Question

Many people use scanners to read documents and store them into pdf file. To help determine which brand of scanner to buy, a student conducts an experiment in which 8 different documents were scanned by each of the two scanners that he is interested in. He records the number of errors made by each. Can he infer that Brand A (the more expensive scanner) is better than Brand B

Answer 1

The data is missing in the question. The data is provided below :

Document : 1     2      3      4      5     6      7      8

Brand A       17  29    18    14    21   25    22    29

Brand B       21  38    15    19    22   30    31   37

Solution :

State of the hypothesis of the null hypothesis and alternate hypothesis.

Null hypothesis : $h_A = h_B$

Alternate hypothesis : $h_A > h_B$

These hypothesis is a one tailed test. The null hypothesis will get rejected when the mean difference between the sample means is very small.

Significance level = 0.05

Therefore the standard error is :  $SE = \sqrt{(\frac{s^2_1}{n_1})+(\frac{s^2_2}{n_2})}$

                                                         = 3.602

And the degree of freedom, DF = 14

$t=\frac{(x_1-x_2)-d}{SE}$

 = -1.319

Here, $s_1$ = standard deviation of the sample 1

        $s_2$ = standard deviation of the sample 2

         $n_1$ = size of the sample 1

        $n_2$ = size of the sample 2

         $x_1$ = mean of the sample 1

        $x_2$ = mean of the sample 2      

          d = the hypothesis difference between the population mean

The observed difference in a sample means t static of -1.32. From t distribution calculator to determine P($t \leq -1.32$) = 0.1042  

Since the P value of 0.1042 is greater than significance level o 0.05, we therefore cannot reject the null hypothesis.

But from the test, we have no sufficient evidence that supports that Brand A is better than Brand B.