Question

On the basis of a survey of 25 random typists, a confidence interval for the mean time needed to complete typing a page was (5.588; 6.412). Knowing that the distribution of the time needed to type a page is normal, and the sample standard deviation was equal to 1, determine the confidence level used for the calculations

Answer 1

95%

Explanation:

Confidence interval:

Mean (m) ± Tcrit * s/sqrt (n)

s = standard deviation = 1

Lower boundary = m - Tcrit * s/sqrt (n) - - - (1)

Upper boundary = m + Tcrit * s/sqrt (n) - - - (2)

m - Tcrit * s/sqrt (n) = 5.588 - - - (1)

m + Tcrit * s/sqrt (n) = 6.412 - - (2)

Subtracting (1) and (2)

-2Tcrit * s/sqrt (n) = - 0.824

Tcrit = [0.824 * sqrt(n)] / 2 * s

Recall:

s = 1

[0.824 * sqrt(n)] / 2 * 1

Sample size = n = 25

[0.824 * sqrt(25)] / 2

[0.824 * 5] ÷ 2

Tcrit = 2.06

Hence, we use the t table to obtain the significance level which gives a t value of 2.06 at (n-1) = (25 - 1) = 24 degrees of freedom.

Significance level α = 0.05

(1 - 0.05) * 100%

0.95 * 100% = 95%