A random sample of 400 voters in a certain city are asked if they favor an additional 4% gasoline tax to provide badly needed revenues for street repairs. If more than 220 but fewer than 260 favor the sales tax, we shall conclude that 60% of the voters are for it.a.) Find the probability of committing a type I error if 60% of the voters favor the increased tax.b.) What is the probability of committing a type II error using this test procedure if actually 48% of the voters are in favor of the additional gasoline tax?

Answer:

A) α = 0.04136

B) β = 0.00256

Step-by-step explanation:

We are given;

Sample size; n = 400

Proportion; p = 60% = 0.6

Formula for mean is;

μ = np

μ = 400 × 0.6 = 240

Standard deviation is given by;

σ = √npq

Where q = 1 - p = 1 - 0.6 = 0.4

σ = √(400 × 0.6 × 0.4)

σ = √96

σ = 9.8

A) our null hypothesis is at p = 0.6

Probability of making a type I error means we reject the null hypothesis when it is true.

This can be expressed in reference to the question as;

α = P(x < 220) + P(x > 260) all at p = 0.6

Now,

P(x < 220) = z = (x¯ - μ)/σ = (220 - 240)/9.8 = -2.04

Also;

P(x > 260) = z = (260 - 240)/9.8 = 2.04

Now, from z-distribution table probability of a z-score of -2.04 is 0.02068.

Also, probability of z-score of 2.04 is (1 - P(z < 2.04) = 1 - 0.97932 = 0.02068

Thus;

α = 0.02068 + 0.02068

α = 0.04136

B) Type II error occurs when we fail to reject the null hypothesis even though it's false.

In this case our alternative hypothesis is at p = 48% = 0.48

Thus;

μ = np

μ = 400 × 0.48 = 192

Standard deviation is given by;

σ = √npq

Where q = 1 - p = 1 - 0.48 = 0.52

σ = √(400 × 0.48 × 0.52)

σ = √99.84

σ = 9.992

Type II error would be given by;

β = [((x1¯ - μ)/σ) < z > ((x2¯ - μ)/σ)]

β = [((220 - 192)/9.992) < z > ((260 - 192)/9.992)]

β = (2.8 < z > 6.81)

Rearranging this gives us;

β = P(z < 6.81) - P(z < 2.8)

From z-distribution tables, we have;

β = 1 - 0.99744

β = 0.00256

In this exercise, we have to calculate the gasoline rates, and the answer will be:

A)[tex]\alpha= 0.04136[/tex]

B)[tex]\beta= 0.00256[/tex]

We are given that the sample size; n = 400, proportion; p = 60% = 0.6 so the formula for mean is;

[tex]\mu= np\\\mu=(400)(0.6)= 240[/tex]

Standard deviation is given by;

[tex]\sigma= \sqrt{npq}[/tex]

Where:

[tex]q = 1 \\p= 0.6\\q-p=0.4[/tex]

[tex]\sigma= \sqrt{(400)(0.6)(0.4)} \\=\sqrt{96} = 9.8[/tex]

A)Our null hypothesis is at p = 0.6. Probability of making a type I error means we reject the null hypothesis when it is true. This can be expressed in reference to the question as:

[tex]\alpha= P(x<220)+P(x>260)[/tex]

Now,

[tex]P(x < 220) = z = (x^-- \mu)/\sigma = (220 - 240)/9.8 = -2.04\\P(x > 260) = z = (260 - 240)/9.8 = 2.04[/tex]

Now, from z-distribution table probability of a z-score of -2.04 is 0.02068. Also, probability of z-score of 2.04 is:

[tex](1 - P(z < 2.04) = 1 - 0.97932 = 0.02068[/tex]

Thus;

[tex]\alpha= 0.02068+0.02068\\\alpha= 0.04136[/tex]

B) Type II error occurs when we fail to reject the null hypothesis even though it's false. In this case our alternative hypothesis is at p = 48% = 0.48. Thus:

[tex]\mu= np\\\mu= (400)(0.48)=192[/tex]

Standard deviation is given by;

[tex]\sigma= \sqrt{npq} \\=\sqrt{(400)(0.48)(0.52)} = 9.992[/tex]

Type II error would be given by:

[tex]\beta= [((x^{-1} - \mu)/\sigma) < z > ((x^{-2} - \mu)/\sigma)]\\= [((220 - 192)/9.992) < z > ((260 - 192)/9.992)]\\= \beta = (2.8 < z > 6.81)\\[/tex]

Rearranging this gives us:

[tex]\beta = P(z < 6.81) - P(z < 2.8)\\= 0.00256[/tex]

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