Question

In: Statistics and Probability

. Based on past data, it is believed that on Thanksgiving, 70% of people eat pumpkin...

. Based on past data, it is believed that on Thanksgiving, 70% of people eat pumpkin pie. Suppose we take a sample of 100 people and find that 75 of the people in that sample eat pumpkin pie on Thanksgiving. We want to see if there is evidence that the percentage of people who eat pumpkin pie on Thanksgiving is increasing. a. If I wanted to control my margin of error and set it to 3% with 99% confidence, what sample size would I need to take instead of the 100? b. Using my original sample size of 100, what would be the 99% confidence interval for the population proportion? c. What are the null and alternative hypotheses? d. What is the critical value at 99% confidence? e. Calculate the test statistic (using the sample of 100). f. Find the p-value. g. What conclusion would be made here at the 99% confidence level?

Solutions

Expert Solution

p = 0.70

n = 100

x = 75

a)

Margin of error = E = 0.03

Confidence level = C = 0.99

Zc = 2.58   ( Using z table)

p = 0.70

We have to find the sample size (n)

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b)

Sample size = n = 100

x = 75

Sample proportion is

We have to construct a 99% confidence interval for the population proportion.

Formula is

Here E is a margin of error.

Zc = 2.58

So confidence interval is ( 0.75 - 0.1117 , 0.75 + 0.1117) => ( 0.6383 , 0.8617)

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c)

Claim:  The percentage of people who eat pumpkin pie on Thanksgiving is increasing.

The null and alternative hypothesis is

H0: P = 0.70

H1: P > 0.70

d)

Level of significance = 0.01

Critical value = 2.33

e)

Sample size = n = 100

x = 75

Sample proportion is

Test statistic is

f)

P-value = P(Z > 1.09) = 1 - P(Z < 1.09) = 1 - 0.8624 = 0.1376

g)

Level of significance = 0.01

P-value > 0.01 we fail to reject null hypothesis.

Conclusion:

The percentage of people who eat pumpkin pie on Thanksgiving is NOT increasing.


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