# Two sample hypothesis test with unknown standard deviation problem help

The problem:

The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airports. The maximum possible rating is 10. Suppose a simple random sample of 50 business travelers is selected and each traveler is asked to provide a rating for the Miami International Airport, and another simple random sample of 50 traveling agents that qualified at the Los Angeles airport.

The exercise provided the data but I will just give you the summary of each sample:

Miami (1):

Size: $$n_1=50$$

Mean: $$\overline{x}_1=6.34$$

Standard deviation: $$s_1=2.1629$$

Los Ángeles (2):

Size: $$n_2=50$$

Mean: $$\overline{x}_1=6.72$$

Standard deviation: $$s_1=2.3737$$

There are two questions:

1. With α = 0.025, perform hypotheses test to determine that the two airports are highly competitive.

2. Using the information of the first problem, perform a hypothesis test, with α = 0.025, to determine if there is a significant difference between the two airports.

This is what I did.

$$H_0: μ_1=μ_2$$ $$H_1: μ_1≠μ_2$$

For the critical value:

Freedom degrees: $$n_1+n_2-2=50+50-2=98$$

Since the α = 0.025 and I believe this is two-tailed $$\frac{α}{2}=0.0125$$ But this number does not appear in the table of t. So I used a website to give me the critical value.

Critical Value: $$t=±2.276$$

Since there is unknown the population standard deviation or variance, I looked for the pooled sample variance.

This is the formula I used:

$$s_p^2=\frac{(n_1-1)s_1^2+(n_1-1)s_2^2}{n_1+n_2-2}$$

$$s_p^2=\frac{(50-1)(2.1629)^2+(50-1)(2.3737)^2}{50+50-2}=5.1563$$

$$t$$-statistic:

$$t=\frac{\overline{x}_1-\overline{x}_2}{\sqrt{\frac{s_p^2}{n_1}+\frac{s_p^2}{n_2}}}$$

$$t=\frac{6.42-6.72}{\sqrt{5.1563(\frac{1}{50}+\frac{1}{50})}}=-0.6609$$

The null hypothesis is not rejected because -0.6609 falls in the region between -2.276 and 2.276.

The data does not show that there is a significant difference between the means of the quality ratings for the airports of Miami and Los Angeles.

Am I right?

If I am, this answers the second question but, what I have to do with hypothesis test to determine that the two airports are highly competitive?

Thanks in advance for any help.

• Thanks for showing what you have tried. I hope my Answer takes care of some of your questions. – BruceET Oct 26 '19 at 5:18

(1) "Highly competitive" is not a statistical term, and I see no guidance in the surrounding information to guess what that means in terms of a test of hypothesis.

(2) "[D]etermine if there is a significant difference between..." is a standard way to suggest testing the null hypothesis that the airports have equal popularity against a two-sided alternative, as you have it at the beginning of your work.

Some of the language about the two groups makes me wonder if comparable subjects were randomly chosen to rate the two airports: "business travelers" for Miami and "traveling agents that qualified" for Los Angeles. That doesn't seem to be parallel terminology, and I wonder why not.

You have chosen to do a pooled two-sample t test. I will show you output from Minitab statistical software based on the information given:

Two-Sample T-Test

Sample   N  Mean  StDev  SE Mean
1       50  6.34   2.16     0.31
2       50  6.72   2.37     0.34

Difference = μ (1) - μ (2)
Estimate for difference:  -0.380
T-Test of difference = 0 (vs ≠):
T-Value = -0.84  P-Value = 0.405  DF = 98
Uses Pooled StDev = 2.2707


I have verified the pooled SD 2.2707 and t statistic -0.84 by direct computation. I think you may have a mistake or two in your computation of the t statistic. (The formula looks OK.)

The P-value for a 2-sided test is the sum of the areas under the density curve for Student's t distribution for 98 degrees of freedom (a) below -0.84 and (b) above 0.84. Ordinarily, you can't find P-values from printed tables of the t disribution. However, the distribution $$\mathsf{T}(98)$$ is very close to standard normal, so you could get a good approximation from printed standard normal tables.

The exact computation in R is shown below; it matches the printout. You can't reject the null hypothesis at the 2.5% level because the P-value exceeds 0.025.

2*pt(-.84, 98)
[1] 0.402953


Critical values for a test at the 2.5% level cut 1.25% of the probability from the upper and lower tails of $$\mathsf{T}(98).$$ From R, critical values are at $$\pm 2.2274.$$ Put in another way, you would reject the null hypothesis at the 2.5% level if $$|T|\ge 2.2274;$$ you don't reject because $$|T| = 0.84.$$

qt(.0125, 98)
[1] -2.276362


The graph below shows the density function of $$\mathsf{T}(98).$$ Vertical dotted lines show the critical values. The observed value of the T statistic is shown by the solid vertical line. The P-value is twice the area under the curve to the left of the solid vertical line.

Notes: (i) In R statistical software pt is a CDF of a t distribution and qt is the inverse CDF or quantile function of a t distribution. (ii) The same procedure in Minitab that shown the t test also shows a confidence interval for the difference in means. I edited out the CI because it does not seem directly relevant to your question.

• I saw my error in the t-statistic: I put 6.42 instead of 6.34. Thank you so much for the time to explain. <3 – gi2302 Oct 28 '19 at 3:35