reported that the average American eats out at a restaurant 5.9 times per week. A random sample of 40 Americans showed that the average amount of times they ate out at a restaurant per week was 5.1 times with a standard deviation of 1.7. Is there enough evidence at the 5% level that the average American eats out less than 5.9 times per week? Find the 95% Confidence Interval and explain: a) 5.9 – (0.95)(0.2688) = 5.65 5.9 + (0.95)(0.2688) = 6.16 We are 95% confident the average amount of times Americans eat out at a restaurant per week is between 5.65 and 6.16
It is reported that the average American eats out at a restaurant 5.9 times per week. A random sample of 40 Americans showed that the average amount of times they ate out at a restaurant per week was 5.1 times with a standard deviation of 1.7. Is there enough evidence at the 5% level that the average American eats out less than 5.9 times per week?
Find the 95% Confidence Interval and explain:
a) 5.9 – (0.95)(0.2688) = 5.65 5.9 + (0.95)(0.2688) = 6.16 We are 95% confident the average amount of times Americans eat out at a restaurant per week is between 5.65 and 6.16
b) 5.1 – (0.95)(0.2688) = 4.85% 5.1 + (0.95)(0.2688) = 5.36% We are 95% confident the average amount of times Americans eat out at a restaurant per week is between 4.85% and 5.36%.
c) 5.9 – (2.02)(0.2688) = 5.36 5.9 + (2.02)(0.2688) = 6.44 We are 95% confident the average amount of times Americans eat out at a restaurant per week is between 5.36 and 6.44.
d) 5.1 – (2.02)(0.2688) = 4.56 5.1 + (2.02)(0.2688) = 5.64 We are 95% confident the average amount of times Americans eat out at a restaurant per week is between 4.56 and 5.64.
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