CALCU-16E Errors in hypothesis testing 715Exercise

Solution by Steps

step 1

A Type I error occurs when we reject the null hypothesis when it is actually true

step 2

In this scenario, the null hypothesis is that the special high-protein feed does not result in a higher average weight gain

step 3

A Type I error would be concluding that the special high-protein feed results in a higher average weight gain when it actually does not

Question 1b

step 1

A Type II error occurs when we fail to reject the null hypothesis when it is actually false

step 2

In this scenario, the null hypothesis is that the special high-protein feed does not result in a higher average weight gain

step 3

A Type II error would be concluding that the special high-protein feed does not result in a higher average weight gain when it actually does

Question 2a

step 1

A 'false positive' in medical testing corresponds to a Type I error

step 2

This is because it indicates that the test has incorrectly identified a patient as having TB when they do not

Question 2b

step 1

A 'false negative' in medical testing corresponds to a Type II error

step 2

This is because it indicates that the test has failed to identify a patient as having TB when they actually do

Question 3a

step 1

We need to find the value of

c

such that

\operatorname{Pr}(\bar{X} \leq c \mid \mu=28.3)=0.01

step 2

Given that the sample mean

\bar{X}

is normally distributed with mean

\mu=28.3

and standard deviation

\sigma/\sqrt{n}

, where

\sigma=5.1

and

n=20

step 3

The standard error of the mean is

\frac{5.1}{\sqrt{20}} \approx 1.14

step 4

Using the standard normal distribution, we find the z-score corresponding to the 1% significance level, which is approximately

-2.33

step 5

We then use the formula

c = \mu + z \cdot \text{SE}

to find

c = 28.3 + (-2.33) \cdot 1.14 \approx 25.655

Question 3b

step 1

We need to find the probability that \bar{X} > c \mid \mu=24.0

step 2

Using the value of

c

from part (a),

c \approx 25.655

step 3

The standard error of the mean is

\frac{5.1}{\sqrt{20}} \approx 1.14

step 4

We calculate the z-score for

\bar{X} = 25.655

with

\mu=24.0

as

z = \frac{25.655 - 24.0}{1.14} \approx 1.45

step 5

Using the standard normal distribution, we find \operatorname{Pr}(Z > 1.45) \approx 0.073

Question 4a

step 1

We need to find the value of

c

such that

\operatorname{Pr}(\bar{X} \leq c \mid \mu=60)=0.05

step 2

Given that the sample mean

\bar{X}

is normally distributed with mean

\mu=60

and standard deviation

\sigma/\sqrt{n}

, where

\sigma=14

and

n=100

step 3

The standard error of the mean is

\frac{14}{\sqrt{100}} = 1.4

step 4

Using the standard normal distribution, we find the z-score corresponding to the 5% significance level, which is approximately

-1.645

step 5

We then use the formula

c = \mu + z \cdot \text{SE}

to find

c = 60 + (-1.645) \cdot 1.4 \approx 57.697

Question 4b

step 1

We need to find the probability that \bar{X} > c \mid \mu=58

step 2

Using the value of

c

from part (a),

c \approx 57.697

step 3

The standard error of the mean is

\frac{14}{\sqrt{100}} = 1.4

step 4

We calculate the z-score for

\bar{X} = 57.697

with

\mu=58

as

z = \frac{57.697 - 58}{1.4} \approx -0.216

step 5

Using the standard normal distribution, we find \operatorname{Pr}(Z > -0.216) \approx 0.586

Answer

Type I error: Concluding that the special high-protein feed results in a higher average weight gain when it actually does not.

Type II error: Concluding that the special high-protein feed does not result in a higher average weight gain when it actually does.

False positive corresponds to a Type I error.

False negative corresponds to a Type II error.

The largest value of the sample mean that would support the conclusion that Gavin's mean travel time is now less than 28.3 minutes is approximately 25.655.

The probability that Gavin concludes that his travel time has not been reduced is approximately 0.073.

The largest value of the sample mean that would support the conclusion that the mean examination score is now less than 60 is approximately 57.697.

The probability that teachers conclude that the new syllabus has not resulted in a lower mean score is approximately 0.586.

