Question

1. A statistician is interested in the gross earnings of several of her favorite bands. She took a random sample of 30 of the Rolling Stones’ North American concerts, and found that the gross earnings averaged $2.27 million with a standard deviation of $0.5 million. One source suggests that the average gross earnings per concert for every stadium performance in North America is $2.11 million. Do the Rolling Stones earn more on average? Test at a 5% level of significance

2. A survey was conducted about the cost for a family of four to visit an amusement park for one day. A sample of 32 families yielded an average cost of $190.28 with a standard deviation of $51.75. Last year, a magazine published that the average cost for a family of four to visit an amusement park was $175. Based on the sample data above, can we conclude that the mean cost is actually higher than this at α=.05

3. According to a large local high school, senior students have a mean GPA of 3.07. A random sample of 38 seniors taking AP courses showed a mean GPA of 3.29 with a standard deviation of 0.42. At the 1% level of significance, can it be stated that seniors taking AP courses have a different GPA than the senior class as a whole?

Answer 1

1)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 2.11
Alternative Hypothesis, Ha: μ > 2.11

Rejection Region
This is right tailed test, for α = 0.05 and df = 29
Critical value of t is 1.699.
Hence reject H0 if t > 1.699

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (2.27 - 2.11)/(0.5/sqrt(30))
t = 1.753

P-value Approach
P-value = 0.0451
As P-value < 0.05, reject the null hypothesis.

2)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 175
Alternative Hypothesis, Ha: μ > 175

Rejection Region
This is right tailed test, for α = 0.05 and df = 31
Critical value of t is 1.696.
Hence reject H0 if t > 1.696

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (190.28 - 175)/(51.75/sqrt(32))
t = 1.67

P-value Approach
P-value = 0.0525
As P-value >= 0.05, fail to reject null hypothesis.

3)
Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 3.07
Alternative Hypothesis, Ha: μ ≠ 3.07

Rejection Region
This is two tailed test, for α = 0.01 and df = 37
Critical value of t are -2.715 and 2.715.
Hence reject H0 if t < -2.715 or t > 2.715

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (3.29 - 3.07)/(0.42/sqrt(38))
t = 3.229

P-value Approach
P-value = 0.0026
As P-value < 0.01, reject the null hypothesis.