An employer wishes to compare typing speeds of graduates from 2 different study programs: A and B. Eight random graduates from course A type at 62, 85, 59, 64, 73, 70, 75, and 72 words per minute. Six random graduates from course B type at 75, 64, 81, 55, 69, and 58 words per minute. You are to find the 90% confidence interval for the mean difference between the typing speeds of course A and B. In order to do this, you will need to complete a 2 sample t-test for the difference in means.

Calculated value t =  0.614< 1.782

we accepted null hypothesis for 12 degrees of freedom at 0.05 level of significance .

Step-by-step explanation:

we will t-test t = x⁻- y⁻ /S√ n₁+n₂-2

Given An employer wishes to compare typing speeds of graduates from 2 different study programs: A and B

course A type at 62, 85, 59, 64, 73, 70, 75, and 72

mean of x is x⁻ = ∑x / n

mean of x = 62+85+59+64+73+70+75+ 72 /8 = 70

x⁻ = 70

course B type at 75, 64, 81, 55, 69, and 58

mean of y is  y⁻ = ∑y / n = 75+64+81+55+69+ 58/6 = 67

y⁻ =67

course A    course B

    x                    y             x- x⁻     (x-x⁻ )^2     y- y⁻     (y-y⁻ )^2    

    62                75            8             64          8            64

    85                64           15            225       -3             9

    59                81           -11            121          14            196      

    64                55           -6            36          -12          144          

    73                69          3                9            2            4

    70               58            0              0           -9            81

    75                              5              25

    72                              2               4

                                            ∑( (x-x⁻ )^2= 484           ∑( (y-y⁻ )^2= 498

Adding ∑( (x-x⁻ )^2+ ∑( (y-y⁻ )^2= 982

n₁+n₂-2 = 8+6-2=12

now S^2 =  982/ 12 =81.833

Null hypothesis :H₀:μ₁=μ₂ ( there is no difference between A and B)

Alternative hypothesis :H₁:μ₁≠μ₂

level of significance ∝=0.05

we will use t-test t =   0.614 ( see attachment )

The degrees of freedom = n₁+n₂-2 = 8+6-2=12

From tabulated value  t = 1.782

Calculated value t = 0.614< 1.782

we accepted null hypothesis for 12 degrees of freedom at 0.05 level of significance .

There is no difference between Course A and Course B

i = 2 / - 5

step-by-step explanation:

evaluation of ( 2 - 4 ) ( p + q ) when p = 2 and q = 5 i, by putting values of p and q,