Twenty graduate students in business were asked how many credit hours they were taking in the current quarter. Their responses are shown as follows:

Student Number Credit Hours Student Number Credit Hours Student Number Credit Hours
1 2 8 8 15 10
2 7 9 12 16 6
3 9 10 11 17 9
4 9 11 6 18 6
5 8 12 5 19 9
6 11 13 9 20 10
7 6 14 13

Required:
a. Determine the mean, median, and mode for this sample of data. Write a sentence explaining what each means.
b. It has been suggested that graduate students in business take fewer credits per quarter than the typical graduate student at this university. The mean for all graduate students is 9.1 credit hours per quarter, and the data are normally distributed. Set up the appropriate null and alternative hypotheses, and determine whether the null hypothesis can be rejected at a 95 percent confidence level.

Since the calculated value of t= -1.016 does not fall in the critical region t ≥ 1.729 we conclude that the mean for all graduate students is 8.5 and fail to reject the null hypothesis.

Explanation:

Student Credit Student Credit Student Credit

Number Hours Number Hours Number Hours

1 2 8 8 15 10

2 7 9 12 16 6

3 9 10 11 17 9

4 9 11 6 18 6

5 8 12 5 19 9

6 11 13 9 20 10

7 6 14 13

The mean is the value which gives the average value of all the data.It is obtained by adding the value and dividing it by the number of values.

x`= ∑x/n

x`=2 +8 +10+ 7+ 12 + 6+ 9 +11+ 9 +9 +6 + 6 +8 + 5+ 9 +11+ 9+ 10+ 6 + 13/20

x`= 8.3

Median is the middle value which divides the data into two equal halves.

2 5,6,6,6,6, 7,8 ,8, 9,9, 9,9 ,9 ,10, 10,11,11, 12 , 13

For even number

Median = n/2=10 th.

Here the 10th value is 9 when the data is arranged in ascending order.

Mode=9

Mode gives the repeated value .

Let the hypotheses be

H0 : u = 8.5 vs Ha: u ≥ 8.5

The mean for all graduate students is 8.5

against the claim that

The mean for all graduate students is greater than 8.5

The sample mean x`= 8.5 and standard deviation s= 2.64

Putting in the test statistic

t= x`- u / s/ √n

t= 8.5-9.1/2.64/√20

t= -0.6/2.64/4.472

t= -1.016

The critical region for alpha = 0.05 for one tailed test with n-1= 2-0-1= 19 degrees of freedom is t ≥ 1.729

Since the calculated value of t= -1.016 does not fall in the critical region t ≥ 1.729 we conclude that the mean for all graduate students is 8.5 and fail to reject the null hypothesis.