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Q.26

Expert-verified

Elementary Statistics

Found in: Page 393

Book edition 9th

Author(s) Weiss, Neil

Pages 590 pages

ISBN 9780321989505

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Short Answer

Cheese Consumption. Refer to Problem $24$ . The following table provides last year's cheese consumption: in pounds, 35 randomly selected Americans.
$\begin{array}{l} 46 & 29 & 33 & 38 & 42 & 40 & 34 \\ 33 & 32 & 36 & 28 & 47 & 26 & 42 \\ 36 & 32 & 45 & 24 & 39 & 28 & 33 \\ 44 & 33 & 26 & 37 & 27 & 31 & 36 \end{array} \begin{array}{l} 37 & 37 & 36 & 22 & 44 & 36 & 29 \end{array}$
At the $10 %$ significance level, do the data provide sufficient evidence to conclude that last year's mean cheese consumption for all Americans has increased over the 2010 mean? Assume that $σ = 6.9 lb$ . Use a z-test. (Note: The sum of the data is $1218 lb$ .)
Given the conclusion in part (a), if an error has been made, what type must it be? Explain your answer.

(a) There is sufficient data to support the assumption that total American cheese consumption increased last year compared to $2010$ .

(b) Type I is error

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Part (a) Step 1: Given information

Given in the question that, last year's cheese consumption, in pounds, for 35 randomly selected Americans

$\begin{array}{l} 46 & 29 & 33 & 38 & 42 & 40 & 34 \\ 33 & 32 & 36 & 28 & 47 & 26 & 42 \\ 36 & 32 & 45 & 24 & 39 & 28 & 33 \\ 44 & 33 & 26 & 37 & 27 & 31 & 36 \\ 37 & 37 & 36 & 22 & 44 & 36 & 29 \end{array}$

Part (a) Step 2: Explanation

The given information states that the mean has increased from 33. The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis states that the population mean is equal to the value mentioned in the claim. If the null hypothesis is the claim, then the alternative hypothesis states the opposite of the null hypothesis.

$H_{0} : μ = 33 H_{0} : μ > 33$

The mean is calculated by dividing the total of all values by the number of values:

x = \frac{\sum_{} x i}{n} = \frac{1218}{35} = 34.8

The sample mean's sampling distribution has a mean

μ

and a standard deviation of

\frac{σ}{\sqrt{n}}

. The sample mean is subtracted from the population mean and divided by the standard deviation to get the

z

value:

z = \frac{x - μ}{\frac{σ}{\sqrt{n}}} = \frac{34.8 - 33}{6.9 \sqrt{35}} \approx 1.54

The test is left-tailed if the alternative hypothesis

H_{a}

contains

<

.The test is right-tailed if the alternative hypothesis

H_{a}

contains

>

.The test is two-tailed if the alternative hypothesis

H_{a}

contains

\neq

.If the null hypothesis is true, the

P

-value is the probability of getting a value that is more extreme or equal to the standardized test statistic

z

.

Part (a) Step 3: P-value

Here, the test is two-tailed, and the

P

-value represents the likelihood that the

z

-score is greater than

z = 1.54

.

P (z > 1.54)

The probability

P (z > 1.54)

is shown in the table's row beginning with "

1.5

" and column beginning with ".04 ".

P = P (z > 1.54) = 1 - P (z < 1.54) = 1 - 0.9382 = 0.0618

The null hypothesis is rejected if the

P

-value is less than the significance level

a

.

0.0618 < 0.10

,Therefore, Reject

H_{0}

There is sufficient data to support the assumption that total American cheese consumption increased last year compared to 2010.

Part (b) Step 1: Given information

In order to define the type of error, we must first describe the error type (a).

Step 2: Explanation

Type I error is When the null hypothesis $H_{0}$ is true, reject the null hypothesis localid="1654764107290" $H_{0}$ .
Type Il error is When the null hypothesis localid="1654764112725" $H_{0}$ is untrue, the null hypothesis localid="1654764119120" $H_{0}$ is not rejected.They could have made a Type I error because they rejected the null hypothesis localid="1654764123114" $H_{0}$ in part (a).

Most popular questions for Math Textbooks

In Exercise 8.146 on page 345, we introduced one-sided one mean-t-intervals. The following relationship holds between hypothesis test and confidence intervals for one-mean t-procedures: For a right-tailed hypothesis test at the significance level $α$ , the null hypothesis $H_{0} : μ = μ_{0}$ will be rejected in favor of the alternative hypothesis data-custom-editor="chemistry" $H_{a} : μ > μ_{0}$ if and only if $μ_{0}$ is less than or equal to the $(1 - α)$ -level lower confidence bound for $μ$ . In each case, illustrate the preceding relationship by obtaining the appropriate lower confidence bound and comparing the result to the conclusion of the hypothesis test in the specified exercise.

Part (a): Exercise 9.114 (both parts)

Part (b) Exercise 9.115

The normal probability curve and histogram of the data are shown in figure; $σ$ is known.

Perform Hypothesis test for mean of the population from which data is obtained and decide whether to use z-test, t-test or neither. Explain your answer.

Determine the critical value(s) for a one-mean z-test at the 1 % significance level if the test is

a. right tailed.

b. left tailed.

c. two tailed.

The normal probability curve and stem-to-leaf diagram of the data are shown in figure; $σ$ is unknown.

Perform Hypothesis test for mean of the population from which data is obtained and decide whether to use z-test, t-test or neither. Explain your answer.

Ankle Brachial Index. The ankle brachial index (ABI) compares the blood pressure of a patient's arm to the blood pressure of the patient's leg. The ABI can be an indicator of different diseases, including arterial diseases. A healthy (or normal) ABI is 0.9 or greater. In a study by M. McDermott et al. titled "Sex Differences in Peripheral Arterial Disease: Leg Symptoms and Physical Functioning" (Journal of the American Geriatrics Society, Vol. 51, No. 2, Pp. 222-228), the researchers obtained the ABI of 187 women with peripheral arterial disease. The results were a mean ABI of 0.64 with a standard deviation of 0.15 At the 1 % significance level, do the data provide sufficient evidence to conclude that, on average, women with peripheral arterial disease have an unhealthy ABI?

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