A soft-drink producing company believes that 30 % of its customers prefer the mango flavor. Out

A soft-drink producing company believes that 30 % of its...

Key Concepts

Hypothesis Testing

Hypothesis testing is a statistical method used to make decisions about a population parameter based on sample data. It involves formulating a null hypothesis, which represents a default or accepted claim, and an alternative hypothesis, which represents a contrary claim. The process includes calculating a test statistic from the data and comparing it against a critical value or computing a p?value to determine whether to reject the null hypothesis.

Null and Alternative Hypotheses

The null hypothesis typically states that there is no effect or no difference – in this context, that the proportion of customers who prefer mango flavor is 30%. The alternative hypothesis suggests that the true proportion differs from this claim. Defining these hypotheses is crucial as they form the basis for the decision-making process in hypothesis testing.

Significance Level

The significance level, often denoted by ?, is the threshold for rejecting the null hypothesis. It represents the probability of making a Type I error, which is rejecting the null hypothesis when it is actually true. In hypothesis testing, a significance level of 0.05 is commonly used to determine whether the observed evidence is strong enough to reject the null hypothesis.

Binomial Distribution

The binomial distribution is used when the outcome of each trial can be classified into one of two categories (e.g., success or failure). In this context, the number of customers who prefer mango flavor follows a binomial distribution if the trials are independent and the probability of success is constant for each customer. This distribution is useful for modeling the number of successes out of a fixed number of trials in the hypothesis testing scenario.

Transcript

00:01 So this question of us is saying that a soft drink producing company believes that 30 % of its customer prefers the mango flavor.

00:08 Out of 24 customer tested, 10 bought mango flavored items.

00:13 What can you the statistician conclude about the company's claim? and we are asked to use a 5 % level of significance.

00:21 So let's move into our worksheet.

00:24 So i've written out the parameters.

00:26 The population proportion is 30 % that is 0 .3.

00:30 The sample size is 24.

00:31 The fraction of the sample that is the number of customers that bought mango flavored goods that are 10 and the sample proportion is the number of customers that bought mango flavored items divided by the total number of sample that is 24 so we have 10 over 24 gives us 0 .42 so we are supposed to perform an hypothesis test so that simply our null hypothesis is that p is equal to 0 .3 that is according to the manufacturer the proportion of customers that preferred mango flavored items they are 0 .3 they are 30 percent but we are going to go against this as the as the statistician we don't know if it is either greater than or lesser than but we know it's not equals to 0 .3 so we have that our alternative hypothesis is that p is not equals to 0 .3 so i'll be making use of the pay value method to make my decision here and the decision rule states that if my p value is greater than my alpha i accept the null hypothesis and if my p value is lesser than my alpha i reject the null hypothesis our alpha in this case is 0 .05 so before we can be able to get a p value we need our test statistics and our test statistics is going to be a t test test.

02:03 So we have that and that's because the sample size here is lesser than 30 so we are going to be using a t test in this case of us so we have t is equal to p cap minus p divided by the square root of p times q divided by n so our p cap in this case is 0 .42 minus our p is 0 .3 divided by the square root of so our p is also 0 .42 and our q is going to be 1 minus p that gives us 1 minus 0 .42 and we will do the math we have 0 .58 i guess yeah so i think that's correct okay so we have 0 .58 divided by 24 so let's do our math we have 0 .42 minus 0 .3 we have 0 .12 and let us work on everything we have at the denominator we have 0 .42 times 0 .58 divided by 24 for us to have 0 .107 so when we do our math we have that 0 .12 divided by 0 .107 and we have 1 .190 and we have 1 .19 to 2.

03:41 So our test statistics which is a t test is 1 .19 so let's go to our calculator to get a p value so this is my calculator here and our test is a two -tail test so let me quickly refresh so our test is a two -tail test our test is this t score so the degree of freedom is n minus one that is the sample size minus one so our sample size is 24 so 24 minus 1 gives us 23 and our t score here is 1 .19...

Please give Ace some feedback