Math / Data & Statistics

QuestionI In a survey on 50 gas stations in an area, the mean cost of regular gas was $\$ 1.95$ per gallon. Was this cost less than the national average cost of $\$ 2.00$ where the assumed known standard deviation is $\$ 0.20$ per gallon? Use a $1 \%$ significance level. What is the P-VALUE from Table A2 for testing HO? See two entry formats below.
Note: Two formats possible: 1) Enter LT01 if P-value is less than 0.01 2) 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? We're checking if the average gas price of $\$1.95$ per gallon at 50 gas stations is significantly less than the national average of $\$2.00$ , using a significance level of $1\%$ . Watch out! Don't forget to use the correct significance level and standard deviation in your calculations!

STEP 2

1. Define the null and alternative hypotheses
2. Calculate the test statistic
3. Find the P-value
4. Determine if the result is significant

STEP 3

Alright, let's get started by setting up our hypotheses!
The null hypothesis $H_0$ is that the mean cost of gas in our area is equal to the national average, so $H_0: \mu = 2.00$ .

STEP 4

The alternative hypothesis $H_a$ is that the mean cost is less than the national average, so $H_a: \mu < 2.00$ .

STEP 5

Next, we need to calculate the test statistic using the formula for the z-score in hypothesis testing:
$z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ where: - $\bar{x} = 1.95$ (the sample mean), - $\mu = 2.00$ (the population mean), - $\sigma = 0.20$ (the standard deviation), - $n = 50$ (the sample size).

STEP 6

Plug in the numbers:
$z = \frac{1.95 - 2.00}{\frac{0.20}{\sqrt{50}}}$

STEP 7

Calculate the denominator:
$\frac{0.20}{\sqrt{50}} = \frac{0.20}{7.071} \approx 0.0283$

STEP 8

Now, calculate the z-score:
$z = \frac{-0.05}{0.0283} \approx -1.77$

STEP 9

With our z-score of $-1.77$ , we need to find the P-value.
This tells us the probability of observing a sample mean as extreme as ours, assuming the null hypothesis is true.

STEP 10

Using a standard normal distribution table (Table A2), find the P-value corresponding to $z = -1.77$ .
The P-value is approximately $0.0384$ .

STEP 11

Compare the P-value to the significance level.
Our significance level is $0.01$ or $1\%$ .

STEP 12

Since $0.0384$ is greater than $0.01$ , we fail to reject the null hypothesis.
This means there's not enough evidence to say the gas price is significantly less than the national average.

STEP 13

The P-value is $0.0384$ , which is greater than the significance level of $0.01$ .
Therefore, we enter 0.0384 as the answer, rounded to two decimal places: 0.04.

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Studdy Solution

STEP 1

STEP 2

1. Define the null and alternative hypotheses
2. Calculate the test statistic
3. Find the P-value
4. Determine if the result is significant

STEP 3

Alright, let's get started by setting up our hypotheses!
The null hypothesis $H_0$ is that the mean cost of gas in our area is equal to the national average, so $H_0: \mu = 2.00$ .

STEP 4

The alternative hypothesis $H_a$ is that the mean cost is less than the national average, so $H_a: \mu < 2.00$ .

STEP 5

STEP 6

Plug in the numbers:
$z = \frac{1.95 - 2.00}{\frac{0.20}{\sqrt{50}}}$

STEP 7

Calculate the denominator:
$\frac{0.20}{\sqrt{50}} = \frac{0.20}{7.071} \approx 0.0283$

STEP 8

Now, calculate the z-score:
$z = \frac{-0.05}{0.0283} \approx -1.77$

STEP 9

With our z-score of $-1.77$ , we need to find the P-value.
This tells us the probability of observing a sample mean as extreme as ours, assuming the null hypothesis is true.

STEP 10

Using a standard normal distribution table (Table A2), find the P-value corresponding to $z = -1.77$ .
The P-value is approximately $0.0384$ .

STEP 11

Compare the P-value to the significance level.
Our significance level is $0.01$ or $1\%$ .

