Essential Statistics
Essential Statistics
2nd Edition
ISBN: 9781259570643
Author: Navidi
Publisher: MCG
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Chapter 9.1, Problem 22E

a.

To determine

State the null and alternate hypotheses.

a.

Expert Solution
Check Mark

Answer to Problem 22E

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

That is, there is no significant difference between the mean weight loss of two diets (low-carbohydrate diet and low-fat diet).

Alternate hypothesis:

H1:μ1μ2

That is, there is a significant difference between the mean weight loss of two diets (low-carbohydrate diet and low-fat diet).

Explanation of Solution

Calculation:

It is given that, the statistics regarding the weight loss of candidates under low-carbohydrate diet is n1=77, x¯1=4.7, s1=7.16 and the statistics regarding the weight loss of candidates under low-fat diet is n2=79, x¯2=2.6, s2=5.9. Furthermore, it is given that α=0.01.

Hypothesis:

Hypothesis is an assumption about the parameter of the population, and the assumption may or may not be true.

Let μ1 be the population mean weight loss subjects under low-carbohydrate diet and μ2 be the population mean weight loss subjects under low-fat diet.

Claim:

Here, the claim is whether the mean weight loss differs between the two diets or not.

The hypotheses are given below:

Null hypothesis:

Null hypothesis is a statement which is tested for statistical significance in the test. The decision criterion indicates whether the null hypothesis will be rejected or not in the favor of alternate hypothesis.

H0:μ1=μ2

That is, there is no significant difference between the mean weight loss of two diets (low-carbohydrate diet and low-fat diet).

Alternate hypothesis:

Alternate hypothesis is contradictory statement of the null hypothesis

H0:μ1μ2

That is, there is a significant difference between the mean weight loss of two diets (low-carbohydrate diet and low-fat diet).

b.

To determine

Find the value of t-test statistic.

b.

Expert Solution
Check Mark

Answer to Problem 22E

The value of t-test statistic is 1.996454.

Explanation of Solution

Calculation:

Requirements for a small sample t-test:

  • The samples must be independent and drawn randomly from the populations.
  • Either the sample size must be greater than 30 or the population must be approximately normal.

The test statistic for the small sample t is obtained as follows:

t=(x¯1x¯2)(μ1μ2)s12n1+s22n2

Under the null hypothesis, (μ1μ2)=0.

Therefore the test statistic is,

t=(x¯1x¯2)0s12n1+s22n2=4.72.67.16277+5.9279=2.11.051865=1.996454

Thus, the test statistic is 1.996454.

c.

To determine

Find the number of degrees of freedom for the test statistic.

c.

Expert Solution
Check Mark

Answer to Problem 22E

The number of degrees of freedom for the test statistic is 76.

Explanation of Solution

Calculation:

Degrees of freedom:

The degrees of freedom for t using computer package is,

Δ=[(s12n1+s22n2)]2(s12n1)2n11+(s22n2)2n21

The degrees of freedom, when computing by hand is smaller of n11 and n21.

The degrees of freedom is,

d.f=min(n11,n21)=min(771,791)=min(76,78)=76

Thus, the degree of freedom is 76.

d.

To determine

Check whether the null hypothesis is rejected at α=0.01.

d.

Expert Solution
Check Mark

Answer to Problem 22E

There is not enough evidence to reject the null hypothesis at α=0.01.

There is no significant difference between the mean weight loss of two diets (low-carbohydrate diet and low-fat diet).

Explanation of Solution

P-value:

Step-by-step procedure to obtain the P-value using the MINITAB software:

  • Choose Graph > Probability Distribution Plot
  • Choose View Probability > OK.
  • From Distribution, choose‘t’ distribution.
  • In Degrees of freedom, enter 76.
  • Click the Shaded Area tab.
  • Choose X value and Both Tail for the region of the curve to shade.
  • In X-value enter 1.996454.
  • Click OK.

Output using the MINITAB software is given below:

Essential Statistics, Chapter 9.1, Problem 22E

From the MINITAB output, the P-value is 2×0.02473=0.04946.

Thus, the P-value is 0.04946.

Decision rule based on P-value:

If Pvalueα, then reject the null hypothesis H0.

If Pvalue>α, then fail to reject the null hypothesis H0.

Here, the level of significance is α=0.01.

Conclusion based on P-value approach:

The P-value is 0.04946 and α value is 0.01.

Here, P-value is greater than the α value.

That is, 0.04946(=Pvalue)>0.01(=α).

By the rejection rule, fail to reject the null hypothesis.

Hence, it can be concluded that there is no significant difference between the mean weight loss of two diets (low-carbohydrate diet and low-fat diet).

Thus, it can be concluded that there is sufficient evidence to infer the equality of two population means at α=0.01_.

