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.

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.

d. Repeating the calculation with n=400, we get a probability of 0.172, which is not very small. The survey result therefore does not cast doubt on the population estimate. SD= !×# $ = %.’()×%.*+) 4%% = 0.019 ; ࠵? = -! / ! 01 = %.’23/%.’() %.%’2 = 0.947 ; ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵? ࠵?࠵? ࠵? > 0.193 = ࠵?(࠵? > 0.947) = 1 − ࠵?(࠵? < 0.947) = 1 − %.*+,45%.*+*2 + = 1 − 0.828 = 0.172

10- 61. BROADBAND IN CANADA. Broadband to the home is widely regarded as important to Canadian households. a. We ask 1000 randomly selected Canadian households whether they have broadband internet access to their homes. In response, 865 of them say yes and 135 say no, so that the proportion of households having broadband is 0.865. What is the standard error of this estimate? b. We ask 1000 households how many broadband internet access lines they have to their homes. In response, 865 of them say one and 135 say none , so that the mean number of broadband lines is 0.865 . What is the standard error of this estimate? ( Hint : First calculate the standard deviation of the number of broadband lines in the sample.) c. Comment on the relationship between (a) and (b). a. np = 865 > 10; nq = 135 > 10; SE= (0.865∗0.135)/1000 =0.0108

b. ࠵?࠵? = s/ (1000) ࠵? = ࠵?࠵?࠵? ࠵? = ࠵?[ ࠵? − ࠵? ! ] = 1 "#$ % ࠵? − ࠵? ! . ࠵?(࠵? = ࠵?) = (0 − 0.135) ! ∗ 0.135 + (1 − 0.865) ! ∗ 0.865 = 0.3419 ࠵?࠵? = $.’(%) %$$$ = 0.0108 ࠵? ࠵? = 1 "#$ % ࠵?. ࠵? ࠵? = ࠵? = 1 ∗ ࠵? ࠵? = 1 + 0 ∗ ࠵? ࠵? = 0 = 1 ∗ 0.865 + 0 ∗ 0.135 = 0.865 c. parts a and b address the same issue using different approaches. Part a uses the sampling distribution of a proportion ; part b uses the sampling distribution of a mean .

11-56. Carbon Tax in British Columbia. It’s not often that taxes are popular, but when Environics surveyed 1023 adult British Columbians in 2011, it found 54 % in favor of the carbon tax in that province. At $25/ton of carbon, the tax adds only 5₵ to the price of a liter of gasoline. a. Calculate a 90 % confidence interval for the percentage of British Columbians in favor of the carbon tax in 2011 and interpret the meaning of your interval in words. b. Phrase your confidence interval in the form commonly used by the media . For example, “x% of Ontarians support the X political party. This result is accurate to plus or minus y%, n times out of N.” c. How many British Columbian adults would Environics need to survey in order to reduce the width of this confidence interval by 25%? d. How much wider is the 95% confidence interval than the 90% interval you calculated in (a)? Give your answer as the ratio between the width of the two confidence intervals.

p±z*SE a. SE(p)= √( pq/n)= √( 0.54×0.46/1023)=0.01558 Now let’s find the critical value for a 90% confidence interval (z-critical value) from the z table: We must find a two-sided CI when alpha is 0.1 = 10%, so z score must be associated with the probability of: 1- alpha/2 =1-0.1/2= 1-0.05=0.95 Then we will find 0.95 in the body of the below z-table. The corresponding Z is 1.6 and for the second decimal number we will find it between 1.64 and 1.65 à 1.645 The critical value for a 90% confidence interval: z*=1.645 Confidence interval: p±z*SE=0.54±1.645×0.01558=0.54±0.0256 . It means that if we collect samples 100 times, we expect that for 90 times the sample proportion fall within this 0.54+0.0256 and 0.54-0.0256 interval.

b. 54% of adult British Columbians are in favor of the Carbon tax. This result is accurate to plus or minus 2.6%, 9 times out of ten = 90%. c. 1-25% =75%= 0.75 ̂࠵? ± ࠵? ∗ ࠵?࠵? ࠵?࠵? + ࠵?࠵? ’ = ࠵? ∗ ࠵?࠵? ࠵? + ࠵? ∗ ࠵?࠵? ࠵? ’ = ࠵? ’ ࠵? + → ࠵?࠵? + ࠵?࠵? ’ = 0.75 ∗ ࠵?࠵? ’ ࠵?࠵? ’ = 0.75 = ࠵? ’ ࠵? + → ࠵? ’ ࠵? + = 0.75 + = 0.5625 → ࠵? + ࠵? ’ = 1 0.5625 = 1.778 → ࠵? + = 1.778 ∗ ࠵? ’ = 1.778 ∗ 1023 ≈ 1819 The width of the confidence interval is proportional to 1/ (࠵?࠵?࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵?) , so to reduce the width by 25% we need a sample size that is ( ’ %.() ) + = 1.778 times the first sample size. 1023×1.778 = 1819

11-57. Canadian Senate. Suppose we surveyed 1000 adult Canadians about reform of the Canadian Senate and 34 % answered “Yes” to the question, “Do you support abolishing the Senate of Canada?” a. Construct a 95 % confidence interval for this proportion and give a verbal explanation of what your interval means. b. Phrase your confidence interval in the form commonly used by the media . For example, “x% of Ontarians support the X political party. This result is accurate to plus or minus y%, n times out of N.” ࠵? ± ࠵? ∗ ࠵?࠵? a. ࠵?࠵?( ̂࠵?) = -!-# $ = %.34×%.,, ’%%% = 0.01498 The critical value for a 95% confidence interval would be ࠵? ∗ = 1.96 . (a two-sided CI for 95% is found with a z score associated with 1-alpha/2). Thus, the confidence interval would be ࠵? ± ࠵? ∗ ࠵?࠵? = 0.34 ± 1.96×0.01498 = 0.34 ± 0.0294

b. 34% of adult Canadians support abolishing the Senate. This result is accurate to plus or minus 2.9%, 19 times out of 20 (=a 95% confidence interval).

a. • Number of non-GMO farmers= 85%*420= 357 Number of GMO farmers=420-357= 63 Both are >10. • We assume the farmers were selected at random. • We assume that 420 is less than 10% of soybean farmers in Manitoba. ࠵?࠵?( ̂࠵?) = -!-# $ = %.*)×%.’) 4+% = 0.01742 ; ࠵? ∗ = 1.645 ; (two-sided CI with an alpha=0.1) A 90% confidence interval: ࠵?࠵? = ࠵? ± ࠵? ∗ ࠵?࠵? = 0.85 ± 1.645×0.01742 = (0.821,0.879) We are 90% confident that the proportion of Manitoba soybean farmers growing non-GMO soybeans was between 0.821 and 0.879.