Math 227s Statistics

Question

In a random sample of 35 visitors to a certain theme park, it was determined that the mean amount of money spent per person at the park (including ticket price) was $93.43 per day, and the standard deviation of the amount spent per person is $12.

a. (6) Construct a 95% confidence interval for the mean amount spend daily per person at the theme park.

b. (8) At a 0.10 significance level, test the claim that the mean amount spend daily per person at the theme park is more than $95.

Answer 1

(a) For constructing a two-sided confidence interval on the mean of one sample, there are two equations to consider, based on how confident we are in the standard deviation. If the number of samples (n) < 30, we want to use a t-distribution to construct the confidence interval. When n ≥ 30, we can use a z table for the standard normal distribution.

The formula for a (1-alpha)% two-sided confidence interval about the mean when n ≥ 30:

µ ∈ x̄ ± Z_α/2(S_x/√n)

Where x̄ is the sample mean, Z_α/2 is essentially how many standard deviations away from the mean (1-α/2)% of the data will be less than, S_x is the sample standard deviation, and n is the sample size.

From the problem, we know:

x̄ = 93.43, S_x = 12, and n = 35

To get Z_α/2, we need to look at a z-table (example: http://www.z-table.com/). For a 95% confidence interval, α = 1 - .95 = 0.5. Therefore α/2 = 0.025.

So we're looking for Z_0.025 on the Z-table. This is looking for where 2.5% of the data will be larger than the rest. Put another way, it's where 100 - 2.5 = 97.5% of the data will be less than the rest. Looking for 0.975, we find Z_0.025 = 1.96.

Now we have everything we need to construct the two-sided 95% confidence interval.

µ ∈ x̄ ± Z_α/2(S_x/√n)

µ ∈ 93.43 ± 1.96(12/√35)

µ ∈ [$89.454, $97.406]

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(b) Here we are running a hypothesis test to see if the mean amount spent daily per person at the theme part is > $95. The null and alternative hypotheses are:

H₀: µ = $95

H₁: µ > $95

We need to compute the test statistic. The test statistic in this case is Z₀ = (x̄ - µ₀)/(S_x/√n). We will reject the null hypothesis H₀ in favor of the alternative hypothesis H₁ if Z₀ > Z_α.

Z₀ = (93.43 - 95)/(12/√35) = -0.774

Z_α = Z_0.1 = 1.28

Since Z₀ <= Z_α, we fail to reject the null hypothesis. At the 0.1 significance level, given the sample data collected, we cannot claim the mean amount spent daily per person is greater than $95.