Question

The following data from a random sample represents the average daily energy intake in kJ for each of eleven healthy women:

5260    5470     5640     6180     6390     6515     6805     7515     7515    8230     8770     

Interest centred on comparing these data with an underlying mean daily energy intake of 7725 kJ This was the recommended daily intake. Departures from this mean in either direction were considered to be of interest. Assuming that the population is normal and the population variance is unknown. An appropriate two-tailed hypothesis test at the 5% level of significance was conducted. The null hypothesis is H₀: m = 7725. In order to make a decision as to whether to accept or reject the null hypothesis, we need to compare the observed test statistic with the critical value. What is this critical value |t_a/2| ?

Answer 1

Step 1:

Ho :

Ha:

Step 2:

n = 11

sample mean = sum of all terms / no of terms = 74290 / 11 = 6753.64

sample standard deviation = s

data	data-mean	(data - mean)2
5260	-1493.6364	2230949.695405
5470	-1283.6364	1647722.407405
5640	-1113.6364	1240186.031405
6180	-573.6364	329058.71940496
6390	-363.6364	132231.43140496
6515	-238.6364	56947.33140496
6805	51.3636	2638.21940496
7515	761.3636	579674.53140496
7515	761.3636	579674.53140496
8230	1476.3636	2179649.479405
8770	2016.3636	4065722.167405

As the data is normally distributed and population sd is not given, we will calculate t statistics

t statistics = -2.821

Step 3:

Level of significane = 0.05

df = 10

The t-critical values for a two-tailed test, for a significance level of α=0.05

tc = − 2.228 and tc = 2.228

As t stat (-2.821) falls in the rejection area, we reject the Null hypothesis.