For Exercises 5 through 20, assume that the variables are...

For Exercises 5 through $20,$ assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. Soda Bottle Content A machine fills 12 -ounce bottles with soda. For the machine to function properly, the standard deviation of the population must be less than or equal to 0.03 ounce. A random sample of 8 bottles is selected, and the number of ounces of soda in each bottle is given. At $\alpha=0.05,$ can we reject the claim that the machine is functioning properly? Use the $P$ -value method. $$ \begin{array}{cccc}{12.03} & {12.10} & {12.02} & {11.98} \\ {12.00} & {12.05} & {11.97} & {11.99}\end{array} $$

For Exercises 5 through $20,$ assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified.
Soda Bottle Content A machine fills 12 -ounce bottles with soda. For the machine to function properly, the standard deviation of the population must be less than or equal to 0.03 ounce. A random sample of 8 bottles is selected, and the number of ounces of soda in each bottle is given. At $\alpha=0.05,$ can we reject the claim that the machine is functioning properly? Use the $P$ -value method.
$$
\begin{array}{cccc}{12.03} & {12.10} & {12.02} & {11.98} \\ {12.00} & {12.05} & {11.97} & {11.99}\end{array}
$$

Key Concepts

Degrees of Freedom

Degrees of freedom refer to the number of independent values in a calculation that are free to vary. In the context of variance testing with the chi-square test, the degrees of freedom are equal to the sample size minus one (n-1). This concept is important because it affects the shape of the chi-square distribution and the subsequent determination of critical values or p-values.

Normal Distribution Assumption

Assuming a normal distribution is crucial when using the chi-square test for variance because the derivation of the chi-square distribution for the test statistic relies on the normality of the data. When this assumption is met, the properties of the chi-square distribution can be properly applied to test hypotheses about the population variance.

Chi-Square Test for Variance

The chi-square test for variance is used when testing claims about the population variance or standard deviation based on sample data from a normally distributed population. The test statistic is calculated using the formula (n-1)S²/??², where S² is the sample variance, ??² is the hypothesized population variance, and n is the sample size. This statistic follows a chi-square distribution with (n-1) degrees of freedom.

Null and Alternative Hypotheses

The formulation of null and alternative hypotheses is central to hypothesis testing. The null hypothesis (H?) typically represents a statement of no effect or a baseline assumption, while the alternative hypothesis (H?) represents a claim that contradicts the null. In variance testing, the null hypothesis often states that the population variance is equal to or less than a specified value, whereas the alternative hypothesis states that it exceeds that value.

Significance Level

The significance level, denoted by alpha (?), is the probability threshold below which the null hypothesis will be rejected. It defines the maximum allowable probability of making a Type I error, which is rejecting a true null hypothesis. Common levels are 0.05, 0.01, or 0.10, with 0.05 being a widely used standard in many tests.

p-value

The p-value is a measure that helps determine the strength of evidence against the null hypothesis. It represents the probability of obtaining a test statistic as extreme as, or more extreme than, the observed value, assuming that the null hypothesis is true. If the p-value is less than the predetermined significance level, the null hypothesis is rejected.

Hypothesis Testing

Hypothesis testing is a statistical framework used to decide whether there is enough evidence to reject a null hypothesis in favor of an alternative hypothesis. It involves calculating a test statistic from sample data, comparing it to a critical value, or evaluating a p-value, while controlling for Type I error at a predetermined significance level.

Transcript

00:09 In this problem, our null hypothesis h0 is sigma equal to 0 .03, which is our claim.

00:20 So, our alternative hypothesis h1 is sigma greater than 0 .03.

00:28 Since we have used greater than, it is a right -tailed test with value of alpha given to be 0 .05.

00:38 Now degrees of freedom equal to n minus 1 where n is the size of the sample.

00:46 If this problem the sample given is of the size 8.

00:51 So degrees of freedom equal to 8 minus 1 which is equal to 7.

00:56 We have to now find the critical value for alpha equal to 0 .05 and degrees of freedom equal to 7.

01:06 So the required critical value is 14 .067 as seen in the table for the kai square distribution.

01:28 We will now find the sample standard deviation.

01:33 For that we will require submission x, that is submission over all individual values, which can be computed using the given data.

01:43 So, summation x is equal to 96.

01:46 We will also require summation x square.

02:02 Summation x square is equal to 1155 .375.

02:21 Now sample variance as square is equal to n multiplied by summation x square minus summation x square divided by n multiplied by n multiplied by and minus 1.

03:02 After putting all these values and computing the value of s square, we get s square is equal to 1 .793 multiplied by 10 rise 2 minus 3...

Please give Ace some feedback