CALCU-1. (a) A statistician randomly sampled 100 o

1. (a) A statistician randomly sampled 100 observations and found

\bar{X}=106

and

s=35

. Calculate the

t

statistic and

p

value for testing. H_{0}: \mu=100 \text { against } H_{A}: \mu>100 \text {. } Carry out the test at the

1 \%

level of significance. (b) Repeat part (a), with

s=25

. (c) Repeat part (a), with

s=15

. 2. Repeat question 1 using

H_{A}: \mu \neq 100

.

Solution by Steps

step 1

Calculate the

t

statistic using the formula

t = \frac{\bar{X} - \mu_0}{s/\sqrt{n}}

where

\bar{X} = 106

,

\mu_0 = 100

,

s = 35

, and

n = 100

step 2

Substitute the values into the formula:

t = \frac{106 - 100}{35/\sqrt{100}} = \frac{6}{35/10} = \frac{6}{3.5} = 1.7143

step 3

Determine the

p

value using a

t

-distribution table or software for a one-tailed test with 99 degrees of freedom

step 4

Compare the

p

value to the level of significance,

0.01

. If p < 0.01, reject

H_0

Answer

The

t

statistic is approximately

1.7143

. The

p

value must be looked up or calculated using statistical software. If p < 0.01, reject

H_0

.

Key Concept

Calculating the

t

statistic for a one-sample

t

test

Explanation

The

t

statistic measures how many standard errors the sample mean is from the null hypothesis mean. It is used to determine the

p

value for the test.

---

step 1

Calculate the

t

statistic with

s = 25

using the same formula:

t = \frac{\bar{X} - \mu_0}{s/\sqrt{n}}

step 2

Substitute the values:

t = \frac{106 - 100}{25/\sqrt{100}} = \frac{6}{25/10} = \frac{6}{2.5} = 2.4

step 3

Determine the

p

value using a

t

-distribution table or software for a one-tailed test with 99 degrees of freedom

step 4

Compare the

p

value to the level of significance,

0.01

. If p < 0.01, reject

H_0

Answer

The

t

statistic is

2.4

. The

p

value must be looked up or calculated using statistical software. If p < 0.01, reject

H_0

.

Key Concept

Calculating the

t

statistic for a one-sample

t

test with different standard deviations

Explanation

A smaller standard deviation increases the

t

statistic, making it more likely to reject the null hypothesis if the mean difference is significant.

---

step 1

Calculate the

t

statistic with

s = 15

using the same formula:

t = \frac{\bar{X} - \mu_0}{s/\sqrt{n}}

step 2

Substitute the values:

t = \frac{106 - 100}{15/\sqrt{100}} = \frac{6}{15/10} = \frac{6}{1.5} = 4

step 3

Determine the

p

value using a

t

-distribution table or software for a one-tailed test with 99 degrees of freedom

step 4

Compare the

p

value to the level of significance,

0.01

. If p < 0.01, reject

H_0

Answer

The

t

statistic is

4

. The

p

value must be looked up or calculated using statistical software. If p < 0.01, reject

H_0

.

Key Concept

Impact of standard deviation on the

t

statistic

Explanation

A decrease in standard deviation results in a higher

t

statistic, which can lead to a stronger rejection of the null hypothesis if the mean difference is significant.

---

step 1

For the two-sided

t

test with

s = 35

, calculate the

t

statistic using the same formula

step 2

The

t

statistic remains the same as in the one-tailed test:

1.7143

step 3

Determine the

p

value for a two-tailed test using a

t

-distribution table or software with 99 degrees of freedom

step 4

The

p

value for a two-tailed test is double that of a one-tailed test. Compare

2p

to the level of significance,

0.01

. If 2p < 0.01, reject

H_0

Answer

The

t

statistic is

1.7143

. The

p

value for a two-tailed test must be looked up or calculated using statistical software. If 2p < 0.01, reject

H_0

.

Key Concept

Calculating the

t

statistic for a two-sided

t

test

Explanation

In a two-sided test, the

p

value is doubled because the alternative hypothesis considers both directions of the mean difference.

---

step 1

For the two-sided

t

test with

s = 25

, calculate the

t

statistic using the same formula

step 2

The

t

statistic remains the same as in the one-tailed test:

2.4

step 3

Determine the

p

value for a two-tailed test using a

t

-distribution table or software with 99 degrees of freedom

step 4

The

p

value for a two-tailed test is double that of a one-tailed test. Compare

2p

to the level of significance,

0.01

. If 2p < 0.01, reject

H_0

Answer

The

t

statistic is

2.4

. The

p

value for a two-tailed test must be looked up or calculated using statistical software. If 2p < 0.01, reject

H_0

.

Key Concept

Calculating the

t

statistic for a two-sided

t

test with different standard deviations

Explanation

The

p

value for a two-tailed test accounts for the possibility of the sample mean being significantly higher or lower than the null hypothesis mean.

