Question

2. Sales at a fast-food restaurant average $6,000 per day. The restaurant decided to introduce an advertising campaign to increase daily sales. To determine the effectiveness of the advertising campaign, a sample of 49 days of sales were taken. They found that the average daily sales were $6,300 per day. From past history, the restaurant knew that its population standard deviation is about $1,000. If the level of significance is 0.05, have sales increased as a result of the advertising campaign?

Multiple Choice

• Reject the null hypothesis and conclude the mean is higher than $6,000 per day.

• Fail to reject the null hypothesis.

• Reject the null hypothesis and conclude the mean is lower than $6,000 per day.

• Reject the null hypothesis and conclude that the mean is equal to $6,000 per day.

4. A hypothesis regarding the weight of newborn infants at a community hospital is that the mean is 6.6 pounds. A sample of seven infants is randomly selected and their weights at birth are recorded as 9.0, 7.3, 6.0, 8.8, 6.8, 8.4, and 6.6 pounds. What is the sample standard deviation?

Multiple Choice

• 1.177

• 1.090

• 1.386

• 1.188

6. A hypothesis regarding the weight of newborn infants at a community hospital is that the mean is 6.6 pounds. A sample of seven infants is randomly selected and their weights at birth are recorded as 9.0, 7.3, 6.0, 8.8, 6.8, 8.4, and 6.6 pounds. What is the alternate hypothesis?

Multiple Choice

• H₁: µ = 6.6

• H₁: µ ≠ 6.6

• H₁: µ > 7.6

• H₁: µ ≥ 6.6

7. The probability of a Type II error is represented by _______.

Multiple Choice

• β

• σ

• α

• the Type I error

8. A random sample of size 15 is selected from a normal population. The population standard deviation is unknown. Assume the null hypothesis indicates a two-tailed test and the researcher decided to use the 0.10 significance level. For what values of t will the null hypothesis not be rejected?

Multiple Choice

• Between −1.761 and 1.761

• To the left of −1.345 or to the right of 1.345

• To the left of −1.282 or to the right of 1.282

• To the left of −1.645 or to the right of 1.645

Answer 1

2) H₀: = 6000

Ha: > 6000

The test statistic is

= 2.1

P-value = P(Z > 2.1)

= 1 - P(Z < 2.1)

= 1 - 0.9821 = 0.0179

Since the P-value is less than , so we should reject H₀.

Reject the null hypothesis and conclude the mean is higher than $6000 per day.

4) = (9 + 7.3 + 6 + 8.8 + 6.8 + 8.4 + 6.6)/7 = 7.557

s = sqrt(((9 - 7.557)^2 + (7.3 - 7.557)^2 + (6 - 7.557)^2 + (8.8 - 7.557)^2 + (6.8 - 7.557)^2 + (8.4 - 7.557)^2 + (6.6 - 7.557)^2)/6) = 1.177

6) H1:

7)

8) df = 15 - 1 = 14

At 0.10 significance level, the critical values are +/- t_{0.05, 14} = +/- 1.761

Between -1.761 and 1.761