.0054

Thus p-value = 0.0054