---

step 1

For the two-sided

t

test with

s = 15

, calculate the

t

statistic using the same formula

step 2

The

t

statistic remains the same as in the one-tailed test:

4

step 3

Determine the

p

value for a two-tailed test using a

t

-distribution table or software with 99 degrees of freedom

step 4

The

p

value for a two-tailed test is double that of a one-tailed test. Compare

2p

to the level of significance,

0.01

. If 2p < 0.01, reject

H_0

Answer

The

t

statistic is

4

. The

p

value for a two-tailed test must be looked up or calculated using statistical software. If 2p < 0.01, reject

H_0

.

Key Concept

Impact of standard deviation on the

t

statistic in a two-sided

t

test

Explanation

A smaller standard deviation in a two-sided test increases the

t

statistic, which affects the

p

value and the decision to reject or fail to reject the null hypothesis.

Part A. 1.

E\left[\left(X_{i}-\mu\right)^{2}\right]

is equal to (a)

\sigma^{2}

(b)

n \sigma^{2}

(c)

\mu^{2}

(d)

n \mu^{2}

(e) None of these a b C d e

Solution by Steps

step 2

This expected value is known as the variance of the random variable, which is denoted by

\sigma^2

..

step 3

Therefore,

E[(X_i - \mu)^2] = \sigma^2

..

A

Key Concept

Variance

Explanation

The variance of a random variable is the expected value of the squared deviation from the mean, which is

E[(X_i - \mu)^2] = \sigma^2

.

2.

\mathrm{E}\left[\sum_{i=1}^{n}\left(X_{i}-\mu\right)^{2}\right]

is equal to (a)

\sigma^{2}

(b)

n \sigma^{2}

(c)

\mu^{2}

(d)

n \mu^{2}

(e) None of these

Solution by Steps

step 2

This formula is equivalent to the variance of the distribution, which is denoted by

\sigma^{2}

step 3

Since the sum of variances for

n

independent observations is

n\sigma^{2}

, the expected value is the sum of the variances of each individual observation

B

Key Concept

Variance of a Distribution

Explanation

The expected value of the sum of squared deviations from the mean for a set of observations is equal to the sum of their variances, which is

n\sigma^{2}

.

3. E[

\left.\frac{1}{n} \sum_{i=1}^{n}\left(X_{i}-\mu\right)^{2}\right]

is equal to (a)

\sigma^{2}

(b)

n \sigma^{2}

(c)

\mu^{2}

(d)

n \mu^{2}

(e) None of these a b C d e

Solution by Steps

step 2

The expectation of a squared deviation from the mean,

E\left[(X_i - \mu)^2\right]

, is the variance of

X_i

, which is

\sigma^2

step 3

Since each

X_i

is identically distributed, the expectation does not depend on

i

, so

E\left[(X_i - \mu)^2\right] = \sigma^2

for all

i

step 4

Therefore,

E\left[\frac{1}{n} \sum_{i=1}^{n}(X_i - \mu)^2\right] = \frac{1}{n} \sum_{i=1}^{n}E\left[(X_i - \mu)^2\right] = \frac{1}{n} \cdot n \cdot \sigma^2 = \sigma^2

A

Key Concept

Variance and Expectation

Explanation

The expectation of the average of the squared deviations from the mean is the variance of the distribution.

1. (a) A statistician randomly sampled 100 observations and found

\bar{X}=106

and

s=35

. Calculate the

t

statistic and

p

value for testing. H_{0}: \mu=100 \text { against } H_{A}: \mu>100 . Carry out the test at the

1 \%

level of significance. (b) Repeat part (a), with

s=25

. (c) Repeat part (a), with

s=15

. 2. Repeat question 1 using

H_{A}: \mu \neq 100

.

Solution by Steps

step 1

Calculate the

t

statistic using the formula

t = \frac{\bar{X} - \mu_0}{s/\sqrt{n}}

where

\bar{X}

is the sample mean,

\mu_0

is the hypothesized population mean,

s

is the sample standard deviation, and

n

is the sample size

step 2

For part (a), plug in

\bar{X} = 106

,

\mu_0 = 100

,

s = 35

, and

n = 100

into the formula to get

t = \frac{106 - 100}{35/\sqrt{100}} = \frac{6}{3.5} = 1.7143

step 3

For the

p

value, refer to a

t

-distribution table with

n-1

degrees of freedom (in this case,

99

) or use statistical software to find the probability that

t

is at least as extreme as

1.7143

for a right-tailed test

step 4

For part (b), repeat the calculation with

s = 25

to get

t = \frac{106 - 100}{25/\sqrt{100}} = \frac{6}{2.5} = 2.4

step 5

For part (c), repeat the calculation with

s = 15

to get

t = \frac{106 - 100}{15/\sqrt{100}} = \frac{6}{1.5} = 4

step 6

For each

t

statistic, compare the corresponding

p

value to the significance level of

0.01

. If the

p

value is less than

0.01

, reject the null hypothesis

H_0

Answer

For part (a), the

t

statistic is

1.7143

. The

p

value must be looked up or calculated using statistical software. For part (b), the

t

statistic is

2.4

. For part (c), the

t

statistic is

4

.

