In: Statistics and Probability
Old | New |
16 | 12 |
9 | 9 |
17 | 15 |
10 | 8 |
9 | 5 |
4 | 4 |
30 | 25 |
29 | 21 |
A researcher believes she has designed a keyboard that is more efficient to use than a standard keyboard. In order to help decide if this is the case, typing speeds were taken for 16 different people. Eight of them are assigned to use old key board, and the other eight people are assigned to use the new key board. The lengths of time, in minutes, for each of the people to type a pre-selected manuscript are listed below. Assume the two population distributions are normal. Use the data in the Table to determine if the original keyboard yields slower times. Use a significance level of α = 0.05. Assume the two group has the same unknown population variance
: Population Mean length of time to type a pre-selected manuscript with old(original) key board
: Population Mean length of time to type a pre-selected manuscript with new key board
Claim : Original key board yields lower times : Mean length of Time taken with old key board is more than new key board or - > 0
Null hypothesis : Ho : - = 0
Alternate Hypothesis : Ha : - > 0
Right Tailed test :
The population variances of the two groups assumed equal.
Sample mean
Sample Standard deviation s:
For Old Key Board
n1 : Sample size =8
Old:x1 | |||
16 | 0.5 | 0.25 | |
9 | -6.5 | 42.25 | |
17 | 1.5 | 2.25 | |
10 | -5.5 | 30.25 | |
9 | -6.5 | 42.25 | |
4 | -11.5 | 132.25 | |
30 | 14.5 | 210.25 | |
29 | 13.5 | 182.25 | |
Total | 124 | 642 | |
Sample Mean: | 124/8 = 15.5 |
Sample Mean length time to type a pre-selected manuscript : = 15.5
Sample standard deviation to type a pre-selected manuscript : s1
For new Key board
New: x2 | |||
12 | -0.375 | 0.140625 | |
9 | -3.375 | 11.39063 | |
15 | 2.625 | 6.890625 | |
8 | -4.375 | 19.14063 | |
5 | -7.375 | 54.39063 | |
4 | -8.375 | 70.14063 | |
25 | 12.625 | 159.3906 | |
21 | 8.625 | 74.39063 | |
99 | 395.875 | ||
Sample Mean | 99/8 = 12.375 |
Sample Mean length time to type a pre-selected manuscript : = 12.375
Sample standard deviation to type a pre-selected manuscript : s2
Degrees of freedom : df = n1 +n2 -2 = 8+8-2 =14
=0.05
For right tailed test :
As P-Value i.e. is greater than Level of significance i.e
(P-value:0.2399 > 0.05:Level of significance); Fail to Reject
Null Hypothesis
There is not sufficient evidence to conclude that original keyboard
yields slower times