Hypothesis Test for Proportion | Examples | P-value | Z table

[@Mvd0x]

Thank you for this learning material, and I wish you the best 👍

[@joshemman]

I wish you the best too, buddy.

[@Aristiq]

You explain it better than any teacher, thanks!

[@velmurugan279]

Very good explanation

[@joshemman]

Thanks and welcome

[@vespereneidk]

you really saved my exams

[@andrewk8817]

Hello sir, could you please do a video on EOQ and EPLS Models in Inventory. thank you

[@princesssheeragamotinan3125]

how did you get the answer 2.32 was

[@naveenpaudel3382]

Best

[@abuda7em20]

Hi Sir , Thank you for your valuable video, but I have a question please. At 2:00 min there is number 5 shows as ( 162.5 >= 5 ) from where we got this 5 ?

[@joshemman]

It is a required condition for the test to be valid. The test requires np and n(1-p) to be >=5. I didn't focus on it in the video because some people use 10 (rather than 5), some use other values, and some don't mention it at all.

[@abuda7em20]

@@joshemman so is it called as ( significant level ) or it has another name ? and thank you for your replay

[@joshemman]

@@abuda7em20No, it (the 5) is not the significance level. alpha is the significance level.

[@jda817]

The conditions you see in the image (np ≥ 5 and 𝑛(1−𝑝) ≥ 5) are known as the normality conditions or success-failure conditions. These conditions are part of the requirements for using a normal approximation to a binomial distribution when conducting hypothesis tests or constructing confidence intervals for proportions. Here's what these conditions mean and why they are used: What is Being Evaluated? 1. Ensuring Approximation Accuracy: These conditions are checking whether the sample size 𝑛 and the expected number of successes (𝑛𝑝) and failures (𝑛(1−𝑝)) are large enough for the sampling distribution of the sample proportion to be approximately normal. This allows us to use the normal distribution (and thus, z-tests) for hypothesis testing or confidence interval estimation. 2. Why 𝑛𝑝 and 𝑛(1−𝑝) Specifically? 𝑛𝑝: This represents the expected number of successes in the sample. 𝑛(1−𝑝): This represents the expected number of failures in the sample. Both of these values need to be sufficiently large to ensure that the distribution of sample proportions is close to a normal distribution. If either is too small, the approximation may not hold, and the results of any hypothesis test could be misleading. Common Thresholds: 5 or 10 as a Threshold: The thresholds ≥ 5 or ≥10 are commonly used rules of thumb. Using 10 is often considered more conservative and ensures a better approximation, but 5 is also frequently used and accepted in many textbooks and studies. In the image, the condition 𝑛𝑝=162.5 ≥ 5 and 𝑛(1−𝑝)=87.5 ≥ 5 are both satisfied, meaning the sample size and proportions are sufficient to proceed with a normal approximation. Summary: The use of 5 (or 10) is to check whether the sample size is large enough for the normal approximation to the binomial distribution to be valid. This ensures that using the z-score and normal distribution methods for hypothesis testing or constructing confidence intervals is appropriate and accurate. If these conditions are met, we can use the normal distribution to estimate probabilities and make inferences about the population proportion.

[@lavishedgaming2140]

how did u sove the hypothesis test because im getting a different answer i got 5.80

[@joshemman]

The numerator is 0.72 - 0.65 = 0.07 Content of denominator square root: =0.65*0.35/250 = 0.00091 Taking the square root gives 0.0302. Z = 0.07/0.0302 = 2.32

[@theshariqsartwork]

why did we minus it at 3:15. (1-0.9898)?

[@joshemman]

The area in the table is the "less than area". Since out Z is positive, we need the area in the right tail which is 1 - "less than area". If Z were negative, we will use the "less than area" without subtracting from 1.

[@shamma6466]

thank you sooo muchhhh

[@osama_statimath]

👍

[@MensahB]

Please sir can you help me with your email address thanks 🙏