Question

1. It is known that the average American orders food from a restaurant in person or as take out 4.2 times a week on average. Suppose we take a sample of 30 Americans and find that the average amount of meals ordered out is 3.2 per week with a sample standard deviation of 2.82 meals. We are interested in testing if the average amount of meals ordered out is actually lower than 4.2 now.
   1. Using the sample of 30 adults, what would the 95% confidence interval be for the population mean?
   2. What are the null and alternative hypotheses?
   3. What is the critical value at 95% confidence?
   4. Calculate the test statistic.
   5. Find the p-valu
   6. What conclusion would be made here at the 95% confidence level?
   7. Would my conclusion change if I changed alpha to .01? Show reasonin

Answer 1

(a)

n = 30

s = 2.82

SE = s/

= 2.82/ = 0.5149

= 0.05

ndf = 30 - 1 = 29

From Table, critical values of t = 2.0452

Confidence Interval:

3.2 (2.0452 X 0.5149)

= 3.2 1.0530

= (2.1470 , 4.2540)

So,

Answer is:

2.1470 < < 4.2540

(b)

H0: Null Hypothesis: 4.2

HA: Alternative Hypothesis: 4.2

(c)

= 0.05

ndf = 30 - 1 = 29

One Tail - Left Side Test

From Table, critical value of t = - 1.6991

(d)

n = 30

s = 2.82

SE = s/

= 2.82/ = 0.5149

Test statistic is:

t = (3.2 - 4.2)/0.5149 = - 1.9421

(e)

t score = - 1.9421

ndf = 29

One Tail Left Side Test

By Technology, p - value = 0.0311

(f) Since p value= 0.0311 is less than = 0.05, the difference is significant. Reject null hypothesis.

Conclusion:

The data support the claim that the average amount of meals ordered out is actually lower than 4.2 now.

(g)

For = 0.01:

Since p value= 0.0311 is greater than = 0.05, the difference is not significant. Fail to reject null hypothesis.

Conclusion:

The data do not support the claim that the average amount of meals ordered out is actually lower than 4.2 now.

Thus our conclusion would change if we changed alpha to .01.