1. Suppose you have a list of observations and you find the list of z scores by using the formula z score = (raw - mean of list) /(stdev of list) And one of the observations has a z score of -18

a.Very few or none of the other observations on the list will have a z score more than -18 so the percentile of the observation with a zscore of -18 is the 100th percentile or close to the 100th percentile

b.The observation with a z score of -18 is the median (the 50th percentile)

c.Very few or none of the other observations on the list will have a z score less than -18 so the percentile of the observation with a zscore of -18 is the 0th percentile or close to the 0th percentile

d.None of the options is correct

2. If you have a list of observations and you find the list of z scores by using the formula z score = (raw - mean of list) /(stdev of list)

and one of the observations has a z score of 14.9

a.None of the options is correct

b.Very few or none of the other observations on the list will have a z score less than 14.9. so the observation with a zscore or 14.9 will be close to the 0th percentile

c.The observation with a z score of 14.9 is the median (the 50th percentile)

d.Very few or none of the other observations on the list will have a z score more than 14.9, so the observation with a zscore or 14.9 is the 100th percentile or is close to the 100th percentile

3. Which of the assumptions given below is one of the assumptions required for the test of proportion and is also one of the required assumptions for the test of the mean. Select the BEST answer

Select one:

a.

Independent observations

b.

A positive slope

c.

Equal variance

d.

All of the other options are correct

4. Perform the Hypothesis test

Suppose a sample of 23 chocolate bars has a mean weight of 1006.4 with standard deviation 38. Testthe claim the mean weight is above 1006 usea 5% level of significance

The test stat is not above the critical value from the table so do not reject H0 there is not strong evidence the mean is above 1006

the test stat is 14.05

H0: =1006 H1: more than 1006

the test stat is 0.05

The test stat is not above the critical value from the table so reject H0 there is strong evidence the mean is above 1006.40

the test stat is -31.95

H0: p=1006 H0: p more than 1006