Question

Suppose you sample six people on their number of visits to Chick-fil-A in a month given by variable Y.

Y

3

6

7

10

13

15

A) What is the mean of Y? Standard Deviation of Y?

(B) What is the Standard Error of Y?

(C) Construct 90% and 95% Confidence Intervals for Y.

(D) Test the hypothesis that the mean of Y is 5. Show the null hypothesis, your test statistic, and the critical statistic. Interpret your findings.

Answer 1

a)

sample mean, xbar = 9
sample standard deviation, s = 4.5166

b)
stanadard error = (4.5166/sqrt(6))
= 1.8439

c)

sample size, n = 6
degrees of freedom, df = n - 1 = 5

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 2.015

ME = tc * s/sqrt(n)
ME = 2.015 * 4.5166/sqrt(6)
ME = 3.715

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (9 - 2.015 * 4.5166/sqrt(6) , 9 + 2.015 * 4.5166/sqrt(6))
CI = (5.28 , 12.72)

sample mean, xbar = 9
sample standard deviation, s = 4.5166
sample size, n = 6
degrees of freedom, df = n - 1 = 5

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.571

ME = tc * s/sqrt(n)
ME = 2.571 * 4.5166/sqrt(6)
ME = 4.741

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (9 - 2.571 * 4.5166/sqrt(6) , 9 + 2.571 * 4.5166/sqrt(6))
CI = (4.26 , 13.74)

d)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 5
Alternative Hypothesis, Ha: μ ≠ 5

Rejection Region
This is two tailed test, for α = 0.1 and df = 5
Critical value of t are -2.015 and 2.015.
Hence reject H0 if t < -2.015 or t > 2.015

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (9 - 5)/(4.5166/sqrt(6))
t = 2.169

P-value Approach
P-value = 0.082
As P-value < 0.1, reject the null hypothesis.