Question

1.The hypothesis is Ho: U>= 220 and H1: U<220. A sample of 64 observations is selected...

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.