Math / Data & Statistics

QuestionA health expert advertised that people in their their weight loss program would lose MORE than 10 pounds in two months. Twelve (12) customers had the following weight losses: $13,17,9,15,14,9,13,17,15,12,10,8$
Assume a normal distribution for these data. Test whether the mean weight loss was GREATER than 10 lbs . Use $1 \%$ significance level.
What is the ALTERNATIVE HYPOTHESIS, H1 number. 1) H1: mean (greater than)> 10 2) H 1 : mean (greater than) $>12.67$ 3) H1: mean (less than) < 12.67 4) H1: mean (less than) < 10 5) H1: mean Not Equal To 10 5) H1: mean (less than) > 3.11 6) No answer is correct.

Studdy Solution

STEP 1

What is this asking? We're checking if people in a weight loss program lost more than 10 pounds on average, using a 1% significance level. Watch out! Don't confuse the mean weight loss with individual weight losses.
Remember, we're testing the mean!

STEP 2

1. Define the hypotheses
2. Calculate the sample mean and standard deviation
3. Perform the hypothesis test
4. Determine the alternative hypothesis

STEP 3

Alright, let's kick things off by defining our hypotheses!
The null hypothesis $H_0$ is that the mean weight loss is 10 pounds.
The alternative hypothesis $H_1$ is that the mean weight loss is greater than 10 pounds.
This is what we're trying to prove!

STEP 4

First, let's find the sample mean.
Add up all the weight losses and divide by the number of customers:
$\text{Sample Mean} = \frac{13 + 17 + 9 + 15 + 14 + 9 + 13 + 17 + 15 + 12 + 10 + 8}{12}$

STEP 5

Calculate that sum:
$\text{Sum} = 152$

STEP 6

Now, divide by the number of customers:
$\text{Sample Mean} = \frac{152}{12} = 12.67$

STEP 7

Next, let's find the sample standard deviation.
First, calculate the squared differences from the mean:
$(13 - 12.67)^2, (17 - 12.67)^2, ..., (8 - 12.67)^2$

STEP 8

Sum these squared differences and divide by $n-1$ (where $n$ is the number of customers):
$\text{Variance} = \frac{(13 - 12.67)^2 + (17 - 12.67)^2 + \ldots + (8 - 12.67)^2}{11}$

STEP 9

Take the square root to find the standard deviation:
$\text{Standard Deviation} = \sqrt{\text{Variance}}$

STEP 10

Now, let's calculate the test statistic using the formula:
$t = \frac{\text{Sample Mean} - \text{Population Mean}}{\frac{\text{Standard Deviation}}{\sqrt{n}}}$

STEP 11

Substitute the values:
$t = \frac{12.67 - 10}{\frac{\text{Standard Deviation}}{\sqrt{12}}}$

STEP 12

The alternative hypothesis $H_1$ we're testing is that the mean weight loss is greater than 10 pounds.
So, the correct option is 1) H1: mean (greater than)> 10.

STEP 13

The alternative hypothesis $H_1$ is that the mean weight loss is greater than 10 pounds.
Therefore, the correct answer is 1) H1: mean (greater than)> 10.

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Studdy Solution

STEP 1

What is this asking? We're checking if people in a weight loss program lost more than 10 pounds on average, using a 1% significance level. Watch out! Don't confuse the mean weight loss with individual weight losses.
Remember, we're testing the mean!

STEP 2

1. Define the hypotheses
2. Calculate the sample mean and standard deviation
3. Perform the hypothesis test
4. Determine the alternative hypothesis

STEP 3

Alright, let's kick things off by defining our hypotheses!
The null hypothesis $H_0$ is that the mean weight loss is 10 pounds.
The alternative hypothesis $H_1$ is that the mean weight loss is greater than 10 pounds.
This is what we're trying to prove!

STEP 4

First, let's find the sample mean.
Add up all the weight losses and divide by the number of customers:
$\text{Sample Mean} = \frac{13 + 17 + 9 + 15 + 14 + 9 + 13 + 17 + 15 + 12 + 10 + 8}{12}$

STEP 5

Calculate that sum:
$\text{Sum} = 152$

STEP 6

Now, divide by the number of customers:
$\text{Sample Mean} = \frac{152}{12} = 12.67$

STEP 7

Next, let's find the sample standard deviation.
First, calculate the squared differences from the mean:
$(13 - 12.67)^2, (17 - 12.67)^2, ..., (8 - 12.67)^2$

STEP 8

Sum these squared differences and divide by $n-1$ (where $n$ is the number of customers):
$\text{Variance} = \frac{(13 - 12.67)^2 + (17 - 12.67)^2 + \ldots + (8 - 12.67)^2}{11}$

STEP 9

Take the square root to find the standard deviation:
$\text{Standard Deviation} = \sqrt{\text{Variance}}$

STEP 10

Now, let's calculate the test statistic using the formula:
$t = \frac{\text{Sample Mean} - \text{Population Mean}}{\frac{\text{Standard Deviation}}{\sqrt{n}}}$

STEP 11

Substitute the values:
$t = \frac{12.67 - 10}{\frac{\text{Standard Deviation}}{\sqrt{12}}}$

STEP 12

The alternative hypothesis $H_1$ we're testing is that the mean weight loss is greater than 10 pounds.
So, the correct option is 1) H1: mean (greater than)> 10.

STEP 13

The alternative hypothesis $H_1$ is that the mean weight loss is greater than 10 pounds.
Therefore, the correct answer is 1) H1: mean (greater than)> 10.

