Answer :
In conclusion, based on the given data and at a significance level of 0.10, there is not enough evidence to support the claim that more than 55% of adults skip breakfast at least three times a week according to the sample data.
To test the magazine's claim that more than 55% of adults skip breakfast at least three times a week, we can set up a hypothesis test.
Let's define the null hypothesis (H0) and the alternative hypothesis (Ha):
H0: The proportion of adults who skip breakfast at least three times a week is 55% or less.
Ha: The proportion of adults who skip breakfast at least three times a week is greater than 55%.
Next, we need to determine the test statistic and the critical value to make a decision. Since we have a sample proportion, we can use a one-sample proportion z-test.
Given that we have a random sample of 80 adults and 45 of them responded that they skip breakfast at least three days a week, we can calculate the sample proportion:
p = 45/80 = 0.5625
The test statistic (z-score) can be calculated using the sample proportion, the claimed proportion, and the standard error:
z = (p - P) / sqrt(P * (1 - P) / n)
where P is the claimed proportion (55%), and n is the sample size (80).
Let's calculate the test statistic:
z = (0.5625 - 0.55) / sqrt(0.55 * (1 - 0.55) / 80)
≈ 0.253
To make a decision, we compare the test statistic to the critical value. Since the significance level (α) is given as 0.10, we look up the critical value for a one-tailed test at α = 0.10.
Assuming a normal distribution, the critical value at α = 0.10 is approximately 1.28.
Since the test statistic (0.253) is less than the critical value (1.28), we fail to reject the null hypothesis.
To know more about significance level,
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