Question

1)A hot chocolate machine adds a certain amount of chocolate into. 25 cups of hot
chocolate were investigated and found to have a mean chocolate content of 1.10
grams and a standard deviation of 0.015 grams. Find a 95% CI on the mean weight
of chocolate dispensed.(WE SHOULD USE T OR Z?WHY?)

(WE SHOULD USE T OR Z?WHY?)

It is believed that when cars go faster, their fuel consumption change. A car
company randomly assigned cars to go with 120 km/h or 90 km/h, then measured
their fuel consumption. The 180 cars that went with 120 km/h had a mean fuel
consumption of 9 liters, with a standard deviation of 0.3 liters. The 240 cars that
went with 90 km/h had a mean fuel consumption of 7 liters, with a standard
deviation of 0.4 liters. Compute a 99% confidence interval estimate for the
difference between the population means?

Answer 1

1)

We have to use t distribution because n = 25

A small sample of 25 units has a mean and a standard deviation o of 0.015. Find a 95% confidence interval of the mean Confidence Interval Formula for u is as follows: X-tscorea * s/n < <X + tscorea * s/n where: X = sample mean, s = sample standard deviation, tscore = statistic with (n - 1) Degrees of Freedom and a = 1 - confidence Percentage First find degrees of freedom: Degrees of Freedom = n - 1 Degrees of Freedom = 25 - 1 Degrees of Freedom = 24 Calculate a: a = 1 - Confidence% a = 1 -0.95 a = 0.05 Find a spread range: a = 12(a) a = 12(0.05) a = 0.025

Calculate high end confidence interval total: High End = X + tscorea * s/vn High End = 1.10+ 2.0639 x 0.015/125 High End = 1.10+ 2.0639 x 0.003 High End = 1.10+ 0.0061917 High End = 1.1062 Calculate low end confidence interval total: Low End = X-tscorea * s/n Low End = 1.10-2.0639 0.015/125 Low End = 1.10 - 2.0639 x 0.003 Low End = 1.10 - 0.0061917 Low End = 1.0938 Now we have everything, display our 95% confidence interval: 1.0938 < < 1.1062

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