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Studdy Solution

STEP 1

What is this asking? We're checking if people in a weight loss program lost more than **10 pounds** on average, using a **1% significance level**. Watch out! Don't confuse the **mean weight loss** with individual weight losses.Remember, we're testing the **mean**!

STEP 2

1. Define the hypotheses2. Calculate the sample mean and standard deviation3. Perform the hypothesis test4. Determine the alternative hypothesis

STEP 3

Alright, let's kick things off by defining our hypotheses!The **null hypothesis** H0 H_0 H0​ is that the mean weight loss is **10 pounds**.The **alternative hypothesis** H1 H_1 H1​ is that the mean weight loss is **greater than 10 pounds**.This is what we're trying to prove!

STEP 4

First, let's find the **sample mean**.Add up all the weight losses and divide by the number of customers:Sample Mean=13+17+9+15+14+9+13+17+15+12+10+812\text{Sample Mean} = \frac{13 + 17 + 9 + 15 + 14 + 9 + 13 + 17 + 15 + 12 + 10 + 8}{12}Sample Mean=1213+17+9+15+14+9+13+17+15+12+10+8​

STEP 5

Calculate that sum:Sum=152\text{Sum} = 152Sum=152

STEP 6

Now, divide by the number of customers:Sample Mean=15212=12.67\text{Sample Mean} = \frac{152}{12} = 12.67Sample Mean=12152​=12.67

STEP 7

Next, let's find the **sample standard deviation**.First, calculate the squared differences from the mean:(13−12.67)2,(17−12.67)2,...,(8−12.67)2(13 - 12.67)^2, (17 - 12.67)^2, ..., (8 - 12.67)^2(13−12.67)2,(17−12.67)2,...,(8−12.67)2

STEP 8

Sum these squared differences and divide by n−1 n-1 n−1 (where n n n is the number of customers):Variance=(13−12.67)2+(17−12.67)2+…+(8−12.67)211\text{Variance} = \frac{(13 - 12.67)^2 + (17 - 12.67)^2 + \ldots + (8 - 12.67)^2}{11}Variance=11(13−12.67)2+(17−12.67)2+…+(8−12.67)2​

STEP 9

Take the square root to find the standard deviation:Standard Deviation=Variance\text{Standard Deviation} = \sqrt{\text{Variance}}Standard Deviation=Variance​

STEP 10

Now, let's calculate the **test statistic** using the formula:t=Sample Mean−Population MeanStandard Deviationnt = \frac{\text{Sample Mean} - \text{Population Mean}}{\frac{\text{Standard Deviation}}{\sqrt{n}}}t=n​Standard Deviation​Sample Mean−Population Mean​

STEP 11

Substitute the values:t=12.67−10Standard Deviation12t = \frac{12.67 - 10}{\frac{\text{Standard Deviation}}{\sqrt{12}}}t=12​Standard Deviation​12.67−10​

STEP 12

The alternative hypothesis H1 H_1 H1​ we're testing is that the mean weight loss is **greater than 10 pounds**.So, the correct option is **1) H1: mean (greater than)> 10**.

STEP 13

The alternative hypothesis H1 H_1 H1​ is that the mean weight loss is **greater than 10 pounds**.Therefore, the correct answer is **1) H1: mean (greater than)> 10**.

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What is this asking? We're checking if people in a weight loss program lost more than 10 pounds on average, using a 1% significance level. Watch out! Don't confuse the mean weight loss with individual weight losses.
Remember, we're testing the mean!

1. Define the hypotheses
2. Calculate the sample mean and standard deviation
3. Perform the hypothesis test
4. Determine the alternative hypothesis

Alright, let's kick things off by defining our hypotheses!
The null hypothesis $H_0$ is that the mean weight loss is 10 pounds.
The alternative hypothesis $H_1$ is that the mean weight loss is greater than 10 pounds.
This is what we're trying to prove!

First, let's find the sample mean.
Add up all the weight losses and divide by the number of customers:
$\text{Sample Mean} = \frac{13 + 17 + 9 + 15 + 14 + 9 + 13 + 17 + 15 + 12 + 10 + 8}{12}$

Calculate that sum:
$\text{Sum} = 152$

Now, divide by the number of customers:
$\text{Sample Mean} = \frac{152}{12} = 12.67$

Next, let's find the sample standard deviation.
First, calculate the squared differences from the mean:
$(13 - 12.67)^2, (17 - 12.67)^2, ..., (8 - 12.67)^2$

Sum these squared differences and divide by $n-1$ (where $n$ is the number of customers):
$\text{Variance} = \frac{(13 - 12.67)^2 + (17 - 12.67)^2 + \ldots + (8 - 12.67)^2}{11}$

Take the square root to find the standard deviation:
$\text{Standard Deviation} = \sqrt{\text{Variance}}$

Now, let's calculate the test statistic using the formula:
$t = \frac{\text{Sample Mean} - \text{Population Mean}}{\frac{\text{Standard Deviation}}{\sqrt{n}}}$

Substitute the values:
$t = \frac{12.67 - 10}{\frac{\text{Standard Deviation}}{\sqrt{12}}}$

The alternative hypothesis $H_1$ we're testing is that the mean weight loss is greater than 10 pounds.
So, the correct option is 1) H1: mean (greater than)> 10.

The alternative hypothesis $H_1$ is that the mean weight loss is greater than 10 pounds.
Therefore, the correct answer is 1) H1: mean (greater than)> 10.