A group of women tried five brands of fingernail polish and ranked them according to preference. What level of measurement is this? ​

1)An auditor for American Health Insurance reports that 20% of policyholders submit a claim during the year. 15 policyholders are selected randomly. What is the probability that at least 3 of them submitted a claim the previous year?    

.200

​.708

​.250

​.602

2) A research firm conducted a survey to determine the mean amount steady smokers spend on cigarettes during a week.  A sample of 64 smokers revealed that  = $20 and S = $5.  What is the 95% confidence interval for μ? ​

$18.78 to $21.23

​18.16 to 21.84

​$18.60 to $21.40

​18.37 to 21.63

3) When a class interval is expressed as 100 up to 200, _________________________. ​

Observations with values of 200 are included in the class

​Observations with values of 200 are excluded from the class

​Observations with values of 100 are excluded from the class

​The class interval is 99

4) A group of women tried five brands of fingernail polish and ranked them according to preference. What level of measurement is this? ​

Ordinal

​Nominal

​Interval

​Ratio

5) A time series trend equation for Hammer Hardware is Y’ = 5.6 + 1.2t, where sales are in millions of dollars and t increases by one unit for each year. If the value of sales in the base year of 2016 is $5.6 million, what would be the estimated sales amount for 2018?

$8 million

​$6.8 million

​Unable to determine from given information

​$5.6 million

6) A weight-loss company wants to statistically prove that its methods work. They randomly selected 10 clients who had been on the weight loss program for between 55 and 65 days. They looked at their beginning weights and their current weight. The statistical test they should utilize is:​

t test for difference in paired samples

​z test for two population proportions

​z test for two population means

​t test for two population means

7) A coffee manufacturer is interested in whether the mean daily consumption of regular-coffee drinkers is less than that of decaffeinated-coffee drinkers.  A random sample of 50 regular-coffee drinkers showed a mean of 4.35 cups per day, with a standard deviation of 1.2 cups per day.  A sample of 40 decaffeinated coffee drinkers showed a mean of 5.84 cups per day, with a standard deviation of 1.36 cups per day.  What is your computed z-statistic?​

z = -2.45

​z = -5.44

​z = -1.81

​z = – 3.90