Question

(1 point) The owner of a dry cleaning store believes that the mean amount a customer spends on a dry cleaning order exceeds $21.93. A statistician has said to him that he can be statistically sure that his belief is true if the mean amount of 100 randomly chosen customer bills is greater than $23.13, or X > 23.13. It is assumed that the standard deviation in the bill amounts is $12.69 (o = 12.69). (a) Choose the correct statistical hypotheses. A. H: X = 21.93, HA:X #21.93 B. H. : X = 23.13, HA: X > 23.13 C. H : X 21.93, HA: X > 21.93 D. H :u= 21.93 HA: > 21.93 E. H. : X > 23.13, HA: X < 23.13 F. H : M > 21.93, HA:H <21.93 G. H:n = 21.93, HA: +21.93 (b) What is the probability of making a Type I error, using the statistician's criterion? Use at least three decimals in your answer. P(Type 1) = .0500 (C) Unknown to anyone, suppose the mean amount spent by all his customers is $22.17. Find the probability that the owner will conclude that the mean amount a customer spends on a dry cleaning order exceeds $21.93 (his original belief) is correct. Use at least three decimals in your answer. Answer = .508 (d) Suppose the statistician decides to change the sample size to n = 200 and regulate P(Type 1) = 0.05. For what values of X should the null hypothesis in (a) be rejected? A. Reject the null hypothesis if p > 23.41 B. Reject the null hypothesis if X > 24.02 C. Reject the null hypothesis if X > 23.13 D. Reject the null hypothesis if X > 42.8 E. Reject the null hypothesis if X > 21.93 F. Reject the null hypothesis if u > 24.02 G. Reject the null hypothesis if u > 42.8 H. Reject the null hypothesis if X > 23.41

Answer 1

a)

Null and Alternative Hypothesis:

H₀: µ = 21.93

H₁ : µ > 21.93

b)

Test statistic:

z = (x̅- µ)/(σ/√n) = (23.13 - 21.93)/(12.69/√100) = 0.9456

p-value = 1- NORM.S.DIST(0.9456, 1) = 0.1722

Probability of Type I error = 0.1722

c)

1 - β = P(X̅ ≥ 23.13 | µ₁ = 22.17)

= P(Z ≥ (23.13 - 22.17)/(12.69/√100) )

= P(Z ≥ 0.76)

= 1 - P(Z < 0.76)

Using excel function :

= 1 - NORM.S.DIST(0.76, 1)

= 1 - 0.775

= 0.225

d)

µ = 21.93, σ = 12.69, n = 200

Right tailed critical value, z crit = ABS(NORM.S.INV(0.05)) = 1.645

Z = (X̅ - µ) / (σ/√n)

Reject if Z ≥ 1.645

X̅ ≥ µ + z*(σ/√n)

X̅ ≥ 21.93 + (1.645) * 12.69/√200

X̅ ≥ 23.41

The researchers will reject H₀ if the sample mean is greater than 23.41

Answer H.