Math / Data & Statistics

Question1 A six-węek weight loss program of 19 people resulted in a mean weight loss of 6.1 pounds. Weight loss can be assumed normally distributed with known standard deviation of 9.8 lbs. Test whether the weight loss can be claimed to be greater than 0 ? Use the $1 \%$ level of significance. I What is the value of the TEST STATISTIC? $\qquad$ Note: Enter XXX.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 134.784 is entered as $134.78,35.295$ is entered as $35.30, .2749$ is entered as $0.27,-.5650$ is entered as -0.57 $\square$ A)

Studdy Solution

STEP 1

What is this asking? Did the program actually help people lose weight, and can we be super confident about it? Watch out! Don't mix up the sample standard deviation with the population standard deviation!
We're given the population standard deviation here, which simplifies things a bit.
Also, make sure to use the right significance level!

STEP 2

1. Set up the Hypothesis Test
2. Calculate the Test Statistic

STEP 3

Alright, so we're dealing with a hypothesis test!
We want to see if the weight loss is really greater than zero, or if it's just random chance.
Our null hypothesis ( $H_0$ ) is that the true mean weight loss is zero or less: $\mu \le 0$ .
The alternative hypothesis ( $H_1$ ) is what we're trying to prove: that the true mean weight loss is greater than zero: $\mu > 0$ .

STEP 4

Since we know the population standard deviation ( $\sigma = 9.8$ ), we can use a z-test.
Our significance level ( $\alpha$ ) is $\alpha = 0.01$ or 1%.
This means we want to be really sure about our results!

STEP 5

The test statistic for a z-test is calculated using this formula: $z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}}$ Where: * $\bar{x}$ is the sample mean weight loss. * $\mu_0$ is the hypothesized population mean (from our null hypothesis). * $\sigma$ is the population standard deviation. * $n$ is the sample size.

STEP 6

Let's plug in the values!
We have $\bar{x} = 6.1$ , $\mu_0 = 0$ (since we're testing if the weight loss is greater than 0), $\sigma = 9.8$ , and $n = 19$ . $z = \frac{6.1 - 0}{\frac{9.8}{\sqrt{19}}}$

STEP 7

First, let's calculate the denominator: $\frac{9.8}{\sqrt{19}} \approx \frac{9.8}{4.3589} \approx 2.2485$

STEP 8

Now, divide the numerator by this result: $z = \frac{6.1}{2.2485} \approx 2.7128$

STEP 9

We need to round our test statistic to two decimal places, so $z \approx 2.71$ .

STEP 10

The test statistic is $2.71$ .

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Studdy Solution

STEP 1

STEP 2

1. Set up the Hypothesis Test
2. Calculate the Test Statistic

STEP 3

STEP 4

Since we know the population standard deviation ( $\sigma = 9.8$ ), we can use a z-test.
Our significance level ( $\alpha$ ) is $\alpha = 0.01$ or 1%.
This means we want to be really sure about our results!

STEP 5

STEP 6

Let's plug in the values!
We have $\bar{x} = 6.1$ , $\mu_0 = 0$ (since we're testing if the weight loss is greater than 0), $\sigma = 9.8$ , and $n = 19$ . $z = \frac{6.1 - 0}{\frac{9.8}{\sqrt{19}}}$

STEP 7

First, let's calculate the denominator: $\frac{9.8}{\sqrt{19}} \approx \frac{9.8}{4.3589} \approx 2.2485$

STEP 8

Now, divide the numerator by this result: $z = \frac{6.1}{2.2485} \approx 2.7128$

STEP 9

We need to round our test statistic to two decimal places, so $z \approx 2.71$ .

STEP 10

The test statistic is $2.71$ .

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Studdy Solution

STEP 1

STEP 2

1. Set up the Hypothesis Test2. Calculate the Test Statistic

STEP 3

STEP 4

Since we know the **population standard deviation** (σ=9.8\sigma = 9.8σ=9.8), we can use a **z-test**.Our **significance level** (α\alphaα) is α=0.01\alpha = 0.01α=0.01 or **1%**.This means we want to be *really* sure about our results!

STEP 5

STEP 6

Let's plug in the values!We have xˉ=6.1\bar{x} = 6.1xˉ=6.1, μ0=0\mu_0 = 0μ0​=0 (since we're testing if the weight loss is *greater* than 0), σ=9.8\sigma = 9.8σ=9.8, and n=19n = 19n=19. z=6.1−09.819 z = \frac{6.1 - 0}{\frac{9.8}{\sqrt{19}}} z=19​9.8​6.1−0​

STEP 7

First, let's calculate the denominator: 9.819≈9.84.3589≈2.2485 \frac{9.8}{\sqrt{19}} \approx \frac{9.8}{4.3589} \approx 2.2485 19​9.8​≈4.35899.8​≈2.2485

STEP 8

Now, divide the numerator by this result: z=6.12.2485≈2.7128 z = \frac{6.1}{2.2485} \approx 2.7128 z=2.24856.1​≈2.7128

STEP 9

We need to round our **test statistic** to two decimal places, so z≈2.71z \approx 2.71z≈2.71.

STEP 10

The **test statistic** is 2.712.712.71.

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1. Set up the Hypothesis Test
2. Calculate the Test Statistic

Since we know the population standard deviation ( $\sigma = 9.8$ ), we can use a z-test.
Our significance level ( $\alpha$ ) is $\alpha = 0.01$ or 1%.
This means we want to be really sure about our results!

Let's plug in the values!
We have $\bar{x} = 6.1$ , $\mu_0 = 0$ (since we're testing if the weight loss is greater than 0), $\sigma = 9.8$ , and $n = 19$ . $z = \frac{6.1 - 0}{\frac{9.8}{\sqrt{19}}}$

First, let's calculate the denominator: $\frac{9.8}{\sqrt{19}} \approx \frac{9.8}{4.3589} \approx 2.2485$

Now, divide the numerator by this result: $z = \frac{6.1}{2.2485} \approx 2.7128$

We need to round our test statistic to two decimal places, so $z \approx 2.71$ .

The test statistic is $2.71$ .