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NISA [10]
3 years ago
13

A recent taste test showed that 85% of people liked a new chicken sandwich recently added to a restaurant’s menu. The restaurant

manager claimed that only 68% liked the new sandwich. To determine whether this sample supports the population proportion of 0.85, a simulation of 100 trials is run, each with a sample size of 30 and a point estimate of 0.85. The minimum sample proportion from the simulation was 0.66 and the maximum sample proportion from the simulation was 0.88.
a) Estimate the margin of error using the half the range method.


b) Use your margin of error to find the interval estimate.


c) Based on the computed interval estimate, is the restaurant manager’s claim that 68% of people liked the new sandwich valid? Explain your answer.
Mathematics
1 answer:
Vesna [10]3 years ago
3 0

Answer:

a) The margin of error = 0.11

b) The interval estimate = (0.74, 0.96)

c) Not valid because 68% = 0.68 is less than the values in the interval estimate

Step-by-step explanation:

The statistical data from the population and the simulation are;

The percentage of the people that liked the new chicken sandwich = 85% = The population proportion

The percentage the restaurant manager claims liked the new sandwich = 68%

The number of trials of the simulation = 100 trials

The sample size of each trial = 30

The point estimate of each trial = 0.85

The minimum sample proportion = 0.66

The maximum sample proportion = 0.88

a) The confidence interval = (0.66, 0.88)

By the half the range method, the margin of error is given by half the width of the confidence interval

∴ The margin of error, MOE = (0.88 - 0.66)/2 = 0.11

b) The interval estimate are the expected range of numbers for a parameters

The interval estimate = The point estimate ± MOE

∴ The interval estimate = 0.85 ± 0.11 = (0.74, 0.96)

Given that the manager's claim of 68% is not within the interval estimate, the manager's claim is not valid based on the expected values for the people that liked the new sandwich found with the point estimate and the margin of error

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