// Part (a) 204.01 203.73 202.02 197.76 196.75 197.57 202.43 201.87 202.25 196.0
ID: 3222627 • Letter: #
Question
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Part (a) 204.01 203.73 202.02 197.76 196.75 197.57 202.43 201.87 202.25 196.06 197.41 199.34 203.58 201.59 200.51 197.68 195.86 200.55 205.49 200.56 200.47 196.88 195.87 202.06 204.45 202.27 198.95 195.64 195.51 203.78 205.90 200.94 197.20 193.76 197.19 202.64 205.62 201.66 196.68 193.94 196.33 203.82 Part (b) 206.13 200.00 198.61 195.11 196.90 204.16 204.27 201.86 197.04 194.88 196.64 205.84 Question 1 A manufacturer of breakfast cereal makes two claims concerning the weight of the packets they produce. The manufacturer claims that a) The mean is 200g b) The variance is 5g2. To investigate the first claim, the weight of a sample of packets produced in a given shift was measured. The values found are listed in part (a) of worksheet "Dataset 1 with grams (g) as the unit of measurement. Use this data to investigate the claim (a), with an appropriate statistical test. To investigate the second claim, the weight of a second sample of packets was measured. The values found are listed in part (b) of worksheet "Dataset 1', With grams (g as the unit of measurement. Use this data to investigate the claim (b), with an appropriate statistical testExplanation / Answer
(a) The mean of the weights (204.01,203.73,202.02,197.76,196.75,197.57,202.43,201.87,202.25,196.06,197.41,199.34,203.58,201.59,200.51,197.68,195.86,200.55,205.49,200.56,200.47,196.88,195.87,202.06,204.45,202.27,198.95,195.64,195.51,203.78,205.90,200.94,197.20,193.76,197.19,202.64,205.62,201.66,196.68,193.94,196.33,203.82)
= 199.9662
We conduct two-tail t-test to test for the sample mean.
Standard deviation for the weights = 3.42
Standard error for the weights = 3.42/sqrt(42) = 0.5277175
t = (x - mean)/Std error = (199.9662 - 200)/ 0.5277175
= -0.06404942
P-value for t = -0.06404942 and df = 42-1 = 41 is 0.4746211
P(t < -0.06404942) = 0.4746211 and P(t > -0.06404942) = 0.4746211
As it is two-tail test, p-value = 0.4746211 + 0.4746211 = 0.9492422
As, p-value is > 0.05, we fail to reject the null hypothesis and conclude that the mean of weights is 200 g
(b) We conduct Chi-square test to test for the variance.
The variance of the weights (206.13,200.00,198.61,195.11,196.90,204.16,204.27,201.86,197.04,194.88,196.64,205.84)
is 17.47
N = 12
Chi-square stat, T = (N-1) Sample-Variance/ Hypothesized variance
= (12-1) * 17.47/5
= 38.434
Chi-square stat for alpha = 0.975 and df = N-1 = 11 is 21.92005
As, T > Chi-square stat for alpha = 0.975, we reject the null hypothesis and conclude that the variance of weights is not 5g2.
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