Mba510 Week 5 Problem Set
Lind Chapter 9; Exercise 12
The American Sugar Producers Association wants to estimate the mean yearly sugar consumption.
A sample of 16 people reveals the mean yearly consumption to be 60 pounds with a standard deviation of 20 pounds.
a. What is the value of the population mean? What is the best estimate of this value?
The value of population mean is unknown. The best estimate of this value is the sample mean of 60 pounds.
b. Explain why we need to use the t distribution. What assumption do you need to make?
According to Lind, et al. (2005), when population standard deviation is unknown, and the sample is smaller than 30, a t distribution should be used. We need to assume that the sample is from a normal population (pp. …show more content…
AM shift: n₁ = 54; xbar₁ = 345; s₁ = 21
PM shift: n₂ = 60; xbar₂ = 351; s₂ = 28
At 0.05 significance level, critical value = 1.65 (found by 0.50.05 = 0.4500)
H₀: There is no difference between units produced by day shift and afternoon shift
H₁: Number of units produced on the afternoon shift is larger than day shift
Z = xbar₁  xbar₂ / √ s₁²/ n₁ + s₂²/ n₂ = 345351 / 21²/54 + 28²/60 = 6 / √21.23 = 1.30
Pvalue = 1P(z≤1.30) = 1(0.09644) = 0.90356
Since 1.30 does not fall in the rejection region, H₀ is retained; or since pvalue is larger than significance level, H₀ is retained. At the 0.05 significance level, the number of units produced on the afternoon shift is the same as number of units produced by day shift.
Lind Chapter 11; Exercise 38
Two boats, the Prada (Italy) and the Oracle (U.S.A.), are competing for a spot in the upcoming America’s Cup race. They race over a part of the course several times. Below are the sample times in minutes. At the .05 significance level, can we conclude that there is a difference in their mean times?
Boat Times

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