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Chapter 9 Solutions

Essential Statistics

Ch. 9.1 - Prob. 11ECh. 9.1 - Prob. 12ECh. 9.1 - Prob. 13ECh. 9.1 - Prob. 14ECh. 9.1 - Prob. 15ECh. 9.1 - Prob. 16ECh. 9.1 - Prob. 17ECh. 9.1 - Prob. 18ECh. 9.1 - Prob. 19ECh. 9.1 - Prob. 20ECh. 9.1 - Prob. 21ECh. 9.1 - Prob. 22ECh. 9.1 - Prob. 23ECh. 9.1 - Prob. 24ECh. 9.1 - Prob. 25ECh. 9.1 - Prob. 26ECh. 9.1 - 27. Does this diet help? A group of 78 people...Ch. 9.1 - Prob. 28ECh. 9.1 - Prob. 29ECh. 9.1 - Prob. 30ECh. 9.1 - Prob. 31ECh. 9.1 - Prob. 32ECh. 9.1 - Prob. 33ECh. 9.1 - Prob. 34ECh. 9.1 - Prob. 35ECh. 9.1 - Prob. 36ECh. 9.1 - Prob. 37ECh. 9.1 - 38. Interpret calculator display: The following...Ch. 9.1 - Prob. 39ECh. 9.1 - Prob. 40ECh. 9.1 - Prob. 41ECh. 9.2 - Prob. 1CYUCh. 9.2 - Prob. 2CYUCh. 9.2 - Prob. 3CYUCh. 9.2 - Prob. 4CYUCh. 9.2 - Prob. 5ECh. 9.2 - Prob. 6ECh. 9.2 - Prob. 7ECh. 9.2 - Prob. 8ECh. 9.2 - Prob. 9ECh. 9.2 - Prob. 10ECh. 9.2 - Prob. 11ECh. 9.2 - Prob. 12ECh. 9.2 - Prob. 13ECh. 9.2 - Prob. 14ECh. 9.2 - Prob. 15ECh. 9.2 - Prob. 16ECh. 9.2 - Prob. 17ECh. 9.2 - Prob. 18ECh. 9.2 - Prob. 19ECh. 9.2 - Prob. 20ECh. 9.2 - Prob. 21ECh. 9.2 - Prob. 22ECh. 9.2 - Prob. 23ECh. 9.2 - Prob. 24ECh. 9.2 - Prob. 25ECh. 9.2 - Prob. 26ECh. 9.2 - Prob. 27ECh. 9.2 - Prob. 28ECh. 9.2 - Prob. 29ECh. 9.2 - Prob. 30ECh. 9.2 - Prob. 31ECh. 9.2 - Prob. 32ECh. 9.2 - Prob. 33ECh. 9.2 - Prob. 34ECh. 9.2 - Prob. 35ECh. 9.2 - Prob. 36ECh. 9.2 - Prob. 37ECh. 9.3 - Prob. 1CYUCh. 9.3 - Prob. 2CYUCh. 9.3 - Prob. 3CYUCh. 9.3 - Prob. 4CYUCh. 9.3 - Prob. 5ECh. 9.3 - Prob. 6ECh. 9.3 - Prob. 7ECh. 9.3 - Prob. 8ECh. 9.3 - Prob. 9ECh. 9.3 - Prob. 10ECh. 9.3 - Prob. 11ECh. 9.3 - Prob. 12ECh. 9.3 - Prob. 13ECh. 9.3 - Prob. 14ECh. 9.3 - Prob. 15ECh. 9.3 - SAT coaching: A sample of 32 students took a class...Ch. 9.3 - Prob. 17ECh. 9.3 - Prob. 18ECh. 9.3 - Prob. 19ECh. 9.3 - Tires and fuel economy: A tire manufacturer is...Ch. 9.3 - Prob. 21ECh. 9.3 - Prob. 22ECh. 9.3 - Prob. 23ECh. 9.3 - Prob. 24ECh. 9.3 - Prob. 25ECh. 9.3 - Prob. 26ECh. 9.3 - Prob. 27ECh. 9.3 - Prob. 28ECh. 9.3 - Prob. 29ECh. 9.3 - Prob. 30ECh. 9.3 - Advantage of matched pairs: Refer to Exercise...Ch. 9.3 - Prob. 32ECh. 9 - A sample of 15 weight litters is tested to see how...Ch. 9 - Prob. 2CQCh. 9 - Prob. 3CQCh. 9 - Prob. 4CQCh. 9 - Prob. 5CQCh. 9 - Prob. 6CQCh. 9 - Prob. 7CQCh. 9 - Prob. 8CQCh. 9 - Prob. 9CQCh. 9 - Prob. 10CQCh. 9 - Prob. 11CQCh. 9 - In a survey of 300 randomly selected female and...Ch. 9 - Prob. 13CQCh. 9 - Prob. 14CQCh. 9 - Prob. 15CQCh. 9 - Prob. 1RECh. 9 - Prob. 2RECh. 9 - Prob. 3RECh. 9 - Prob. 4RECh. 9 - Prob. 5RECh. 9 - Prob. 6RECh. 9 - Prob. 7RECh. 9 - Prob. 8RECh. 9 - Prob. 9RECh. 9 - Polling results: A simple random sample of 400...Ch. 9 - Treating bean plants: In a study to measure the...Ch. 9 - Prob. 12RECh. 9 - Prob. 13RECh. 9 - Prob. 14RECh. 9 - Prob. 15RECh. 9 - Prob. 1WAICh. 9 - Prob. 2WAICh. 9 - Describe the differences between performing a...Ch. 9 - Prob. 4WAICh. 9 - In what ways is the procedure for constructing a...Ch. 9 - Prob. 6WAICh. 9 - Prob. 7WAICh. 9 - Prob. 1CSCh. 9 - Prob. 2CSCh. 9 - Prob. 3CSCh. 9 - Prob. 4CSCh. 9 - Prob. 5CSCh. 9 - Prob. 6CS
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