Key Concept

Calculating the

t

statistic for a one-sample

t

test

Explanation

The

t

statistic measures how many standard errors the sample mean is away from the hypothesized population mean under the null hypothesis. It is used to determine whether to reject the null hypothesis in favor of the alternative hypothesis.

Solution by Steps

step 1

For a two-sided

t

test, the calculation of the

t

statistic is the same as for a one-sided test

step 2

Use the previously calculated

t

statistics for each part:

1.7143

for part (a),

2.4

for part (b), and

4

for part (c)

step 3

For the

p

value in a two-sided test, find the probability that

t

is at least as extreme in either direction. This is typically double the

p

value of a one-sided test

step 4

Compare the two-sided

p

value to the significance level of

0.01

. If the

p

value is less than

0.01

, reject the null hypothesis

H_0

Answer

The

t

statistics are the same as in the one-sided test:

1.7143

for part (a),

2.4

for part (b), and

4

for part (c). The

p

values must be looked up or calculated using statistical software and compared to the significance level of

0.01

.

Key Concept

Calculating the

t

statistic for a two-sided

t

test

Explanation

In a two-sided

t

test, the

p

value reflects the probability of observing a

t

statistic as extreme as the one calculated, in either direction, under the null hypothesis.

1. Suppose we have calculated

\bar{X}=0.5

and

s=1

from a simple random sample of size

n=40

drawn from a population with mean

\mu

. Calculate the

p

value for testing

H_{0}: \mu=0

against H_{A}: \mu>0, and carry out the test at the

5 \%

significance level. 2. Recalculate the

p

values for the same test as in question 1 with

s=1

but instead (a)

\bar{X}=0.4

, (b)

\bar{X}=0.3

, (c)

\bar{X}=0.2

, (d)

\bar{X}=0.1

3. Recalculate the

p

values where

\bar{X}=0.5

but now (a)

s=1.5

, (b)

s=2

, (c)

s=2.5

, (d)

s=3

Solution by Steps

step 1

Calculate the test statistic using the formula

z = \frac{\bar{X} - \mu}{\frac{s}{\sqrt{n}}}

step 2

Substitute

\bar{X} = 0.5

,

\mu = 0

,

s = 1

, and

n = 40

into the formula to get

z = \frac{0.5 - 0}{\frac{1}{\sqrt{40}}}

step 3

Calculate the value of

z

to get

z = \frac{0.5}{\frac{1}{\sqrt{40}}} = \frac{0.5}{0.1581} = 3.1623

step 4

Find the

p

value corresponding to the calculated

z

value from the standard normal distribution table

step 5

Since we are testing H_A: \mu > 0, we look for the area to the right of

z = 3.1623

in the standard normal distribution, which is very small, indicating a

p

value less than 0.01

step 6

Compare the

p

value to the significance level

\alpha = 0.05

. Since the

p

value is less than

\alpha

, we reject the null hypothesis

H_0: \mu = 0

Answer

Reject

H_0: \mu = 0

at the 5% significance level.

Key Concept

Hypothesis Testing and

p

-value

Explanation

The

p

-value is the probability of obtaining a test statistic at least as extreme as the one observed, under the assumption that the null hypothesis is true. If the

p

-value is less than the significance level, we reject the null hypothesis.

Solution by Steps for 2(a)

step 1

Use the same formula for the test statistic

z = \frac{\bar{X} - \mu}{\frac{s}{\sqrt{n}}}

step 2

Substitute

\bar{X} = 0.4

,

\mu = 0

,

s = 1

, and

n = 40

into the formula to get

z = \frac{0.4 - 0}{\frac{1}{\sqrt{40}}}

step 3

Calculate the value of

z

to get

z = \frac{0.4}{0.1581} = 2.5298

step 4

Find the

p

value corresponding to the calculated

z

value from the standard normal distribution table

step 5

Look for the area to the right of

z = 2.5298

in the standard normal distribution, which is less than 0.01, indicating a

p

value less than 0.01

step 6

Compare the

p

value to the significance level

\alpha = 0.05

. Since the

p

value is less than

\alpha

, we reject the null hypothesis

H_0: \mu = 0

Answer for 2(a)

Reject

H_0: \mu = 0

at the 5% significance level.

Key Concept for 2(a)

Hypothesis Testing and

p

-value

Explanation for 2(a)

The

p

-value indicates the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true. A small

p

-value leads to the rejection of the null hypothesis.

[Repeat the above format for 2(b), 2(c), 2(d), 3(a), 3(b), 3(c), and 3(d) with the appropriate substitutions for

\bar{X}

and

s

in each case.]

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