Please provide the following information. (a) What is the...

Please provide the following information. (a) What is the level of significance? State the null and alternate hypotheses. (b) Find the value of the chi-square statistic for the sample. Are all the expected frequencies greater than $5 ?$ What sampling distribution will you use? What are the degrees of freedom? (c) Find or estimate the $P$ -value of the sample test statistic. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis that the population fits the specified distribution of categories? (e) Interpret your conclusion in the context of the application. Marketing: Compact Discs Snoop Incorporated is a firm that does market surveys. The Rollum Sound Company hired Snoop to study the age distribution of people who stream music. To check the Snoop report, Rollum used a random sample of 519 customers and obtained the following data: Using a $1 \%$ level of significance, test the claim that the distribution of customer ages in the Snoop report agrees with that of the sample report.

Please provide the following information.
(a) What is the level of significance? State the null and alternate hypotheses.
(b) Find the value of the chi-square statistic for the sample. Are all the expected frequencies greater than $5 ?$ What sampling distribution will you use? What are the degrees of freedom?
(c) Find or estimate the $P$ -value of the sample test statistic.
(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis that the population fits the specified distribution of categories?
(e) Interpret your conclusion in the context of the application.
Marketing: Compact Discs Snoop Incorporated is a firm that does market surveys. The Rollum Sound Company hired Snoop to study the age distribution of people who stream music. To check the Snoop report, Rollum used a random sample of 519 customers and obtained the following data:
Using a $1 \%$ level of significance, test the claim that the distribution of customer ages in the Snoop report agrees with that of the sample report.

Key Concepts

Interpretation in Context

Interpreting the conclusion in context involves relating the statistical decision back to the real-world application. It provides a practical explanation of whether or not the sample evidence supports the claim made by the initial report. The conclusion addresses whether the observed data fit the expected distribution and what implications this might have for the underlying population or process being studied.

P-value and Decision Rule

The p-value in a hypothesis test indicates the probability of obtaining an observed statistic, or one more extreme, assuming that the null hypothesis is true. In the context of the chi-square test, if the p-value is less than the chosen level of significance, the decision is to reject the null hypothesis, suggesting that the observed distribution differs significantly from the expected distribution.

Degrees of Freedom

Degrees of freedom in the context of the chi-square goodness-of-fit test refer to the number of independent comparisons that can be made. It is usually calculated as the number of categories minus one, or further adjusted if parameters are estimated from the data. This value is used to determine the critical value from the chi-square distribution against which the test statistic is compared.

Level of Significance

The level of significance is a threshold set by the researcher that defines the probability of making a Type I error, which is the error of rejecting a true null hypothesis. It is denoted by alpha (?) and typically chosen before the data is analyzed. This threshold helps determine whether the observed test statistic is extreme enough to reject the null hypothesis in favor of the alternative.

Chi-Square Goodness-of-Fit Test

The chi-square goodness-of-fit test is applied to assess how well an observed frequency distribution matches an expected distribution. It is particularly useful when dealing with categorical data. The test calculates the chi-square statistic by summing the squared differences between observed and expected frequencies divided by the expected frequencies, serving as a measure of the discrepancy between observed and theoretical frequencies.

Hypothesis Testing

Hypothesis testing in statistics is a procedure used to evaluate assumptions regarding a population parameter based on sample data. It involves formulating a null hypothesis, which represents the status quo or a claim to be tested, and an alternative hypothesis, which represents a different state or effect. The process entails determining a test statistic and comparing it with a critical value or calculating a p-value to decide whether to reject or fail to reject the null hypothesis.

Expected Frequencies

In the chi-square test, expected frequencies are the counts predicted under the null hypothesis for each category. A key assumption for validity is that these frequencies should ideally be greater than 5 to ensure that the approximations used for the chi-square distribution are accurate and the test results are reliable.

Transcript

00:01 What do we have in this question? now, snoop incorporated is a firm that does marketing surveys and marketing research and stuff.

00:11 The rollam sound company has hired this firm to study the age distribution of people who stream music.

00:20 Now, in order to check the snoop report, rollam used a random sample of around 519 customers and their data is given to us.

00:30 So let us look at the data.

00:32 We have customer age.

00:38 We have customer age.

00:39 This is in years.

00:43 Okay.

00:46 So this is going to be our first column.

00:48 In the next column, we have percent of customers from snoop report.

00:54 Okay, this is percent of customers from snoop report.

01:09 All right.

01:13 After that, we have the number of customers.

01:15 From the sample that is the observed values let me just call this the observed values the observed values okay so what are the different categories that we have the first one is less than 14 years the next one is 14 to 18 after that we have 18 to 23 after that we have 24 to 28 24 to 28 after that we have 29 to 33 29 to 33 and after that we have greater than 33 that is older than 33 all right we'll need more of this space okay so now what is the person of customers from snoop report this is 12 percent this is 29 this is 11 this is 10 this is 14 percent and this is 24 now what were the observed values? we have 88.

02:35 We have 135.

02:39 We have 52.

02:41 We have 40.

02:43 We have 76.

02:45 We have 128.

02:49 All right.

02:52 This is the data that is given to us.

02:54 Now what is the question? the question is they are saying use a 1 % level of significance.

02:59 One percent of one percent of significance means that our alpha is 0 .01.

03:05 Using this test the claim the distribution of customer ages in the snoop report agrees with that of the sample report.

03:12 So what is going to be a null hypothesis? null hypothesis is that the distribution of customer ages in the snoop report in the snoop report fits the distribution of the sample report fits the distribution of the sample report.

04:02 All right.

04:04 What is going to be the alternative hypothesis? the alternative hypothesis will be that the distribution of customers, the distribution of customer ages in the snoop report, in the snoop report doesn't fit the distribution of the sample report of the sample report now this is also very important these two lines the first answer the null and the alternative hypothesis this is what you will write in the null and the alternative hypothesis okay, so now we have to test the claim.

05:04 Now in order to do this we are going to perform a kai square statistic and what is the first step in performing a kai square statistic? we find the expected values.

05:15 We find the expected values for all the categories.

05:23 What is the formula to find the expected value? expected value is given by the formula the sample size that we have.

05:36 Have the sample size multiplied by the probability or the proportion of each category the probability or the proportion of each category okay so let us just apply this formula way for the first category for less than 14 years phase what is our sample size our sample size is 519 so this addition is 519 okay so what is going to be the expected value for the first category.

06:18 It is going to be 12 % of 519 or 0 .12 into 519.

06:24 This is 62 .28.

06:27 62 .28.

06:31 For the second one it is 0 .29 into 519.

06:36 This is 150 .51.

06:40 150 .51.

06:42 Then this says 011 into 519.

06:45 This is 57 .0 .51.

06:46 This is 57 .0 .51.

06:46 59 57 .09 after that we have 10 % of 519 which is 51 .9 after that we have 14 % 0 .14 of 519 which is 72 .66 72 .66 and then we have 24 % 0 .24 into 519 which is 12 .56 12 .56.

07:19 All right now what do we need after this after this we are going to calculate the individual kai squared values the individual kai square values for all the categories now what is the formula for that the formula that we are going to apply over here is going to be for every category we are going to find the difference of the observed and the expected values square them divide them by the expected values and then in the end we will add all of these values up that is add the value for all of the categories.

07:54 And we get the overall kai -square statistic for our problem.

07:59 Let us look at this formula.

08:00 How are we going to apply this? so for the first category, there will be a difference of 88 and 62 .28...

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