Question

Answer 1

Answer

(A) mean = (sum of all data values)/(total number of data values)

= (28+27+25+26+28+26+29+25)/8

= 107/4

= 26.75

Standard deviation = $\sqrt{\sum(x_i-\bar{x})^2/(n-1)}$

where x(bar) = 26.75, n(sample size) = 8 and xi are given data values

V[(28 - 26.75)2 (27- 26.75)2 +. (29 - 26.75)2 (25 - 26.75)21/(8 - 1)

15.5/7

2.2143

1.488

(B) Company is advertising that the new tire's life expectancy has increased, i.e. mean value is more than 26 (thousand miles)

So, null hypothesis is

Ho : μ = 26

And alternate hypothesis is

H1 : μ > 26

(C) degree of freedom = n- 1= 8 - 1 = 7

alpha level is 0.01

using t distribution table for right tailed hypothesis, t critical value =2.998

test statistics t $= (\bar{x}-\mu)/(s/\sqrt{n})$

(26.75 -26)/(1.488/V8)

0.75/0.5261

1.43

Using t critical value approach, it is clear that the t calculated is less than t critical value, this means that it falls in the null hypothesis acceptance region

Therefore, we failed to reject the null hypothesis and we can conclude that the new tire's life expectancy has not increased.

(D) Using t calculated value of 1.43 and df = 7

check t calculated in the column and df = 7 in the row in the t distribution table, select the intersecting cell

this implies, p value = 0.0979

p value is greater than 0.01 significance level, failed to reject the null hypothesis as p value insignificance

Therefore, we failed to reject the null hypothesis and we can conclude that the new tire's life expectancy has not increased.