Question

The amount of water consumed each day by a healthy adult follows a normal distribution with a mean of 1.38 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:

1.48	1.50	1.42	1.40	1.56	1.80	1.62	1.84	1.42	1.39

At the 0.050 significance level, can we conclude that water consumption has increased? Calculate and interpret the p-value.

1. State the null hypothesis and the alternate hypothesis. (Round your answers to 2 decimal places.)
2. State the decision rule for 0.050 significance level. (Round your answer to 3 decimal places.)
3. Compute the value of the test statistic. (Round your intermediate and final answer to 3 decimal places.)
4. At the 0.050 level, can we conclude that water consumption has increased?
5. Estimate the p-value.

Answer 1

(a)

Ho: = 1.38

Ha: > 1.38

Null hypotheis sstates that the mean water consumption each day by a helthy adult is 1.38 lts.

Alternative hypotheis sstates that the mean water consumption each day by a helthy adult is greater than 1.38 lts.

(b) Decision rule

level of significance = 0.5

As the population sd is not given, we will calculate t stat.

The t-critical value for a right-tailed test,  df = 9,  significance level of α=0.05 : 1.833

IF t stat is > tcritical (1.833) we will reject the Null hypothesis.

Also if p value for the z stat if less than 0.05, we Reject the Ho. And if p value is > 0.05, we fail to reject the Ho.

(c) Test statistics

As the population sd is given, we will calculate z stat.

sample mean = sum of all terms / no of terms = 15.43 / 10 = 1.543

sample sd = s

data	data-mean	(data - mean)2
1.48	-0.063	0.003969
1.50	-0.043	0.001849
1.42	-0.123	0.015129
1.40	-0.143	0.020449
1.56	0.017	0.000289
1.80	0.257	0.066049
1.62	0.077	0.005929
1.84	0.297	0.088209
1.42	-0.123	0.015129
1.39	-0.153	0.023409

s = 0.163

As the population sd is not give, we will calculate t stat.

(d) As the t stat (3.162) > t critical (1.833), we reejct the Null hypothesis

(e) P value using excel formula = P value = TDIST (t statistics, df, tails)

p value = TINV(3.162, 9 ,1) = 0.0058

As the p value is less than 0.05, we reject the Null hypothesis