15.10.2022

Please help me with this #8

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Step-by-step answer

09.07.2023, solved by verified expert

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Mathematics
Step-by-step answer
P Answered by PhD

The values of the box-plot are: B = {8.2, 8.3, 8.4, 8.5 and 8.9}.

Step-by-step explanation:

The data provided is as follows:

X     Frequency

8        0

8.1        0

8.2        2

8.3        2

8.4        2

8.5        3

8.6        0

8.7         1

8.8        0

8.9         1

 9           0

So, the actual data is:

S = {8.2 , 8.2 , 8.3 , 8.3 , 8.4 , 8.4 , 8.5 , 8.5 , 8.5 , 8.7 , 8.9 }

A boxplot, also known as a box and whisker plot is a method to demonstrate the distribution of a data-set based on the following 5 number summary,

Minimum (shown at the bottom of the chart) First Quartile (shown by the bottom line of the box) Median (or the second quartile) (shown as a line in the center of the box) Third Quartile (shown by the top line of the box) Maximum (shown at the top of the chart).

The data set is arranged in ascending order.

The minimum value is, Min. = 8.2

The lower quartile is,

[\frac{n+1}{4}]^{th}\ obs.=[\frac{11+1}{4}]^{th}\ obs.=3^{rd }\ obs. =8.3,

Q₁ = 8.3.

The median value is,

[\frac{n+1}{2}]^{th}\ obs.=[\frac{11+1}{2}]^{th}\ obs.=6^{th}\ obs.=8.4

Median = 8.4

The upper quartile is,

[\frac{3(n+1)}{4}]^{th}\ obs.=[\frac{3(11+1)}{4}]^{th}\ obs.=9^{th }\ obs. =8.5,

Q₃ = 8.5.

The maximum value is, Max. = 8.9.

So, the values of the box-plot are: B = {8.2, 8.3, 8.4, 8.5 and 8.9}.

Mathematics
Step-by-step answer
P Answered by Master
The min value is the smallest value of data in the list, which is 8.2. The max is 8.9. Ranking all the numbers from smallest to largest, we found that 8.4 is in the 5th place, right in the middle, so it's the median. the upper quartile (bigger than 75% of data) number is 8.5, while the lower quartile (bigger than 25% of data) is 8.3.
Mathematics
Step-by-step answer
P Answered by PhD

The values of the box-plot are: B = {8.2, 8.3, 8.4, 8.5 and 8.9}.

Step-by-step explanation:

The data provided is as follows:

X     Frequency

8        0

8.1        0

8.2        2

8.3        2

8.4        2

8.5        3

8.6        0

8.7         1

8.8        0

8.9         1

 9           0

So, the actual data is:

S = {8.2 , 8.2 , 8.3 , 8.3 , 8.4 , 8.4 , 8.5 , 8.5 , 8.5 , 8.7 , 8.9 }

A boxplot, also known as a box and whisker plot is a method to demonstrate the distribution of a data-set based on the following 5 number summary,

Minimum (shown at the bottom of the chart) First Quartile (shown by the bottom line of the box) Median (or the second quartile) (shown as a line in the center of the box) Third Quartile (shown by the top line of the box) Maximum (shown at the top of the chart).

The data set is arranged in ascending order.

The minimum value is, Min. = 8.2

The lower quartile is,

[\frac{n+1}{4}]^{th}\ obs.=[\frac{11+1}{4}]^{th}\ obs.=3^{rd }\ obs. =8.3,

Q₁ = 8.3.

The median value is,

[\frac{n+1}{2}]^{th}\ obs.=[\frac{11+1}{2}]^{th}\ obs.=6^{th}\ obs.=8.4

Median = 8.4

The upper quartile is,

[\frac{3(n+1)}{4}]^{th}\ obs.=[\frac{3(11+1)}{4}]^{th}\ obs.=9^{th }\ obs. =8.5,

Q₃ = 8.5.

The maximum value is, Max. = 8.9.

So, the values of the box-plot are: B = {8.2, 8.3, 8.4, 8.5 and 8.9}.

Mathematics
Step-by-step answer
P Answered by Specialist
The min value is the smallest value of data in the list, which is 8.2. The max is 8.9. Ranking all the numbers from smallest to largest, we found that 8.4 is in the 5th place, right in the middle, so it's the median. the upper quartile (bigger than 75% of data) number is 8.5, while the lower quartile (bigger than 25% of data) is 8.3.
Mathematics
Step-by-step answer
P Answered by PhD

Explained below.

Step-by-step explanation:

A two-sample test is to be performed to determine whether the mean weight loss for those on low-carbohydrate or Mediterranean diets is greater than the mean weight loss for those on a conventional low-fat diet.

(a)

The hypothesis can be defined as follows:

H₀: μ₁ - μ₂ ≤ 0 vs. Hₐ: μ₁ - μ₂ > 0

(b-1)

Use Excel to perform the t test for independent samples.

