Math / Data & Statistics

QuestionA potato chip manufacturer produces bags of potato chips that are supposed to have a net weight of 326 grams. Because the chips vary in size, it is difficult to fill the bags to the exact weight desired. However, the bags pass inspection so long as the standard deviation of their weights is no more than 4 grams. A quality control inspector wished to test the claim that one batch of bags has a standard deviation of more than 4 grams, and thus does not pass inspection. If a sample of 28 bags of potato chips is taken and the standard deviation is found to be 4.6 grams, does this evidence, at the 0.025 level of significance, support the claim that the bags should fail inspection? Assume that the weights of the bags of potato chips are normally distributed.
Step 3 of 3: Draw a conclusion and interpret the decision.
Answer 2 Points Tables Keypad Keyboard Shortcuts
We reject the null hypothesis and conclude that there is sufficient evidence at a 0.025 level of significance that the bags should fail inspection.
We fail to reject the null hypothesis and conclude that there is insufficient evidence at a 0.025 level of significance that the bags should fail inspection.
We fail to reject the null hypothesis and conclude that there is sufficient evidence at a 0.025 level of significance that the bags should fail inspection.
We reject the null hypothesis and conclude that there is insufficient evidence at a 0.025 level of significance that the bags should fail inspection.

Studdy Solution

STEP 1

What is this asking? Are these potato chip bags too inconsistently weighted, based on a sample, and a specific significance level? Watch out! Don't mix up standard deviation and variance!
Also, remember to interpret the results in the context of the problem, not just statistically.

STEP 2

1. Set up the hypothesis test
2. Calculate the test statistic
3. Determine the critical value and make a decision

STEP 3

We're assuming the standard deviation is acceptable, so our null hypothesis $H_0$ is $\sigma \le 4$ .
This means the standard deviation of the bag weights is less than or equal to 4 grams.

STEP 4

The inspector believes the standard deviation is greater than 4 grams, so our alternative hypothesis $H_1$ is $\sigma > 4$ .

STEP 5

Our significance level, often denoted by $\alpha$ , is given as 0.025.
This means there's a 2.5% chance we'll reject the null hypothesis even if it's true (Type I error).

STEP 6

We'll use the chi-squared test statistic, which is calculated as: $\chi^2 = \frac{(n-1)s^2}{\sigma^2}$ Where $n$ is the sample size, $s$ is the sample standard deviation, and $\sigma$ is the hypothesized standard deviation under the null hypothesis.

STEP 7

We have $n = \textbf{28}$ , $s = \textbf{4.6}$ , and $\sigma = \textbf{4}$ .
Plugging these values into the formula, we get: $\chi^2 = \frac{(28-1) \cdot 4.6^2}{4^2} = \frac{27 \cdot 21.16}{16} = \frac{571.32}{16} = \textbf{35.7075}$ So, our chi-squared test statistic is 35.7075.

STEP 8

The degrees of freedom, denoted as $df$ , are calculated as $n - 1$ .
In our case, $df = 28 - 1 = \textbf{27}$ .

STEP 9

We need to find the critical chi-squared value, $\chi^2_{\alpha, df}$ , corresponding to our significance level $\alpha = \textbf{0.025}$ and degrees of freedom $df = \textbf{27}$ .
Using a chi-squared table or calculator, we find the critical value to be approximately 43.194.

STEP 10

Our calculated test statistic ( $\chi^2 = \textbf{35.7075}$ ) is less than the critical value ( $\chi^2_{0.025, 27} \approx \textbf{43.194}$ ).

STEP 11

Since our test statistic is less than the critical value, we fail to reject the null hypothesis.
This means we don't have enough evidence to support the claim that the standard deviation is greater than 4 grams at the 0.025 significance level.

STEP 12

We fail to reject the null hypothesis and conclude that there is insufficient evidence at a 0.025 level of significance that the bags should fail inspection.

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Studdy Solution

STEP 1

What is this asking? Are these potato chip bags too inconsistently weighted, based on a sample, and a specific significance level? Watch out! Don't mix up standard deviation and variance!
Also, remember to interpret the results in the context of the problem, not just statistically.

STEP 2

1. Set up the hypothesis test
2. Calculate the test statistic
3. Determine the critical value and make a decision

STEP 3

We're assuming the standard deviation is acceptable, so our null hypothesis $H_0$ is $\sigma \le 4$ .
This means the standard deviation of the bag weights is less than or equal to 4 grams.

STEP 4

The inspector believes the standard deviation is greater than 4 grams, so our alternative hypothesis $H_1$ is $\sigma > 4$ .

STEP 5

Our significance level, often denoted by $\alpha$ , is given as 0.025.
This means there's a 2.5% chance we'll reject the null hypothesis even if it's true (Type I error).

STEP 6

We'll use the chi-squared test statistic, which is calculated as: $\chi^2 = \frac{(n-1)s^2}{\sigma^2}$ Where $n$ is the sample size, $s$ is the sample standard deviation, and $\sigma$ is the hypothesized standard deviation under the null hypothesis.

