[SOLVED] ceng222 - hw3

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Answer 1

a)

µ₀ = 7, X = 7.8, σ = 1.4, n = 17 and α = 0.05:

H₀ : µ = 7 and H_A : µ > 7

For level α = 0.05 test with a right-tail alternative, Z_α = Z_0.05 = 1.645:

Then, the acceptance region is (−∞,1.645]

As 2.356 > 1.645, we reject H₀, the null hypothesis. Thus, the service can be regarded as successful. b)

As X = 7.8, the new mean will be:

For the new mean,

As 0.7952 < 1.645, we accept the null hypothesis. Thus, the service can not be regarded as successful.

c)

For 45 customers, the new mean will be:

For the new mean,

As 2.8749 > 1.645, we reject the null hypothesis. Thus, the mistake does not affect the success and the service can be regarded as successful.

d)

For µ₀ = 8, test statistic would be negative. Thus, the test statistic would be in the acceptance region and we accept the null hypothesis.

Answer 2

µ₀ = 5.8, X = 6.2, σ = 1.5, n = 55 and α = 0.05

H₀ : µ = 5.8 and H_A : µ > 5.8

For level α = 0.05 test with a right-tail alternative, Z_α = Z_0.05 = 1.645

As 0.1978 < 1.645, we accept the null hypothesis. We can not state that the new vaccine protects for a longer duration.

Answer 3

a)

At %95 confidence, α = 0.05 and Z_α/2 = 1.96:

The margin of error for ˆp_red = 0.48 at %95 confidence level is:

1.

In other words, ˆp_red is at 48 ± 4.9%

The margin of error for ˆp_blue = 0.37 at %95 confidence level is:

1.

In other words, ˆp_blue is at 37 ± 4.7%

b)

The margin of error for ˆp_red − pˆ_blue = 0.11 at %95 confidence level is:

1.

In other words, ˆp_red − pˆ_blue is at 11 ± 6.8%

c)

Margin of error of Reds is larger.

The reason is, margin of error is at maximum for sample proportion ˆp = 0.5 and ˆp_red is closer to

0.5.

d)

With the new size, margins of error will be multiplied with:

Thus, new margins will be less than margins with 400 participants.