Homework Sourse1Dominos Pizza claims that the average time a customer waits
1.Domino\'s Pizza claims that the average time a customer waits on hold is less than 5 minutes. You conducted a sample of 35 customers that has an average wait time of 4.78 minutes and standard deviation of 1.8 minutes. Test Domino\'s pizza claim at 0.05 significance level by comparing the calculated z-score to the critical z-score.
2.Test the hypothesis that the average grocery shopping in California is not equal to the average in Miami by comparing the sample z-score to the critical z-score with a 0.05 significance level. Use the information below.
CA = sample 35, sample mean $126.15, standard calculation $42.00
MIA = sample 46, sample mean $135.60, standard deviation $39.00
3. Test the hypothesis test that the average pizza delivery man in New York is more than DC by comparing the sample z-score to the critical z-score at a 0.05 significance level. Use the information listed below.
NY = sample 40, sample mean $52,378, standard deviation $6,812
DC = sample 42, sample mean $48,773, standard deviation $7,514
4. States the one-way ANOVA hypothesis, then calculate the Sum of Squares (SST), sum of squares between (SSB), and sum of squares within (SSW) using the information below that was recorded by 15 customers satisfaction ratings for Wal-Mart, Target, and TJ Max using a scale 1 to 10.
Wal-Mart = 7,7,6,5,3
Target = 8,9,7,6,9
TJ Max = 9,7,8,10,8
Then perform a hypothesis test to determine whether there is difference in the customer satisfaction rating of the scores at a 0.05 significance level using the mean square between (MSB) and then the mean square within (MSW).
5. Calculate the number of fish caught per day, assuming the population is uniformly distributed using the information below.
days = 1, 2, 3, 4, 5
# of fish = 15, 17, 12, 10, 6
Conduct a hypothesis testing to test the claim that the distribution of fish caught is uniformly distributed at a 0.05 significance level.
Solution
Formulating the null and alternative hypotheses,
Ho: u >= 5
Ha: u < 5
As we can see, this is a left tailed test.
Thus, getting the critical z, as alpha = 0.05 ,
alpha = 0.05
zcrit = - 1.644853627
Getting the test statistic, as
X = sample mean = 4.78
uo = hypothesized mean = 5
n = sample size = 35
s = standard deviation = 1.8
Thus, z = (X - uo) * sqrt(n) / s = -0.723076418
Comparing z > zcrit, we FAIL TO REJECT THE NULL HYPOTHESIS.
Thus, there is no significant evidence that a customer waits on hold is less than 5 minutes. [CONCLUSION]
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