Key Concept

Type I and Type II errors in hypothesis testing

Explanation

Type I error is rejecting a true null hypothesis, while Type II error is failing to reject a false null hypothesis.

Solution by Steps

step 1

To determine the values of the sample mean that would support the conclusion that the average age at graduation has increased, we need to perform a hypothesis test. The null hypothesis (

H_0

) is that the average age at graduation has not increased, and the alternative hypothesis (

H_a

) is that it has increased

step 2

We use the

z

-test for the sample mean. The test statistic is given by

z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}

, where

\bar{x}

is the sample mean,

\mu

is the population mean,

\sigma

is the standard deviation, and

n

is the sample size

step 3

Given that

\mu = 24.5

years,

\sigma = 2.8

years, and

n = 100

, we need to find the critical value of

z

at the

1\%

level of significance. For a one-tailed test at the

1\%

level, the critical value of

z

is approximately 2.33

step 4

Setting up the inequality for the sample mean: \frac{\bar{x} - 24.5}{2.8 / \sqrt{100}} > 2.33

step 5

Solving for

\bar{x}

: \bar{x} > 24.5 + 2.33 \times \frac{2.8}{\sqrt{100}} = 24.5 + 2.33 \times 0.28 = 24.5 + 0.6524 = 25.1524

step 6

Therefore, the sample mean must be greater than 25.1524 years to support the conclusion that the average age at graduation has increased at the

1\%

level of significance

5a Answer

A

Key Concept

Hypothesis Testing

Explanation

Hypothesis testing involves comparing a sample statistic to a population parameter to determine if there is enough evidence to reject the null hypothesis in favor of the alternative hypothesis.

---

step 1

To find the probability of a Type II error, we need to calculate the probability that the sample mean falls within the acceptance region when the true mean is 24.5 years

step 2

The acceptance region is defined by the critical value found in the previous part, which is

\bar{x} \leq 25.1524

step 3

Given that the true mean is 24.5 years, we use the

z

-test formula:

z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}

step 4

Substituting the values:

z = \frac{25.1524 - 24.5}{2.8 / \sqrt{100}} = \frac{0.6524}{0.28} = 2.33

step 5

The probability of a Type II error is the probability that

z \leq 2.33

when the true mean is 24.5 years. Using the standard normal distribution table,

P(z \leq 2.33) \approx 0.9901

step 6

Therefore, the probability of a Type II error is approximately 0.9901

5b Answer

B

Key Concept

Type II Error

Explanation

A Type II error occurs when the null hypothesis is not rejected when it is false. The probability of a Type II error is denoted by

\beta

.

---

step 1

To determine the values of the sample mean that would support the conclusion that sales have increased after the marketing campaign, we need to perform a hypothesis test. The null hypothesis (

H_0

) is that the mean daily sales have not increased, and the alternative hypothesis (

H_a

) is that they have increased

step 2

We use the

z

-test for the sample mean. The test statistic is given by

z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}

, where

\bar{x}

is the sample mean,

\mu

is the population mean,

\sigma

is the standard deviation, and

n

is the sample size

step 3

Given that

\mu = \$2000

,

\sigma = \$500

, and

n = 10

, we need to find the critical value of

z

at the

5\%

level of significance. For a one-tailed test at the

5\%

level, the critical value of

z

is approximately 1.645

step 4

Setting up the inequality for the sample mean: \frac{\bar{x} - 2000}{500 / \sqrt{10}} > 1.645

step 5

Solving for

\bar{x}

: \bar{x} > 2000 + 1.645 \times \frac{500}{\sqrt{10}} = 2000 + 1.645 \times 158.11 = 2000 + 260.83 = 2260.83

step 6

Therefore, the sample mean must be greater than \

2260.83 to support the conclusion that sales have increased after the marketing campaign at the

5\%$ level of significance

6a Answer

C

Key Concept

Critical Value

Explanation

The critical value is the threshold value that the test statistic must exceed in order to reject the null hypothesis. It is determined by the significance level of the test.