The output is attached below.

The test statistic value is, t = 7.58.

(b-2)

The degrees of freedom is, df = 58.

The critical value is, t₅₈ = 1.672.

The rejection region is:

Rejected H₀ if t > t₅₈ = 1.672.

(c)

Decision:

t = 7.58 > t₅₈ = 1.672

The null hypothesis will be rejected at 5% level of significance.

Thus, the nutritionist can conclude that overweight people on low-carbohydrate or Mediterranean diets lost more weight than people on a conventional low-fat diet.

The correct option is Yes.


According to a study published in the New England Journal of Medicine, overweight people on low-carb
Mathematics
Step-by-step answer
P Answered by Specialist

Step-by-step explanation:

a)

Here researcher claims that the mean weight loss for those on  low-carbohydrate or Mediterranean diets is greater than the mean  weight loss for those on a conventional low-fat diet

Set the null and alternative hypotheses to test the above claim as below.

H_0:\mu_1\leq\mu_2\\\\=H_0:\mu_1-\mu_2\leq 0 versus H_A:\mu_1  \mu_2\\\\H_A:\mu_1 -\mu_20

b)

Assume that the variance of the two populations are equal.

Under the null hypothesis, the test statistics is defined as

t= \frac{\bar x_1 - \bar x_2}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

where s_p=\sqrt{\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}}

Use the foll owing Excel instructions and conduct the above test.

Step1: Enter the datainto two columns of spreadsheet.

Step2: Select Data Analysis from Data ribbon.

Step3: Select t-test. Two-sample Assuming equal variances.

Step4: Input the data range for variable 1 and variable 2.

Step5: Enter 0 as the Hypothesized Mean Difference.

Step6: Click OK.

Thus, the resultant outout is as follows:

         

                                                       Low-carb                        Low-fat

Mean                                               9.773333333       6.283333333

Variance                                          3.197885057                    3.160747126

Observations                                  30                                      30

Pooled Variance                             3.179316092

Hypothesized Mean Difference     0

df                                                      58

t Stat                                                 7.580610844

P(T < = t) one-tail                              1.54916E - 10

t Critical one — tail                           1.671552762

P(T <2) two-tail                                  3.09831E - 10

t Critical two - tail                              2.001717484

1) From the above output, the test statistics is obtained as t =  7.5806

2) From the above output, the critical value for one-tailed testis  obtained as t_{crit}=1.672

Rejection rule:  Reject H_0 if t_{crit}1.672

c)

At 5% level of significance, the calculated value of test statistics  is greater than the critical value.

Therefore, reject the null hypothesis.

Hence, the nutritionist conclude that overweight people on  low-carbohydrate or Mediterranean diets lost more weight than  people on a conventional low-fat diet.

Mathematics
Step-by-step answer
P Answered by PhD

Explained below.

Step-by-step explanation:

A two-sample test is to be performed to determine whether the mean weight loss for those on low-carbohydrate or Mediterranean diets is greater than the mean weight loss for those on a conventional low-fat diet.

(a)

The hypothesis can be defined as follows:

H₀: μ₁ - μ₂ ≤ 0 vs. Hₐ: μ₁ - μ₂ > 0

(b-1)

Use Excel to perform the t test for independent samples.

The output is attached below.

The test statistic value is, t = 7.58.

(b-2)

The degrees of freedom is, df = 58.

The critical value is, t₅₈ = 1.672.

The rejection region is:

Rejected H₀ if t > t₅₈ = 1.672.

(c)

Decision:

t = 7.58 > t₅₈ = 1.672

The null hypothesis will be rejected at 5% level of significance.

Thus, the nutritionist can conclude that overweight people on low-carbohydrate or Mediterranean diets lost more weight than people on a conventional low-fat diet.

The correct option is Yes.


According to a study published in the New England Journal of Medicine, overweight people on low-carb
Computers and Technology
Step-by-step answer
P Answered by Specialist

#include <stdlib.h>

#include <stdio.h>

void func1(int product[]){

int orders[6]={0};

for(int i=0;i<70;i++){

orders[product[i]]++;

}

printf("Total number of each type of products that were bought\n");

for(int i=1;i<=5;i++){

printf("Product %d = %d\n",i,orders[i]);

}

}

void func2(int product[],int quantity[],float price[]){

float total_cost=0;

for(int i=0;i<70;i++){

total_cost+= price[product[i]]* quantity[i];

}

printf("The total cost of all 70 orders = %.2f\n",total_cost);

}

void func3(int product[],int quantity[],int destination[],float price[]){

float total_cost=0;

for(int i=0;i<70;i++){

if(destination[i]==8){

total_cost+= price[product[i]]* quantity[i];

}

}

printf("The total cost of all products shipped to destination 8 = %.2f\n",total_cost);

