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A transport consultancy company claims that less than 30% of people who both work and live in Kowloon Tong. A study revealed that among a random sample of 200 people working in Kowloon Tong, 62 also lived in Kowloon Tong. Does the sample support the company’s claim? Use a level of significance of 0.01.

Answer :

The sample data does not support the company's claim that less than 30% of people who both work and live in Kowloon Tong.

We can use a hypothesis test to answer this question. The null hypothesis is that the proportion of people who both work and live in Kowloon Tong is less than 30%. The alternative hypothesis is that the proportion is equal to or greater than 30%.

The sample proportion is 62/200 = 0.31, which is greater than 0.3. So, the sample data does not support the company's claim.

We can calculate the p-value for this test as follows:

p-value = P(X > 62)

where X is the number of people in the sample who both work and live in Kowloon Tong.

We can use the Binomial distribution to calculate the p-value. The number of trials is 200, the probability of success is 0.3, and the number of successes is 62.

P(X > 62) = 1 - P(X <= 62)

P(X <= 62) = 10C0(0.3)⁰(0.7)²⁰⁰ + 10C1(0.3)¹(0.7)¹⁹⁹ + ... + 10C62(0.3)⁶²(0.7)³⁸

P(X > 62) = 1 - 0.00000157

The p-value is very small, so we can reject the null hypothesis. This means that there is sufficient evidence to support the alternative hypothesis, which is that the proportion of people who both work and live in Kowloon Tong is equal to or greater than 30%.

Therefore, the sample does not support the company's claim that less than 30% of people who both work and live in Kowloon Tong.

To know more about sample data refer here :

https://brainly.com/question/32823975#

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