Homework Sourse2Test the hypothesis that the average grocery shopping in Ca
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
2.
Formulating the null and alternative hypotheses,
Ho: u1 - u2 = 0
Ha: u1 - u2 =/ 0
At level of significance = 0.05
As we can see, this is a two tailed test.
Calculating the means of each group,
X1 = 126.15
X2 = 135.6
Calculating the standard deviations of each group,
s1 = 42
s2 = 39
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):
n1 = sample size of group 1 = 35
n2 = sample size of group 2 = 46
Thus, df = n1 + n2 - 2 = 79
Also, sD = 9.135930023
Thus, the t statistic will be
t = [X1 - X2 - uD]/sD = -1.03437745
where uD = hypothesized difference = 0
Now, the critical value for t is
tcrit = +/- 1.96
Thus, comparing t and tcrit, as out t is between these two values, WE FAIL TO REJECT THE NULL HYPOTHESIS.
Thus, there is no significant evidence that the average grocery shopping in California is not equal to the average in Miami. [CONCLUSION]
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