The mean length of a candy bar is 43 millimeters. There is concern that the settings of the machine cutting the bars have changed. Test the claim at the 0.02 level that there has been no change in the mean length. The alternate hypothesis is that there has been a change. Twelve bars (n = 12) were selected at random and their lengths recorded. The lengths are (in millimeters) 42, 39, 42, 45, 43, 40, 39, 41, 40, 42, 43, and 42. The mean of the sample is 41.5 and the standard deviation is 1.784. Computed t = -2.913. Has there been a statistically significant change in the mean length of the bars?
Yes, because the computed t lies in the rejection region.
No, because the information given is not complete.
No, because the computed t lies in the area to the right of -2.718.
Yes, because 43 is greater than 41.5.

The correct option is;

Yes because the computed t lies in the rejection region

Step-by-step explanation:

Here we have

The calculated mean of the values given as

= Sample mean = 41.5

μ = Population mean = 43

σ = Standard deviation 1.784

The confidence interval is

At 0.02 the critical t is 2.718 with a p value of 0.014 which is less than the confidence level at 0.02. We therefore reject the null hypothesis and we fail to reject the alternate hypothesis which states that there has been a change in the mean length

Therefore the answer is Yes because the computed t lies in the rejection region.

34.5 cm

step-by-step explanation:

generally for a circular segment of angle, θ° and radius r

arc length

= (θ/360) x circumference

= (θ/360) x 2πr

in our case, θ = 60°, π=3.142 and r = 33 cm

substitute into above formula

arc length

= (60/360) x 2(3.142)( 33) = 34.5 cm