CALCU-Задача 21. Среднее время сборки изделия сост

Solution by Steps

step 1

First, we need to calculate the sample mean

\bar{x}

of the given data:

79, 74, 112, 95, 83, 96, 77, 84, 70, 90

step 2

The sample mean

\bar{x}

is calculated as follows:

\bar{x} = \frac{79 + 74 + 112 + 95 + 83 + 96 + 77 + 84 + 70 + 90}{10} = \frac{860}{10} = 86

minutes

step 3

Next, we calculate the sample standard deviation

s

. The formula for the sample standard deviation is

s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}

step 4

Calculate each

(x_i - \bar{x})^2

:

(79-86)^2 = 49

,

(74-86)^2 = 144

,

(112-86)^2 = 676

,

(95-86)^2 = 81

,

(83-86)^2 = 9

,

(96-86)^2 = 100

,

(77-86)^2 = 81

,

(84-86)^2 = 4

,

(70-86)^2 = 256

,

(90-86)^2 = 16

step 5

Sum these values:

49 + 144 + 676 + 81 + 9 + 100 + 81 + 4 + 256 + 16 = 1416

step 6

Calculate the sample variance:

\frac{1416}{10-1} = \frac{1416}{9} = 157.33

step 7

The sample standard deviation

s

is

\sqrt{157.33} \approx 12.54

minutes

step 8

To construct a 95% confidence interval for the mean, we use the formula

\bar{x} \pm t_{\alpha/2} \frac{s}{\sqrt{n}}

step 9

For a 95% confidence level and

n-1 = 9

degrees of freedom, the critical value

t_{\alpha/2}

from the t-distribution table is approximately 2.262

step 10

Calculate the margin of error:

E = t_{\alpha/2} \frac{s}{\sqrt{n}} = 2.262 \frac{12.54}{\sqrt{10}} \approx 8.98

step 11

The confidence interval is:

86 \pm 8.98

, which gives us

(77.02, 94.98)

Answer

The 95% confidence interval for the new mean assembly time is

(77.02, 94.98)

minutes.

Key Concept

Confidence Interval

Explanation

A confidence interval provides a range of values that is likely to contain the population mean with a certain level of confidence, in this case, 95%.

Solution by Steps

step 1

First, we need to calculate the sample mean

\bar{x}

of the given data:

79, 74, 112, 95, 83, 96, 77, 84, 70, 90

step 2

The sample mean

\bar{x}

is calculated as follows:

\bar{x} = \frac{79 + 74 + 112 + 95 + 83 + 96 + 77 + 84 + 70 + 90}{10} = \frac{860}{10} = 86

minutes

step 3

Next, we calculate the sample standard deviation

s

. The formula for the sample standard deviation is

s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}

, where

x_i

are the sample values and

n

is the sample size

step 4

Calculate each

(x_i - \bar{x})^2

:

(79 - 86)^2 = 49, \quad (74 - 86)^2 = 144, \quad (112 - 86)^2 = 676, \quad (95 - 86)^2 = 81, \quad (83 - 86)^2 = 9,

(96 - 86)^2 = 100, \quad (77 - 86)^2 = 81, \quad (84 - 86)^2 = 4, \quad (70 - 86)^2 = 256, \quad (90 - 86)^2 = 16

step 5

Sum these squared differences:

49 + 144 + 676 + 81 + 9 + 100 + 81 + 4 + 256 + 16 = 1416

step 6

Divide by

n-1

(which is

10-1 = 9

) and take the square root:

s = \sqrt{\frac{1416}{9}} = \sqrt{157.33} \approx 12.54

step 7

To find the 95% confidence interval for the mean, we use the formula:

\bar{x} \pm t_{\alpha/2, n-1} \cdot \frac{s}{\sqrt{n}}

where

t_{\alpha/2, n-1}

is the t-value for 95% confidence and

n-1

degrees of freedom. For

n-1 = 9

,

t_{0.025, 9} \approx 2.262

step 8

Calculate the margin of error:

E = t_{\alpha/2, n-1} \cdot \frac{s}{\sqrt{n}} = 2.262 \cdot \frac{12.54}{\sqrt{10}} \approx 8.98

step 9

The confidence interval is:

86 \pm 8.98

which gives us the interval:

(77.02, 94.98)

Answer

The 95% confidence interval for the new mean assembly time is

(77.02, 94.98)

minutes.

Key Concept

Confidence Interval Calculation

Explanation

The confidence interval provides a range within which we can be 95% confident that the true mean assembly time lies. This is calculated using the sample mean, sample standard deviation, and the t-distribution.