}

void func4(int product[],int quantity[],float price[]){

int total_orders=0;

for(int i=0;i<70;i++){

if(price[product[i]]* quantity[i]>=50){

total_orders++;

}

}

printf("The total number of orders where each order is $50 or more = %d\n",total_orders);

}

void func5(int product[],int quantity[],int origination[],float price[]){

int total_orders=0;

for(int i=0;i<70;i++){

if(origination[i]==3 && price[product[i]]* quantity[i]>=50){

total_orders++;

}

}

printf("The total number of orders that originated from 3 where each order is $50 or more. = %d\n",total_orders);

}

void func6(int product[],int quantity[],int origination[],float price[]){

float total_cost=0;

for(int i=0;i<70;i++){

if(origination[i]==3 && price[product[i]] * quantity[i]>=50){

total_cost += price[product[i]] * quantity[i];

}

}

printf("The total number of orders that originated from 3 where each order is $50 or more. = %.2f\n",total_cost);

}

void func7(int origination[],int destination[]){

int total_orders=0;

for(int i=0;i<70;i++){

if(origination[i]==3 && destination[i]==8){

total_orders++;

}

}

printf("The total number of orders that originated from 3 and shipped to 8. = %d\n",total_orders);

}

void func8(int product[],int quantity[],int origination[],int destination[],float price[]){

float total_cost=0;

for(int i=0;i<70;i++){

if(origination[i]==3 && destination[i]==8){

total_cost += price[product[i]] * quantity[i];

}

}

printf("The total cost of orders that originated from 3 and shipped to 8. = %.2f\n",total_cost);

}

void func9(int destination[]){

int total_orders=0;

for(int i=0;i<70;i++){

if(destination[i]!=8){

total_orders++;

}

}

printf("The total number of orders that was shipped to all destinations except to 8. = %d\n",total_orders);

}

void func10(int product[],int quantity[],int destination[],float price[]){

float total_cost=0;

for(int i=0;i<70;i++){

if(destination[i]!=8){

total_cost += price[product[i]] * quantity[i];;

}

}

printf("The total cost of orders that was shipped to all destinations except to 8. = %.2f\n",total_cost);

}

int main(){

int product[70] = {4, 2, 4, 2, 4, 5, 5, 2, 2, 5, 5, 4, 3, 5, 4, 2, 5, 3, 1, 2, 2, 3, 3, 4, 5, 5, 4, 5, 3, 5, 5, 1, 4, 5, 1, 5, 3, 2, 4, 1, 2, 4, 5, 1, 5, 5, 5, 5, 5, 2, 5, 1, 4, 4, 4, 2, 3, 3, 3, 3, 4, 3, 5, 5, 3, 2, 3, 5, 3, 2};

int quantity[70] = {10, 9, 6, 4, 10, 4, 9, 6, 10, 7, 3, 4, 4, 9, 1, 8, 9, 1, 5, 8, 7, 2, 3, 4, 10, 5, 6, 2, 1, 7, 2, 8, 6, 9, 8, 8, 7, 7, 9, 10, 6, 7, 8, 2, 1, 7, 6, 3, 3, 1, 8, 4, 10, 7, 1, 10, 6, 9, 8, 2, 4, 6, 1, 8, 2, 6, 10, 2, 6, 2};

int origination[70] = {2, 7, 5, 5, 7, 2, 7, 2, 7, 7, 5, 2, 5, 5, 5, 2, 2, 7, 2, 7, 7, 2, 2, 2, 2, 5, 7, 5, 7, 7, 5, 5, 2, 2, 5, 7, 2, 5, 7, 2, 5, 7, 2, 5, 7, 2, 2, 7, 2, 7, 5, 2, 2, 2, 5, 7, 2, 5, 5, 5, 7, 7, 2, 5, 2, 7, 5, 2, 5, 7};

int destination[70] = {8, 7, 3, 10, 2, 6, 4, 5, 1, 3, 5, 9, 5, 8, 6, 4, 3, 7, 1, 2, 7, 2, 8, 2, 2, 1, 2, 6, 10, 2, 7, 7, 8, 6, 8, 8, 4, 8, 3, 10, 6, 9, 4, 9, 5, 1, 7, 3, 1, 7, 5, 5, 4, 9, 3, 10, 8, 1, 1, 1, 1, 8, 10, 3, 5, 2, 8, 7, 4, 10};

float price[6]={0,11.95,7.95,19.95,24.95,15.25};

func1(product);

func2(product,quantity,price);

func3(product,quantity,destination,price);

func4(product,quantity,price);

func5(product,quantity,origination,price);

func6(product,quantity,origination,price);

func7(origination,destination);

func8(product,quantity,origination,destination,price);

func9(destination);

func10(product,quantity,destination,price);

}

Explanation:

The program inputs order, products and cities products are shipped.

After a series of conditional requirements being met, will output the destination each product os going to and the number of products with its associated price.

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