Question

he average movie admission price for a recent year was $7.18. The population variance was 3.81. A random sample of 15 theater admission prices had a mean of $8.02 with a standard deviation of $2.08. At α = 0.05, is there sufficient evidence to conclude a difference from the population variance? Assume the variable is normally distributed.

Answer 1

Answer:

Since the calculated value of  z= 1.6279 does not fall in the critical region Z ≥ ±1.96 we conclude that there is no difference between the population variance and the sample variance.

Step-by-step explanation:

The data given is

Population mean μ= $ 7.18

Population variance= σ²= 3.81

Population Standard Deviation = √σ²= √3.81= 1.952

Sample Mean= x`= $ 8.02

Sample Standard Deviation =s = $ 2.08

Sample Size = 15

Significance Level = ∝= 0.05

The null and alternate hypotheses are

H0: σ1=σ2 against the claim that Ha: σ1≠ σ2

where  σ1 is the population variance and

σ2 is the sample variance

The rejection region is Z ≥ ±1.96 for two tailed test  at ∝= 0.05

The test statistic z is used

z= x`-  μ/ σ/√n

Putting the values

Z= 8.02-7.18/1.952/√15

z= 0.84/0.51599

z= 1.6279

Since the calculated value of  z= 1.6279 does not fall in the critical region Z ≥ ±1.96 we conclude that there is no difference between the population variance and the sample variance.