Question

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

   1.78   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. (Round your answers to 2 decimal places.)

H0: μ ≤
H1: μ >

b.	State the decision rule for 0.050 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 0.050 level, can we conclude that water consumption has increased?

    (Click to select)  Reject  Fail to reject  H₀ and conclude that water consumption has  (Click to select)  not increased  increased  .

e.	Estimate the p-value.

p-value is

(Click to select)  less than 0.0005  between 0.01 and 0.025  between 0.005 and 0.01  between 0.05 and 0.10  greater than 0.10  between 0.025 and 0.05  between 0.0005 and 0.005

Answer 1

a)  

H0: μ ≤ 1.46

H1: μ > 1.46 For given data , Sample size:10 Mean (x̄): 1.68 Standard deviation (s):0.21726 b)   Reject H0 if t> 1.833 Degree of freedom = n - 1 = 9 t critical value with degree of freedom 9 and level of signficance 0.05 for right tailed alternative hypothesis is 1.833 t critical value by using t table. c) We can use here t test statistic formula is    =3.202 the value of the test statistic =3.202 d)   Reject H0 and conclude that water consumption has increased e) P value is 0.0054 ..................by using TDIST(3.202,9,1)