STEP 12

Since $0.0384$ is greater than $0.01$ , we fail to reject the null hypothesis.
This means there's not enough evidence to say the gas price is significantly less than the national average.

STEP 13

The P-value is $0.0384$ , which is greater than the significance level of $0.01$ .
Therefore, we enter 0.0384 as the answer, rounded to two decimal places: 0.04.

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Studdy Solution

STEP 1

STEP 2

1. Define the null and alternative hypotheses2. Calculate the test statistic3. Find the P-value4. Determine if the result is significant

STEP 3

Alright, let's get started by setting up our hypotheses!The **null hypothesis** H0H_0H0​ is that the mean cost of gas in our area is equal to the national average, so H0:μ=2.00H_0: \mu = 2.00H0​:μ=2.00.

STEP 4

The **alternative hypothesis** HaH_aHa​ is that the mean cost is less than the national average, so Ha:μ<2.00H_a: \mu < 2.00Ha​:μ<2.00.

STEP 5

STEP 6

Plug in the numbers:z=1.95−2.000.2050z = \frac{1.95 - 2.00}{\frac{0.20}{\sqrt{50}}}z=50​0.20​1.95−2.00​

STEP 7

Calculate the denominator:0.2050=0.207.071≈0.0283\frac{0.20}{\sqrt{50}} = \frac{0.20}{7.071} \approx 0.028350​0.20​=7.0710.20​≈0.0283

STEP 8

Now, calculate the z-score:z=−0.050.0283≈−1.77z = \frac{-0.05}{0.0283} \approx -1.77z=0.0283−0.05​≈−1.77

STEP 9

With our z-score of −1.77-1.77−1.77, we need to find the P-value.This tells us the probability of observing a sample mean as extreme as ours, assuming the null hypothesis is true.

STEP 10

Using a standard normal distribution table (Table A2), find the P-value corresponding to z=−1.77z = -1.77z=−1.77.The P-value is approximately 0.03840.03840.0384.

STEP 11

Compare the P-value to the significance level.Our significance level is 0.010.010.01 or 1%1\%1%.

STEP 12

Since 0.03840.03840.0384 is greater than 0.010.010.01, we **fail to reject** the null hypothesis.This means there's not enough evidence to say the gas price is significantly less than the national average.

STEP 13

The P-value is 0.03840.03840.0384, which is greater than the significance level of 0.010.010.01.Therefore, we enter **0.0384** as the answer, rounded to two decimal places: **0.04**.

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1. Define the null and alternative hypotheses
2. Calculate the test statistic
3. Find the P-value
4. Determine if the result is significant

Alright, let's get started by setting up our hypotheses!
The null hypothesis $H_0$ is that the mean cost of gas in our area is equal to the national average, so $H_0: \mu = 2.00$ .

The alternative hypothesis $H_a$ is that the mean cost is less than the national average, so $H_a: \mu < 2.00$ .

Plug in the numbers:
$z = \frac{1.95 - 2.00}{\frac{0.20}{\sqrt{50}}}$

Calculate the denominator:
$\frac{0.20}{\sqrt{50}} = \frac{0.20}{7.071} \approx 0.0283$

Now, calculate the z-score:
$z = \frac{-0.05}{0.0283} \approx -1.77$

With our z-score of $-1.77$ , we need to find the P-value.
This tells us the probability of observing a sample mean as extreme as ours, assuming the null hypothesis is true.

Using a standard normal distribution table (Table A2), find the P-value corresponding to $z = -1.77$ .
The P-value is approximately $0.0384$ .

Compare the P-value to the significance level.
Our significance level is $0.01$ or $1\%$ .

Since $0.0384$ is greater than $0.01$ , we fail to reject the null hypothesis.
This means there's not enough evidence to say the gas price is significantly less than the national average.

The P-value is $0.0384$ , which is greater than the significance level of $0.01$ .
Therefore, we enter 0.0384 as the answer, rounded to two decimal places: 0.04.