Question 10 0 days 0.72 days 2.5 days 3 days Question 20 The mean length of a ca
ID: 3325865 • Letter: Q
Question
Question 10
0 days
0.72 days
2.5 days
3 days
Question 20
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.
THESE ARE THE ONLY ONES I AM NOT SURE OF PLEASE HELP!! THANK YOU
0 days
0.72 days
2.5 days
3 days
Number of Days Absent Probability 0.60 0.20 0.12 0.04 0.04 0.00Explanation / Answer
10. Mode of the distribution = Number which occurs with the highest probability = 0 days
Option A is correct.
20. Degrees of freedom = n - 1 = 12 - 1 = 11
Hence,
For 11 degrees of freedom and a two tailed test:
p - Value = 2*P(t < -2.913) = 2*0.00705 = 0.0141
Since p - value is less than 0.02, we reject the null hypothesis and conclude a significant change. Therefore,
Yes, because the computed t lies in the rejection region.
Option A is correct
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