Question

A popular online retail website was interested in consumer spending during the holiday shopping weekend after Thanksgiving. They collect data on a random sample of 400 shoppers, particularly noting how much money each shopper spent. The following histogram was constructed using the data: 50 100 200 250 300 150 Spending The retailer determines that this sample had an average of $84.71 with a standard deviation of $46.87. a. (1 point) What is the best point estimate for the true mean amount of money spent by all shoppers at this website during the weekend after Thanksgiving? b. (3 points) Are the conditions for constructing a confidence interval satisfied? Be sure to comment on each condition individually. c. (3 points) Find the margin of error for a 90% confidence interval estimate for the true mean spending. d. (2 points) Construct a 90% confidence interval estimate for the true mean spending- e. (1 point) Interpret the interval in the context of the data. Your answer should reference what "90% confidence" means and how the two endpoints of the interval relate to the estimate. f. (2 points) Last year, shoppers spent an average of $82.24 on this website. The company uses their sample data to claim that the average consumer spending has increased this year. Based on the confidence interval, is this claim justified? Why or why not? R. (1 point) Without doing any calculation, explain why a 95% confidence interval would be a wider interval than the one you computed in part (d). h. (2 points) How would the confidence interval be affected if the sample was of size 200 instead of 400? Explain briefly without calculating the new confidence interval.

please show work and/or sketch of graph. Thanks

Answer 1

Here, n = sample size = 400, sample mean
T= 84.71
and sample standard deviation
0 = 46.87

Ans (a):

The best point estimate to the true mean is sample mean.

Therefore, the best point estimate to the true mean = 84.71.

Ans (b):

Conditions for constructing confidence interval:

i) Randomization Condition: The data must be sampled randomly. This condition is satisfied because, they collected data from 400 random samples.

ii) Independence Assumption: The sample values must be independent of each other. This means that the occurrence of one event has no influence on the next event. Usually, if we know that people or items were selected randomly we can assume that the independence assumption is met.

iii) Sample Size Condition: The sample size must be sufficiently large. Here we have a 400 sample, so this condition is also satisfied.

Ans (c):

The margin of error can be defined by either of the following equation:

Margin of error = Critical value x Standard error of the statistic

Here, standard deviation is given as 46.87 and for sufficiently large sample size we can use z-score as the critical value.

Standard Error, SE =

o 46.87 vñ on = 2.3435

Since, confidence level is 90% i.e.

a=0.10

Now, from standard table  

29 = 20.10 = 20.05 = 1.645

Now, margin of error =

29 X SE= 1.645 x 2.3435 = 3.8551

Therefore the margin of error for 90% confidence interval estimate for true mean spending is 3.8551.

Ans (d):

90% confidence interval for true mean is given as,

i.e. Lower Limit = Sample Mean - Margin of Error and Upper Limit = Sample Mean + Margin of Error.

i za X SE

Therefore, Lower limit = 80.8549 and Upper limit = 88.5651

Hence [80.8549, 88.5651] is 90% confidence interval for true mean spending.