Mathematics : asked on Nov25
 17.10.2020

If 7 is mean of 10,3,4,a,6,9,10. Find the value of a

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09.07.2023, solved by verified expert
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a = 7

Step-by-step explanation:

Given:

7 is mean of 10,3,4,a,6,9,10

To Find:

Value of a

Solve:

Mean → If 7 is mean of 10,3,4,a,6,9,10. Find the value, №18010830, 17.10.2020 19:13

⇒ 42+a=49

Subtract 42 from both sides

x + 42 - 42 = 49 - 42

Simplify

⇒ a=7  

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Mathematics
Step-by-step answer
P Answered by Master

a = 7

Step-by-step explanation:

Given:

7 is mean of 10,3,4,a,6,9,10

To Find:

Value of a

Solve:

Mean → \frac{(4+10+3+6+9+10+a)}{7}=7

⇒ 42+a=49

Subtract 42 from both sides

x + 42 - 42 = 49 - 42

Simplify

⇒ a=7  

~Learn with Lenvy~

Mathematics
Step-by-step answer
P Answered by Master

I'm not really sure because I have not done this in a while but I think True

Step-by-step explanation:

...

Mathematics
Step-by-step answer
P Answered by Specialist

I'm not really sure because I have not done this in a while but I think True

Step-by-step explanation:

...

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 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.

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