Question

Typographic errors in a text are either nonword errors (as when "the" is typed as "teh") or word errors that result in a real but incorrect word. Spell-checking software will catch nonword errors but not word errors. Human proofreaders catch 70% of word errors. You ask a fellow student to proofread an essay in which you have deliberately made 10 word errors. What is the smallest number of misses m with
P(X ≥ m)
no larger than 0.05? You might consider m or more misses as evidence that a proofreader actually catches fewer than 70% of word errors.

Answer 1

You will use the z-test for proportions.

The z score is given by
\[z = \frac{ p - P }{ \sigma }\]
Where p is your actual sample proportion and P is the population proportion
\[\sigma = \sqrt{\frac{ P(1-P) }{ n }}\]

Now as you do not have p the sample proportion, you need to find the z-score using the theoretical alpha (p-value) given of 0.05

Our null hypothesis: human proofreader captures 70% of errors
Alternative hypothesis: human proof reader captures fewer than 70% of errors (so left side of probability distribution)

So find a negative z so that P(z) = 0.05 (use your z tables to convert probability into a z-score). This corresponds to z = -1.65 assuming a one -tailed t-test

Thus for p-value to be 0.05 or less, we require z < - 1.65
This is the same as
\[z = \frac{ p - P }{ \sqrt{\frac{ P(1-P) }{ n }} } = \frac{ p - 0.7 }{ \sqrt{\frac{ 0.7(1-0.7) }{ 10 }} }= \frac{ p-0.7 }{ \sqrt{\frac{ 0.7(0.3) }{ 10 }} }<-1.65 \]
Solve for p.

Then recall p = X/n and find the suitable X that satisfies the inequality.