Chapter 15.1, Problem 11E

Weight loss: A weight loss company claims that the median weight loss for people who follow their program is at least 15 pounds. Weight losses for a random sample of 20 people in the program were recorded, and the results were as follows. Can you conclude that the claim is false? Use the $\alpha =0.05$ level of significance.

1. State the appropriate null and alternate hypotheses.
2. Compute the value of the test statistic.
3. Find the critical value.
4. State a conclusion.

To determine

To find:The value of test statistic.

Answer to Problem 11E

The hypothesis are $H_0\text{:}m\ge 15$ and $H_1\text{:}m<15$ .

Explanation of Solution

Given information:

The given data is, $14,8,12,17,15,7,7,6,8,9,6,11,2,0,11,9,6,7,6,17$ .

Concept used:

The test statistic is,

If the sample size is less than or equal to $25$ then the test statistic is $X$ .

If the sample size is greater than $25$ then the test statics is,

  $z=\frac{x+0.5-\frac{n}{2}}{\frac{\sqrt{n}}{2}}$

Here, the number of times the less frequent sign occur is $x$ .

The conditions for test statistics are shown below in table.

Type of test	Test statistic
Right-tailed test	Number of minus signs
Left-tailed test	Number of plus signs
Two-tailed test	Number of plus or minus signs, whichever is smaller

Calculations:

The hypothesis are,

  $H_0\text{:}$ The median weight loss for people who follow their program is at least $15$ pound.

  $H_1\text{:}$ The median weight loss for people who follow their program is less than $15$ pound.

Therefore, the hypothesis are $H_0\text{:}m\ge 15$ and $H_1\text{:}m<15$ .

To determine

To find:The value of test statistic.

Answer to Problem 11E

The value of test statistic is $2$ .

Explanation of Solution

Given information:

The given data is, $14,8,12,17,15,7,7,6,8,9,6,11,2,0,11,9,6,7,6,17$ .

Concept used:

The test statistic is,

If the sample size is less than or equal to $25$ then the test statistic is $X$ .

If the sample size is greater than $25$ then the test statics is,

  $z=\frac{x+0.5-\frac{n}{2}}{\frac{\sqrt{n}}{2}}$

Here, the number of times the less frequent sign occur is $x$ .

The conditions for test statistics are shown below in table.

Type of test	Test statistic
Right-tailed test	Number of minus signs
Left-tailed test	Number of plus signs
Two-tailed test	Number of plus or minus signs, whichever is smaller

Calculations:

The value of test statistic is shown below.

Sample	Difference	Signs
14	14-15 =-I	-
8	8-15 =-7	-
12	12 -15 =-3	-
17	17 -15=+2	+
15	15-15 =0	0
7	7-15= -8	-
7	7-15= -8	-
6	6-15 =-9	-
8	8-15 =-7	-
9	9-15 =--6	-
6	6-15=-9	-
11	11-15 =-4	-
2	2- 15=-13	-
0	0-15=-15	-
11	11-15 =-4	-
9	9-15 =--6	-
6	6-15 =-9	-
7	7-15= -8	-
6	6-15 =-9	-
17	17 - 15=+2	+

From the above table, the number of minus sign is $17$ and the number of plus sign is $2$ and the sample size is $19$ .

Since, the sample size is less than $25$ .

The test statistic is given by the number of plus or minus sign which are smaller in number.

  $x=2$

Therefore, the value of test statistic for the given sample is $2$ .

To determine

To find:The critical value.

Answer to Problem 11E

The critical value is $5$ .

Explanation of Solution

Given information:

The level of significance is $0.05$ .

Calculations:

The critical value for one-tailed test at of significance of $0.05$ is $5$ .

Therefore, the critical value is $5$ .

To determine

To find:The conclusion for the test.

Answer to Problem 11E

The median weight loss for people who follow their program is less than $15$ pounds.

Explanation of Solution

Given information:

The level of significance is $0.05$ .

Calculations:

The critical value for one-tailed test at of significance of $0.05$ is $5$ .

Since, the test statistics is $2$ which less than the critical value.

Thus, the hypothesis $H_0$ is rejected.

Therefore, the median weight loss for people who follow their program is less than $15$ pounds.