Question

The radar detection distances in miles for 24 targets are shown below. The observation xi are for the standard system and the observations yi are for the new system. An initial look at the data indicates that the detectability of the targets varies from about 45 miles in some cases to over 55 miles in other cases, and this confirms the advisability of a paired experiment. Is the new radar system better than the standard radar system? Use a 0.05

Answer 1

This is a test for the paired-sample t-test.

There is a radar detection in miles of 24 targets.

Here, xi denotes the standard radar system and yi denotes the New radar system.

Now, We have to calculate the mean of two tests.

So, we are to test,

$\fn_cm Ho:\mu_D= 0 \ vs \ H1:\mu_D\neq0$

The differences between the two radar system are given below:-

Standard Radar System(Xi) New Radar System(Yi) |Differences(Di)|

48.40 51.14 2.74

47.73 46.48 -1.25

51.30 50.90    -0.40

50.49 49.82 -0.67

47.06 47.99 0.93

53.02 53.20 0.18

48.96 46.76    -2.20

53.02 54.44 1.42

51.09 49.85 -1.24

47.35 47.45 0.10

50.15 50.66 0.51

46.59 47.92 1.33

52.03 52.37 0.34

51.96 52.90 0.94

49.15 50.67 1.52

48.12 49.50 1.38

51.97 51.29 -0.68

53.24 51.60 -1.64

55.87 54.48 -1.39

45.60 45.62 0.02

   51.80 52.24 0.44

47.64 47.33 -0.31

49.90 51.13    1.23

55.89 57.86 1.97

Now, the sum of the differences is:-

$\sum_{i=1}^{24}Di=4.93$

Here, n=24

Now, the mean of the difference is= 4.93/24 =0.219583 and the standard deviation Sd= 1.22192

Now, the standard error is:-

$S.E.= Sd/\sqrt{n}$

So,

$S.E.= 1.22192/\sqrt{24}$

= 0.249423

So,

$\fn_cm t=\bar{D}/S.E.(\bar{D})$

= 0.219583/0.249423

=0.880364

$\fn_cm Now, \alpha=0.05$

From the test statistics, we see that Observed t is greater than $\fn_cm \alpha$.

So, We will accept Ho i.e. The new radar system and the standard radar system is the same.