Solution by Steps

step 1

To determine if there is a dependency between hair color and region of residence, we will use the Chi-Square Test for Independence

step 2

First, we need to calculate the expected frequencies for each cell in the table. The expected frequency for a cell is given by the formula:

E_{ij} = \frac{(row\ total\ of\ cell\ i) \times (column\ total\ of\ cell\ j)}{grand\ total}

For example, the expected frequency for Region A and Red hair is:

E_{A,Red} = \frac{(total\ of\ row\ A) \times (total\ of\ column\ Red)}{grand\ total} = \frac{20 \times 20}{100} = 4

step 3

Calculate the expected frequencies for all cells:

E_{A,Red} = 4

E_{A,Light} = \frac{20 \times 30}{100} = 6

E_{A,Dark} = \frac{20 \times 50}{100} = 10

E_{B,Red} = \frac{30 \times 20}{100} = 6

E_{B,Light} = \frac{30 \times 30}{100} = 9

E_{B,Dark} = \frac{30 \times 50}{100} = 15

E_{C,Red} = \frac{50 \times 20}{100} = 10

E_{C,Light} = \frac{50 \times 30}{100} = 15

E_{C,Dark} = \frac{50 \times 50}{100} = 25

step 4

Next, we calculate the Chi-Square statistic using the formula:

\chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}}

where

O_{ij}

is the observed frequency and

E_{ij}

is the expected frequency

step 5

Calculate the Chi-Square statistic for each cell:

\chi^2_{A,Red} = \frac{(2 - 4)^2}{4} = 1

\chi^2_{A,Light} = \frac{(9 - 6)^2}{6} = 1.5

\chi^2_{A,Dark} = \frac{(9 - 10)^2}{10} = 0.1

\chi^2_{B,Red} = \frac{(3 - 6)^2}{6} = 1.5

\chi^2_{B,Light} = \frac{(6 - 9)^2}{9} = 1

\chi^2_{B,Dark} = \frac{(21 - 15)^2}{15} = 2.4

\chi^2_{C,Red} = \frac{(15 - 10)^2}{10} = 2.5

\chi^2_{C,Light} = \frac{(15 - 15)^2}{15} = 0

\chi^2_{C,Dark} = \frac{(20 - 25)^2}{25} = 1

step 6

Sum all the Chi-Square values to get the total Chi-Square statistic:

\chi^2_{total} = 1 + 1.5 + 0.1 + 1.5 + 1 + 2.4 + 2.5 + 0 + 1 = 11

step 7

Determine the degrees of freedom for the test. The degrees of freedom for a Chi-Square test for independence is given by:

df = (number\ of\ rows - 1) \times (number\ of\ columns - 1) = (3 - 1) \times (3 - 1) = 4

step 8

Compare the calculated Chi-Square statistic to the critical value from the Chi-Square distribution table at

\alpha = 0.05

and

df = 4

. The critical value is 9.488. Since

\chi^2_{total} = 11

is greater than 9.488, we reject the null hypothesis

Answer

There is a significant dependency between hair color and region of residence at the

\alpha = 0.05

level.

Key Concept

Chi-Square Test for Independence

Explanation

The Chi-Square Test for Independence is used to determine if there is a significant association between two categorical variables. In this case, it was used to test the relationship between hair color and region of residence.

Solution by Steps

step 1

We are given the data for two production processes and need to test the hypothesis of equality in defect rates. The null hypothesis (

H_0

) is that the defect rates are equal, and the alternative hypothesis (

H_a

) is that they are not equal

step 2

Calculate the sample proportions of defects for each process:

p_1 = \frac{5}{200} = 0.025

p_2 = \frac{2}{151} \approx 0.0132

step 3

Calculate the pooled sample proportion:

p = \frac{5 + 2}{200 + 151} = \frac{7}{351} \approx 0.0199

step 4

Calculate the standard error of the difference between the two proportions:

SE = \sqrt{p(1 - p) \left( \frac{1}{n_1} + \frac{1}{n_2} \right)} = \sqrt{0.0199 \times 0.9801 \left( \frac{1}{200} + \frac{1}{151} \right)} \approx 0.011

step 5

Calculate the z-score:

z = \frac{p_1 - p_2}{SE} = \frac{0.025 - 0.0132}{0.011} \approx 1.073

step 6

Determine the critical value for a two-tailed test at

\alpha = 0.01

. The critical z-value is approximately

\pm 2.576

step 7

Compare the calculated z-score with the critical value. Since 1.073 < 2.576, we fail to reject the null hypothesis

Answer

We fail to reject the null hypothesis, meaning there is not enough evidence to conclude that the defect rates are different between the two processes.

Key Concept

Hypothesis Testing for Proportions

Explanation

We used a z-test for two proportions to determine if there is a significant difference in defect rates between two production processes. The test showed no significant difference at the 0.01 significance level.

Solution by Steps

step 1

We are given the observed frequencies of events

A_1, A_2, A_3,

and

A_4

as 2040, 4940, 1037, and 1983 respectively. The total number of observations is 10000

step 2

The null hypothesis

H_0

states that the ratio of the probabilities of events

A_1, A_2, A_3,

and

A_4

is 2:5:1:2

step 3

To test this hypothesis, we first calculate the expected frequencies based on the given ratio. The total ratio is

2 + 5 + 1 + 2 = 10

step 4

The expected frequencies are calculated as follows: -

E(A_1) = \frac{2}{10} \times 10000 = 2000

-

E(A_2) = \frac{5}{10} \times 10000 = 5000

-

E(A_3) = \frac{1}{10} \times 10000 = 1000

-

E(A_4) = \frac{2}{10} \times 10000 = 2000

step 5

We use the Chi-Square test statistic to compare the observed and expected frequencies. The formula for the Chi-Square test statistic is:

\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}

where

O_i

are the observed frequencies and

E_i

are the expected frequencies

step 6

Substituting the values, we get:

\chi^2 = \frac{(2040 - 2000)^2}{2000} + \frac{(4940 - 5000)^2}{5000} + \frac{(1037 - 1000)^2}{1000} + \frac{(1983 - 2000)^2}{2000}

\chi^2 = \frac{40^2}{2000} + \frac{60^2}{5000} + \frac{37^2}{1000} + \frac{17^2}{2000}

\chi^2 = \frac{1600}{2000} + \frac{3600}{5000} + \frac{1369}{1000} + \frac{289}{2000}

\chi^2 = 0.8 + 0.72 + 1.369 + 0.1445

\chi^2 = 3.0335

step 7

The degrees of freedom for this test is

df = k - 1 = 4 - 1 = 3

step 8

We compare the calculated

\chi^2

value with the critical value from the Chi-Square distribution table at the 0.1 significance level and 3 degrees of freedom. The critical value is approximately 6.251

step 9

Since 3.0335 < 6.251, we fail to reject the null hypothesis

Answer

We fail to reject the null hypothesis at the 0.1 significance level.

Key Concept

Chi-Square Test for Goodness of Fit

Explanation

The Chi-Square Test for Goodness of Fit is used to determine if the observed frequencies of events match the expected frequencies based on a specified ratio. In this case, the observed frequencies did not significantly differ from the expected frequencies, so we fail to reject the null hypothesis.

Solution by Steps

step 1

We need to determine if the observed variables are positively correlated using the sample correlation coefficient

r_{\mathrm{B}} = 0.23

and the significance level

\alpha = 0.05

step 2

First, we calculate the test statistic for the correlation coefficient. The test statistic

t

is given by:

t = \frac{r \sqrt{n-2}}{\sqrt{1-r^2}}

where

r = 0.23

and

n = 123

step 3

Substituting the values into the formula:

t = \frac{0.23 \sqrt{123-2}}{\sqrt{1-0.23^2}} = \frac{0.23 \sqrt{121}}{\sqrt{1-0.0529}} = \frac{0.23 \times 11}{\sqrt{0.9471}} \approx \frac{2.53}{0.973} \approx 2.60

step 4

Next, we compare the calculated

t

value with the critical

t

value from the

t

-distribution table for

n-2 = 121

degrees of freedom at

\alpha = 0.05

. For a two-tailed test, the critical

t

value is approximately

1.98

step 5

Since 2.60 > 1.98, we reject the null hypothesis that the correlation coefficient is zero

step 6

Therefore, we can conclude that the observed variables are positively correlated at the

0.05

significance level

Answer

The observed variables are positively correlated at the

0.05

significance level.

Key Concept

Test statistic for correlation coefficient

Explanation

The test statistic for the correlation coefficient helps determine if the observed correlation is statistically significant. By comparing the calculated

t

value with the critical

t

value, we can decide whether to reject the null hypothesis.

Solution by Steps

step 1

To test the hypothesis using the Chi-Square test, we first need to determine the expected frequencies for a uniform distribution. Since there are 150 numbers and 5 intervals, the expected frequency for each interval is

\frac{150}{5} = 30

step 2

Next, we calculate the Chi-Square statistic using the formula:

\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}

, where

O_i

are the observed frequencies and

E_i

are the expected frequencies

step 3

For the interval 0-19:

O_1 = 31

,

E_1 = 30

, so

\frac{(31 - 30)^2}{30} = \frac{1}{30}

step 4

For the interval 20-39:

O_2 = 32

,

E_2 = 30

, so

\frac{(32 - 30)^2}{30} = \frac{4}{30}

step 5

For the interval 40-59:

O_3 = 33

,

E_3 = 30

, so

\frac{(33 - 30)^2}{30} = \frac{9}{30}

step 6

For the interval 60-79:

O_4 = 25

,

E_4 = 30

, so

\frac{(25 - 30)^2}{30} = \frac{25}{30}

step 7

For the interval 80-99:

O_5 = 29

,

E_5 = 30

, so

\frac{(29 - 30)^2}{30} = \frac{1}{30}

step 8

Summing these values, we get the Chi-Square statistic:

\chi^2 = \frac{1}{30} + \frac{4}{30} + \frac{9}{30} + \frac{25}{30} + \frac{1}{30} = 1.3333

step 9

The degrees of freedom for this test is

df = k - 1 = 5 - 1 = 4

step 10

Using a Chi-Square distribution table, we find the critical value for

\alpha = 0.05

and

df = 4

is approximately 9.488

step 11

Since

\chi^2 = 1.3333

is less than 9.488, we fail to reject the null hypothesis

Answer

The observations conform to a uniform distribution on the interval (0;99) at the 0.05 significance level.

Key Concept

Chi-Square Test for Uniform Distribution

Explanation

The Chi-Square test compares observed frequencies with expected frequencies to determine if the data follows a specified distribution. In this case, the test shows that the data conforms to a uniform distribution.

AskSia

Plus