Page 428 - DMTH404_STATISTICS

Statistics

Notes Solution.

We note that s is not given and n = 6 (< 30),  t - test is applicable.
Using sample information we have

+
+
+
+
2.75 5.25 4.50 2.50 4.25 3.25
+
X = = 3.75.
6
X - 3.75
i
To calculate s, we define u = = (X - 3.75 ) 4´
i i
0.25
X 2.75 5.25 4.50 2.50 4.25 3.25
i
u i - 4 6 3 - 5 2 - 2
u 2 16 36 9 25 4 4
i
94
2
From the above table å u = 94 . Therefore, s = 0.25 = 1.085
i
6 1
-
We have to test H : m = 3.50 against H : m ¹ 3.50.
0 a
-
X  0
The test statistic ~ t - distribution with (n - 1) d.f.
s / n

3.75 3.50
-
t = = 0.564
Thus, cal
1.085/ 6
Further, the critical value of t, from table at 5% level of significance and with 5 d.f. is 2.571. Since
t is less than this value, there is no evidence against at 5% level of significance.
cal

Example 15: Daily sales figures of 40 shopkeepers showed that their average sales and
standard deviation were Rs 528 and Rs 600 respectively. Is the assertion that daily sales on the
average is Rs 400, contradicted at 5% level of significance by the sample?
Solution.

Since n > 30, standard normal test is applicable. It is given that n = 40, X = 528 and S = 600.
We have to test H :  = 400 against H :   400.
0 a
-
528 400
z cal = = 1.35.
600/ 40
Since this value is less than 1.96, there is no evidence against H at 5% level of significance.
0
Hence, the given assertion is not contradicted by the sample.

420 LOVELY PROFESSIONAL UNIVERSITY