In: Statistics and Probability
The average daily attendance of a small amusement park is 4,219 people. In order to increase the average daily attendance, the park owners decided to lower the price for admissions. For the first 25 days after the highly publicized price reduction the average daily attendance was 4,537. You can assume that the population standard deviation is 674. Assume that these 25 days can be considered a random sample of the days to come and that daily attendance follows a normal distribution.
a.) Test at the 5% level of significance whether the price reduction was effective. Explain your approach (including your hypotheses and test statistic) and conclusion.
b.) How and why would your answer in Part 1 change if the significance level had been 1%? Explain using 1 or 2 sentences.
thank you! :)
Given that the average daily attendance of a small amusement park is = 4,219 people and the first n=25 days( Sample size) after the highly publicized price reduction the average daily attendance was =4,537 and also the population standard deviation is = 674.
Based on the claim that whether the price reduction was effective the hypotheses are:
Based on the hypothesis we can run an upper tailed test with known population variance at 0.05 level of significance.
Rejection region:
At 0.05 level of significance reject Ho if Z>Z0.05 =1.645 ( Z-critical is computed using the Z table shown below or by excel formula as =NORM.S.INV(0.95).
Test statistic:
The test statistic is calculated as:
P-value:
The P-value is also calculated using the Z table shown below or using excel by formula 1-=NORM.S.DIST(2.359,TRUE)
P-value=0.0092
Conclusion:
Since P-value<0.05 and Z>Z-crit hence we reject the null hypothesis and conclude that there is sufficient evidence to support the claim that the price reduction was effective.
b) If we take the level of significance as 0.01 even then the conclusion will not change since P-value is also less than 0.01 so, the conclusion remains the same.
The Z table: