From a random sample of 20 bars selected at random from those produced, calculations gave a mean weight of = 52.46 grams and standard deviation of s = 0.42 grams. Assuming t distribution is followed, construct a 90% confidence interval for the mean weight of bars produced, giving the limits to two decimal places.

(52.30 ; 52.62)

Step-by-step explanation:

Given :

Sample size, n = 20

Mean, xbar = 52.46

Standard deviation, s = 0.42

We assume a t - distribution

The 90% confidence interval

The confidence interval relation :

C.I = xbar ± Tcritical * s/√n

To obtain the Tcritical value :

Degree of freedom, df = n - 1 ; 20 - 1 = 19 ; α = (1 - 0.90) /2 = 0.1/2 = 0.05

Using the T-distribution table, Tcritical = 1.729

We now have :

C.I = 52.46 ± (1.729 * 0.42/√20)

C. I = 52.46 ± 0.1624

C.I = (52.30 ; 52.62)

-9

step-by-step explanation: