1.The hypothesis is Ho: U>= 220 and H1: U<220. A sample of 64 observations is selected from a normal polulation with a population standard deviation of 15. The sample mean is 215. In conducting a hypothesis test using a .03 significance level what will be the value of your test statistic, i.e. the Z you calculated?
2.The Hypothesis is Ho: U>= 220 and H1: U<220. A sample of 64 observations is selected from a normal polulation with a population standard deviation of 15. The sample mean is 215. Ho: U>= 220 and H1: U<220. In conducting a hypothesis test using a .03 significance level what will be your decision regarding the null hypothesis?
3.A prospective employee was told she could average more than $80 a day in tips. Assume the standard deviation of the population is $3.24. Over the first 35 days the employee worked the mean daily amount of her tips was $84.85. At the .01 significance level what will be the value of the critical value if we are testing to see if the true average daily amount of tips is greater than $80?
4.A prospective employee was told she could average more than $80 a day in tips. Assume the standard deviation of the population is $3.24. Over the first 35 days the employee worked the mean daily amount of her tips was $84.85. At the .01 significance level what is the value of your calculated test statistic if we are testing to see if the true average daily amount of tips is greater than $80?
5.A prospective employee was told she could average more than $80 a day in tips. Assume the standard deviation of the population is $3.24. Over the first 35 days the employee worked the mean daily amount of her tips was $84.85. At the .01 significance level can we (she) conclude that the true average daily amount of tips is greater than $80? yes or no
Solution 1:
Test Statistics
Z = (215 - 220)/ (15/64)
Z = -2.67
Solution 2:
Null Hypothesis (Ho):
220
Alternative Hypothesis (Ha):
< 220
Test Statistics
Z = (215 - 220)/ (15/64)
Z = -2.67
Using Z-tables, the critical value at a = 0.03 is
Z (0.03) = -1.881
Since test statistics is less than the critical value, we reject Ho.
Solution 3:
We know that a = 0.01 and one-tailed test. Using Z-tables, the critical value is
Z (0.01) = 2.33
Solution 4:
Test Statistics
Z = (84.85 - 80)/ (3.24/35)
Z = 10.66
Solution 5:
Since test statistics is greater than the critical value, we reject Ho.
Hence, we can conclude that the true average daily amount of tips is greater than $80.
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