Martys Burger Bam provides a Be A Glutton option for all of

Marty\'s Burger Bam provides a \"Be A Glutton\" option for all of their combo meals. This option is offered to all customers purchasing a combo meal for an additional charge of $0.35. Marty has determined that if a store is marketing the option properly, at least 70% of eligible customers will choose to \"Be A Glutton\". Marty has asked for your help in evaluating the \"Be A Glutton\" performance at one of his locations, so you have correctly formulated the following hypothesis test. You have informed Marty that you intend to use a level of significance of 0.05 and that the appropriate confidence interval on p is guaranteed to have an error of 0.02 or less. Define p. Briefly explain how you determined the alternative hypothesis. What sample size will you use? You may assume that this sample size is used for the remainder of the problem. What is the probability of a Type II error when p = 0.65? Explain your answer to part (d) to Marty. Remember that he has no training in probability and statistics. After collecting appropriate data, you find that 1148 customers chose to \"Be A Glutton\". Compute an appropriate point estimate on p. Compute the test statistic and the P-value, and determine the conclusion of the test. Construct the appropriate confidence interval, and determine the conclusion of the test using this confidence interval. What can you say about the \"Be A Glutton\" performance of this location?

Solution

a) p is the proportion of customers that take the \"Be a Glutton\" option while purchasing a combo meal

b) The alternative hypothesis is the one that is being tested and we are interested to see whether the proportion of ustomers that take the \"Be a Glutton\" option while purchasing a combo meal is less than 70%

c) Margin of Error = 0.02

Critical value for alpha 0.05 is + 1.96

Std Error = sqrt[p * (1 - p)/n]

Margin of Error = Std Error * Critical value

0.02 = sqrt[0.7 * (1 - 0.7)/n] * 1.96

n = 2016.84

Rounding off since sample size cant be a decimal

n = 2017

d) Std Error = sqrt[0.7 * (1 - 0.7)/2017] = 0.01

critical value for alpha 0.05 = 1.65

we fail to reject a null hypothesis if we get a z score > -1.65

-1.65 = (0.65 - p)/ 0.01

p = 0.665

Now, P[Z> (0.665 - 0.70)/0.01] = 0.005

e) The probability that we fail to reject the fact that the proportion of customers that take the \"Be a Glutton\" option while purchasing a combo meal <= 70% is 0.995 when infact it is not true

f) p = 1148/2017 = 0.569

g) test stat = (0.569 - 0.7)/0.01 = -13.1

p value = 0.000001

The test is significant and we reject the null

h) Confidence interval

Margin of error = 0.02

Confidence interval = 0.569 + 0.02 = 0.549 , 0.589

Since the confidence interval does not contain 0.70 we can reject the null

i) the proportion of ustomers that take the \"Be a Glutton\" option while purchasing a combo meal is less than 70%