Analyzing Customer Spending and Family Sizes in Canada

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DGD SESSION 1 Shahryar Moradi smora005@uottawa.ca
10-48. Store Receipts. Grocery store receipts show that customer purchases have a skewed distribution with a mean of $32 and a standard deviation of $20. a. Explain why you cannot determine the probability that the next customer will spend at least $40. b. Can you estimate the probability that the next 10 customers will spend an average of at least $40? Explain. c. Is it likely that the next 50 customers will spend an average of at least $40? Explain. A) Since the distribution is skewed and is not normal, we cannot use a normal model to estimate probabilities. B) It is unlikely to estimate the probability as 10 is not a large sample. It depends on the amount of skewness seen in the distribution.
C) Prob( y >40) = Prob(Z>z 0 ) = 1 - Prob(Z<= z 0 ) Where Z is the normal standard equivalent for y (avg. spent amount) , z 0 is the z-score that we should calculate, and Prob(Z<=z) is the probability that we can find in the Z table. And z-score formula à SD = 20/sqrt (50) = 20/7.07 = 2.83 Z = (40 – 32) / 2.83 = 2.826 Probability of at least $40 = P( "࠵? >40) = 1 - P( "࠵? <40) = 1 - P(Z<2.82)= 1 - 0.9976 = 0.0024 Final Answer: No, with 50 customers, the normal distribution model has a mean of 32 and a standard deviation of 2.83, and the resulting probability is only 0.0024 . This is highly unlikely.
10-58. Large Families in Canada. According to the Canadian Census, the proportion of families with three or more children (aged less than 25) declined from 42% in 1961 to 19% in 2011. Suppose we estimate that this proportion is 17.5% this year . In order to market products to large families, we randomly select 1600 Canadian families for the survey. a. What is the expected proportion of families with three or more children in our sample? b. In our sample , we find 19 .3% of the families have three or more children. What is the probability of a proportion as high as 19.3% or higher in the sample if our estimate of 17 .5% in the population is correct? c. Comment on whether the survey result casts doubt on the population estimate. d. Suppose we had surveyed a random sample of only 400 families instead of 1600 and had gotten the same result: 19.3% of the sample had three or more children. Comment on whether this result from a smaller survey casts doubt on the population estimate.
a. Same as the population estimate 17.5%. b. p= 17.5% =0.175 , q= 1-0.175= 0.825 np>10; nq>10; SD= !×# $ = %.’()×%.*+) ’,%% = 0.009499 ; ࠵? = -! / ! 01 = %.’23/%.’() %.%%2422 = 1.895 ; ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵? ࠵?࠵? ࠵? > 0.193 = ࠵?(࠵? > 1.895) = 1 − ࠵?(࠵? < 1.895) = 1 − %.2(%,5%.2(’3 + = 1 − 0.9709 = 0.0291 Note: Prob of z=1.895 is between 0.9706 and 0.9713, therefore we find the prob value in between which is 0.9709 c. It does cast doubt since if the initial estimate were correct, p=0.175, the chance of getting this rare sample would have been very small, 2.91%. So the the fact that we got this sample in our first try means that we should not be so sure about the estimate.
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