Question

Population Standard Deviation 150.0000 Sample Size 18 Sample Mean 554.6111 Confidence Interval Confidence Coefficient 0.95 Lower...

Population Standard Deviation 150.0000
Sample Size 18
Sample Mean 554.6111
Confidence Interval
Confidence Coefficient 0.95
Lower Limit
Upper Limit
Hypothesis Test
Hypothesized Value 660
Test Statistic
P-value (Lower Tail) 0.0014
P-value (Upper Tail)
P-value (Two Tail)
Sample Size 18
Sample Mean 554.6111
      Sample Standard Deviation 162.8316
Confidence Interval
Confidence Coefficient 0.95
Lower Limit
Upper Limit
Hypothesis Test
Hypothesized Value 660
Test Statistic
P-value (Lower Tail) 0.0069
P-value (Upper Tail)
P-value (Two Tail)

The Hospital Care Cost Institute randomly selected 18 individuals, recorded the amount spent each year on prescription drugs, then put this information into Excel. Assume a population standard deviation of $150. Can you reject the hypothesis that the average amount spent per person each year on prescription drugs is at least $660 at α=.005? Based on this paragraph of text, use the correct excel output above to answer the following question.

For the hypothesis stated above, what is the conclusion?

a.

There is significant evidence to conclude that the average amount spent per person each year on prescription drugs is less than $660.

b.

None of the answers is correct

c.

There is not significant evidence to conclude that the average amount spent per person each year on prescription drugs is more than $660.

d.

There is significant evidence to conclude that the average amount spent per person each year on prescription drugs is more than $660.

e.

There is not significant evidence to conclude that the average amount spent per person each year on prescription drugs is less than $660.

Homework Answers

Answer #1

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 660
Alternative Hypothesis, Ha: μ < 660

Rejection Region
This is left tailed test, for α = 0.005
Critical value of z is -2.576.
Hence reject H0 if z < -2.576

Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (554.6111 - 660)/(150/sqrt(18))
z = -2.98

P-value Approach
P-value = 0.0014
As P-value < 0.005, reject the null hypothesis.

There is significant evidence to conclude that the average amount spent per person each year on prescription drugs is less than $660.

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