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# Jason K.

I am a web applications developer and a past upper-level Mathematics and Computer Science instructor for 8 years. I've taught at the high school and collegiate (remedial) levels. My experience includes teaching Precalculus, AP Calculus, BC Calculus, AP Statistics, AP Computer Science, College Algebra, and other math courses. I currently spend my days developing small, medium, and large scale Angular4/C/T-SQL web application systems used by thousands of staff for a local school district.

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• 2 months ago
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• @jason-k-8 Jason K. commented @Yiu A. It sounds like what you're looking at is performing a one-sample inference confidence interval based upon a sample. From the problem, you appear to have a sample size of 974, each of which is a value from 1 - 5. These are categorical in nature, but assuming you know how many rated a "5", you can determine the same proportion hat(p) by simply dividing the number of "5" responses by 974.

From this, you can get an estimate for the standard error of the sample average SE using the formula SE = sqrt( (hat(p)(1-hat(p)))/n ).

This is useful because you can construct a confidence interval for estimating p (the true population proportion) using the formula CI = hat(p) +- z^{star}SE, where z^{star} is the critical z-score for the confidence interval. For a 95% interval, you split the remaining 5% in half, and reference a z-score table to find the z-score for a left-tail area of 0.025. It turns out this z-score is 1.96.

Once you find the confidence interval estimates for the true population proportion p in this manner, all that would really remain would be to convert those proportions into approximate counts of people, which you could do by multiplying them by the 234,564,000 count of adults. For example (just example!!!), say you find CI = 0.35 +- .12, which comes out to (0.23, 0.47). In that case, you could multiple the 234 million by 0.23 and by 0.47 (round properly so you don't have a half of a person!), and that would be the count.
on How to do this question? Can you tell me the steps? (p.s. I didn't show the sample statistics here but I have those on excel)
2 months ago
• @jason-k-8 Jason K. commented Yes! The null and alternate should, together, cover all possible values. So, if you make you null something like mu = 5, typically your alternate is mu != 5. If you make you null mu &gt; 5, you'd want your alternate to be mu &lt;= 5#. In most cases, I've generally seen people choose to make the alternate be what "someone wants to claim" (like the student in this problem). The thinking is the null should be "the usual understanding", while the alternate is what you claim is somehow different from the usual.

For this problem, the student is claiming that unlike what people think, the student thinks it's more than 5 cases. That's why I originally solved by making the null "less than or equal to 5" - it's what the expected result is, while the alternate is what the student is claiming.

But really, as long as you are careful and use the correct kind of tail probability (left, right, or two-tail), you should be okay.
on How to do this hypothesis testing question?
2 months ago
• @jason-k-8 Jason K.
2 months ago