---

step 1

To find the probability of a Type II error, we need to calculate the probability that the sample mean falls within the acceptance region when the true mean is \$2400

step 2

The acceptance region is defined by the critical value found in the previous part, which is

\bar{x} \leq 2260.83

step 3

Given that the true mean is \

2400, we use the

z

-test formula:

z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}$

step 4

Substituting the values:

z = \frac{2260.83 - 2400}{500 / \sqrt{10}} = \frac{-139.17}{158.11} = -0.88

step 5

The probability of a Type II error is the probability that

z \leq -0.88

when the true mean is \

2400. Using the standard normal distribution table,

P(z \leq -0.88) \approx 0.1894$

step 6

Therefore, the probability of a Type II error is approximately 0.1894

6b Answer

D

Key Concept

Type II Error

Explanation

A Type II error occurs when the null hypothesis is not rejected when it is false. The probability of a Type II error is denoted by

\beta

.

---

step 1

To determine the probability that the researcher rejects the null hypothesis when it is true (a Type I error), we need to find the critical value of the sample mean

step 2

The null hypothesis (

H_0

) is that the population mean time to complete the task is 27.5 seconds, and the alternative hypothesis (

H_a

) is that it is more than 27.5 seconds

step 3

We use the

z

-test for the sample mean. The test statistic is given by

z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}

, where

\bar{x}

is the sample mean,

\mu

is the population mean,

\sigma

is the standard deviation, and

n

is the sample size

step 4

Given that

\mu = 27.5

seconds,

\sigma = 3.2

seconds, and

n = 25

, we need to find the critical value of

z

at the

5\%

level of significance. For a one-tailed test at the

5\%

level, the critical value of

z

is approximately 1.645

step 5

Setting up the inequality for the sample mean: \frac{\bar{x} - 27.5}{3.2 / \sqrt{25}} > 1.645

step 6

Solving for

\bar{x}

: \bar{x} > 27.5 + 1.645 \times \frac{3.2}{\sqrt{25}} = 27.5 + 1.645 \times 0.64 = 27.5 + 1.0528 = 28.5528

step 7

Therefore, the sample mean must be greater than 28.5528 seconds to reject the null hypothesis

step 8

The probability of a Type I error is the probability that the sample mean is greater than 28.5528 seconds when the true mean is 27.5 seconds. Using the standard normal distribution table, P(z > 1.645) \approx 0.05

step 9

Therefore, the probability of a Type I error is approximately 0.05

7a Answer

A

Key Concept

Type I Error

Explanation

A Type I error occurs when the null hypothesis is rejected when it is true. The probability of a Type I error is denoted by

\alpha

.

---

step 1

To find the probability of a Type II error, we need to calculate the probability that the sample mean falls within the acceptance region when the true mean is 29.0 seconds

step 2

The acceptance region is defined by the critical value found in the previous part, which is

\bar{x} \leq 28.5528

step 3

Given that the true mean is 29.0 seconds, we use the

z

-test formula:

z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}

step 4

Substituting the values:

z = \frac{28.5528 - 29.0}{3.2 / \sqrt{25}} = \frac{-0.4472}{0.64} = -0.6981

step 5

The probability of a Type II error is the probability that

z \leq -0.6981

when the true mean is 29.0 seconds. Using the standard normal distribution table,

P(z \leq -0.6981) \approx 0.2420

step 6

Therefore, the probability of a Type II error is approximately 0.2420

7b Answer

B

Key Concept

Type II Error

Explanation

A Type II error occurs when the null hypothesis is not rejected when it is false. The probability of a Type II error is denoted by

\beta

.

---

step 1

To determine if the researcher has set up an effective study, we need to consider the probabilities of Type I and Type II errors

step 2

The probability of a Type I error is 0.05, which is a standard significance level and indicates a low probability of incorrectly rejecting the null hypothesis

step 3

The probability of a Type II error is 0.2420, which is relatively high and indicates a higher probability of failing to reject the null hypothesis when it is false

step 4

An effective study should have a low probability of both Type I and Type II errors. In this case, the probability of a Type II error is relatively high, suggesting that the study may not be very effective in detecting a true effect

step 5

Therefore, the researcher may need to increase the sample size or use a different significance level to reduce the probability of a Type II error and improve the effectiveness of the study

7c Answer

C

Key Concept

Effectiveness of Study

Explanation

An effective study should minimize both Type I and Type II errors to ensure accurate and reliable results.

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