STEP 7

We have $n = \textbf{28}$ , $s = \textbf{4.6}$ , and $\sigma = \textbf{4}$ .
Plugging these values into the formula, we get: $\chi^2 = \frac{(28-1) \cdot 4.6^2}{4^2} = \frac{27 \cdot 21.16}{16} = \frac{571.32}{16} = \textbf{35.7075}$ So, our chi-squared test statistic is 35.7075.

STEP 8

The degrees of freedom, denoted as $df$ , are calculated as $n - 1$ .
In our case, $df = 28 - 1 = \textbf{27}$ .

STEP 9

We need to find the critical chi-squared value, $\chi^2_{\alpha, df}$ , corresponding to our significance level $\alpha = \textbf{0.025}$ and degrees of freedom $df = \textbf{27}$ .
Using a chi-squared table or calculator, we find the critical value to be approximately 43.194.

STEP 10

Our calculated test statistic ( $\chi^2 = \textbf{35.7075}$ ) is less than the critical value ( $\chi^2_{0.025, 27} \approx \textbf{43.194}$ ).

STEP 11

Since our test statistic is less than the critical value, we fail to reject the null hypothesis.
This means we don't have enough evidence to support the claim that the standard deviation is greater than 4 grams at the 0.025 significance level.

STEP 12

We fail to reject the null hypothesis and conclude that there is insufficient evidence at a 0.025 level of significance that the bags should fail inspection.

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Studdy Solution

STEP 1

What is this asking? Are these potato chip bags too inconsistently weighted, based on a sample, and a specific significance level? Watch out! Don't mix up standard deviation and variance!Also, remember to interpret the results in the context of the problem, not just statistically.

STEP 2

1. Set up the hypothesis test2. Calculate the test statistic3. Determine the critical value and make a decision

STEP 3

We're assuming the standard deviation is acceptable, so our **null hypothesis** H0H_0H0​ is σ≤4\sigma \le 4σ≤4.This means the standard deviation of the bag weights is less than or equal to **4 grams**.

STEP 4

The inspector believes the standard deviation is *greater* than 4 grams, so our **alternative hypothesis** H1H_1H1​ is σ>4\sigma > 4σ>4.

STEP 5

Our **significance level**, often denoted by α\alphaα, is given as **0.025**.This means there's a 2.5% chance we'll reject the null hypothesis even if it's true (Type I error).

STEP 6

We'll use the chi-squared test statistic, which is calculated as: χ2=(n−1)s2σ2 \chi^2 = \frac{(n-1)s^2}{\sigma^2} χ2=σ2(n−1)s2​ Where nnn is the **sample size**, sss is the **sample standard deviation**, and σ\sigmaσ is the **hypothesized standard deviation** under the null hypothesis.

STEP 7

STEP 8

The degrees of freedom, denoted as dfdfdf, are calculated as n−1n - 1n−1.In our case, df=28−1=27df = 28 - 1 = \textbf{27}df=28−1=27.

STEP 9

STEP 10

Our calculated test statistic (χ2=35.7075\chi^2 = \textbf{35.7075}χ2=35.7075) is *less than* the critical value (χ0.025,272≈43.194\chi^2_{0.025, 27} \approx \textbf{43.194}χ0.025,272​≈43.194).

STEP 11

Since our test statistic is less than the critical value, we **fail to reject the null hypothesis**.This means we don't have enough evidence to support the claim that the standard deviation is greater than 4 grams at the 0.025 significance level.

STEP 12

We fail to reject the null hypothesis and conclude that there is insufficient evidence at a 0.025 level of significance that the bags should fail inspection.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? Are these potato chip bags too inconsistently weighted, based on a sample, and a specific significance level? Watch out! Don't mix up standard deviation and variance!
Also, remember to interpret the results in the context of the problem, not just statistically.

1. Set up the hypothesis test
2. Calculate the test statistic
3. Determine the critical value and make a decision

We're assuming the standard deviation is acceptable, so our null hypothesis $H_0$ is $\sigma \le 4$ .
This means the standard deviation of the bag weights is less than or equal to 4 grams.

The inspector believes the standard deviation is greater than 4 grams, so our alternative hypothesis $H_1$ is $\sigma > 4$ .

Our significance level, often denoted by $\alpha$ , is given as 0.025.
This means there's a 2.5% chance we'll reject the null hypothesis even if it's true (Type I error).

We'll use the chi-squared test statistic, which is calculated as: $\chi^2 = \frac{(n-1)s^2}{\sigma^2}$ Where $n$ is the sample size, $s$ is the sample standard deviation, and $\sigma$ is the hypothesized standard deviation under the null hypothesis.

The degrees of freedom, denoted as $df$ , are calculated as $n - 1$ .
In our case, $df = 28 - 1 = \textbf{27}$ .

Our calculated test statistic ( $\chi^2 = \textbf{35.7075}$ ) is less than the critical value ( $\chi^2_{0.025, 27} \approx \textbf{43.194}$ ).

Since our test statistic is less than the critical value, we fail to reject the null hypothesis.
This means we don't have enough evidence to support the claim that the standard deviation is greater than 4 grams at the 0.025 significance level.