Answer:
a) ![[-0.134,0.034]](https://tex.z-dn.net/?f=%5B-0.134%2C0.034%5D)
b) We are uncertain
c) It will change significantly
Step-by-step explanation:
a) Since the variances are unknown, we use the t-test with 95% confidence interval, that is the significance level = 1-0.05 = 0.025.
Since we assume that the variances are equal, we use the pooled variance given as
,
where
.
The mean difference
.
The confidence interval is

![= -0.05\pm 1.995 \times 0.042 = -0.05 \pm 0.084 = [-0.134,0.034]](https://tex.z-dn.net/?f=%3D%20-0.05%5Cpm%201.995%20%5Ctimes%200.042%20%3D%20-0.05%20%5Cpm%200.084%20%3D%20%5B-0.134%2C0.034%5D)
b) With 95% confidence, we can say that it is possible that the gaskets from shift 2 are, on average, wider than the gaskets from shift 1, because the mean difference extends to the negative interval or that the gaskets from shift 1 are wider, because the confidence interval extends to the positive interval.
c) Increasing the sample sizes results in a smaller margin of error, which gives us a narrower confidence interval, thus giving us a good idea of what the true mean difference is.
Answer:
hello attached below is the required table that is missing and the completed table as well
A) mean deviation = 0.1645
B) 2/3
Step-by-step explanation:
From the table we calculated the midpoint value (x) and c.f
N = 27
median class = 2.35 to 2.45
median = l + [ (N/2) - cf / F ] * H
= 2.35 + [ (13.5 - 13) / 6 ] *0.1 = 2.3583
hence mean deviation by median
= summation of fi |xi -M| / summation of fi
= 4.4417 / 27 = 0.1645
B ) probability of getting an odd number or prime number or both
the probability of an odd number = 1/2
also the probability of an even number = 1/2
while the probability of getting neither of them = 1/3
hence the probability of getting odd or prime or both
= 1/2 + 1/2 - 1/3 = 2/3
The bookshelf would be 4 feet 3 inches.