(a) identify the expected distribution and state H0 and Ha,...

Key Concepts

Significance Level

The significance level, denoted by alpha, is the threshold probability for rejecting the null hypothesis. It represents the probability of committing a Type I error, that is, rejecting a true null hypothesis.

Interpretation of Decision in Hypothesis Testing

After comparing the computed test statistic with the critical value, a decision is made to either reject or fail to reject the null hypothesis. This decision is then interpreted in the context of the original claim, determining whether there is sufficient evidence to dispute the assumed distribution.

Chi-Square Test Statistic

The test statistic is computed by summing the squared differences between the observed and expected frequencies, divided by the expected frequency for each category. This value measures the overall deviation from the expected distribution.

Critical Value and Rejection Region

The critical value is obtained from the chi-square distribution table based on the degrees of freedom and the significance level (alpha). The rejection region is defined as the set of values of the test statistic that, if exceeded, lead to rejection of the null hypothesis.

Null and Alternative Hypotheses

The null hypothesis (H?) generally states that there is no difference between the observed and expected frequencies, meaning the data follow the specified distribution (uniform in this case). The alternative hypothesis (H?) states that the observed frequencies differ significantly from the expected frequencies, suggesting the real distribution is not uniform.

Uniform Distribution

A uniform distribution in this context means that each category—in this case, each day of the week—has an equal probability of occurring. For the test, it implies that the expected number of births on each day is the same.

Chi-Square Goodness-of-Fit Test

This test is used to determine whether there is a significant difference between the observed frequencies and the expected frequencies under a specific hypothesis. It assesses how well the observed distribution fits with the theoretical distribution proposed by the null hypothesis.

Degrees of Freedom

Degrees of freedom in a chi-square goodness-of-fit test are calculated as the number of categories minus one. They determine the shape of the chi-square distribution used to assess the significance of the test outcome.

Transcript

00:01 Now in this question there is a doctor that claims that the number of births by day of the week is uniformly distributed we're skeptical what we do is we randomly select 700 births from a recent year and record the day of the week on which they took place so we have the table of our observations with us again what we will do is the first step draw the table we have the day okay, so how many days do we have from sunday? to saturday.

00:38 So the first one is sunday.

00:42 Then we have monday, tuesday, wednesday, thursday, thursday, friday and saturday.

00:59 Then we have the frequencies.

01:00 Let me just call these the observed values.

01:06 The observed values.

01:08 Okay.

01:13 We have the observed values.

01:15 For sunday, it is 65.

01:20 Sunday, it is 65, then it is 107, then it is 117, 117, 115, 114, 109, and 73.

01:35 All right.

01:37 Now, what is the null hypothesis? our null hypothesis will be, what is the researcher, the doctor claiming? the doctor claims that the number of birds by the day of the week is uniformly distributed.

01:50 So the null hypothesis will be that, the number.

01:54 Number of birds the number of birds by day of the week is uniformly distributed is uniformly distributed what will be our alternative hypothesis the alternative hypothesis will be that the number of birds by day of the week is not uniformly is not uniformly distributed.

03:09 All right.

03:12 Now one very important point over here is, since they are saying that it is uniformly distributed, what would be the probabilities? this is going to be the column for probabilities.

03:26 This is going to be the column for probabilities.

03:29 The probability has to be the same for all the seven days.

03:33 So this will be one by seven for all of them, okay? 1 by 7 for all of them all right what is the first step the first step in this this is a kaiswell goodness a bit test that we are going to conduct here so the first step is finding the expected values finding the expected value for all the categories so how will we do that expected value for category i will be we use a formula sample size which happens to be n multiplied by the probability of the category multiplied by the probability of the category which is p i in this case okay probability of the iath category so let us go here what is the sample size that we have 700 so this addition is 700 so what will be the expected values these will be the expected values this is the column for the expected values all right since the probabilities are the same for all the categories this will be 700 into 1 by 7 which is nothing but 100 so the expected value will be 100 for all the categories now since this is 100 our calculations are going to become very simple what is the next step the next step is finding the kai square values, the kai square.

05:25 How do we do that? well, for every category, we apply the formula, observed minus the expected, we square it.

05:34 So, since you are squaring it, it does not matter if you do observe minus expected or expected minus observed since there is a square.

05:42 So positive and negative square, they will give you the same answer.

05:47 Then you divide that value that you get by the expected value and in the end you sum them all up and this will give you the overall kai square for the problem so let us go here look at this formula in action for the first category the difference is between hundred and sixty -five so the answer is 35 we square this and divide this by 100 so this is 12 .25 this is 12 .25 similarly the difference here is 7 7 square is 49.

06:23 49 by 100 is 0 .49.

06:27 Now the difference is 17.

06:29 17 square is 289.

06:33 Divided by 100 is 2 .89.

06:35 Difference here is 15.

06:36 15 square is 225 and divided by 100 is 2 .25.

06:41 Now it is 14 and 14 square is 196.

06:47 So this is 1 .96.

06:49 Now this is 9.

06:51 9 square is 81 so this becomes 0 .81 then the difference is 27 the square is 729 and divided by 100 is 7 .29 in the end as we saw that we had a summation sign so we sum them on up 12 .25 plus 0 .49 plus 2 .89 plus 2 .25 plus 1 .96 plus 0 .86 plus 0 .8 1 plus 7 .29.

07:27 This is 27 .94.

07:30 The submission is 27 .94.

07:33 Now we have solved enough problems to actually recognize that this value is very high.

07:39 So we will end up rejecting our null hypothesis.

07:42 But let us just follow all the steps to verify our answer.

07:47 The next step is finding the degrees of freedom.

07:51 This is given by the formula.

07:53 Number of categories might.

07:59 1.

08:02 All right.

08:04 How many categories do i have? i have seven days.

08:06 So this is 7 minus 1 or this is going to give me 6.

08:10 My answer here is 6.

08:13 Now that i have my kai square value and my degrees of freedom, what i need to do is determine whether i have to reject my null hypothesis or not.

08:23 Now there are two ways of doing this...

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