Will 10% of tourists visit a cafe?

[JohnA]

Tourism is one consideration for CoffeeTime’s future. A survey of 1,233 visitors to Mumbai last year revealed that 110 visited a small café during their visit. Laura claims that 10% of tourists will include a visit to a café. Use a 0.05 significance level to test her claim. Would it be wise for her to use that claim in trying to convince management to increase their advertising spending to travel agents? Explain.


The first step I did was calculate the sample proportion:
p = X/n = 110/1233 = 0.0892 = 8.92%

In order to find the z value, I used with the equation:
0.10 = p + z*sqrt((p)*(p-1)/n)

Then solving for z, I found:
z = (0.10 - p)/sqrt((p)*(p-1)/n)
= (0.10 - 0.0892)/sqrt((0.0892)*(0.0892-1)/1233)
= 1.329.

Since the probability that z is greater than 1.329 is 90.80%, there is only a 9.20% probability that 10% of tourists will visit a café. Since 9.20% is significantly less than 99.95% (for the 0.05 significance level), Laura should not claim that 10% of tourists will visit a café.

Is this the correct methodology I should use to solve this problem?

 

[JakeD]

First off, .05 significance level means 5%, not .05%.

I can't find a way to write the null and alternative hypotheses to justify your calculation of the Z score. Anyway, here's how I interpret the test.

Laura's claim is the null hypothesis. Any percentage found greater than 10% would be in her favor.

\(\displaystyle \large H_0: \mu\ \ge\ .10\)
\(\displaystyle \large H_1: \mu\ <\ .10\)

To implement this test, use

\(\displaystyle \large H_0: \mu\ =\ .10\)
\(\displaystyle \large H_1: \mu\ <\ .10\)

Approximate with the normal distribution. Under the null hypothesis the standard deviation is \(\displaystyle \large \sqrt{.10(1-.10)} .\)

The Z score is \(\displaystyle \large (.0892 - .10)/\sqrt{.10(.90)/1233} = -1.264\) which has p-value .1031 = 10.31%. This is greater than the 5% significance level so there is not enough evidence to reject Laura's claim.

That test give's Laura's claim the benefit of the doubt. You could go the opposite way, but this would be a very conservative test. It would force Laura to prove at a 5% level that the mean was greater than 10%.

\(\displaystyle \large H_0: \mu\ \le\ .10\)
\(\displaystyle \large H_1: \mu\ >\ .10\)

To implement this test, use

\(\displaystyle \large H_0: \mu\ =\ .10\)
\(\displaystyle \large H_1: \mu\ >\ .10\)

The Z score would be the same but the p-value would be different: .8968 = 89.68%. No evidence to say the mean is greater than 10%.