Question

A 1980 study was conducted whose purpose was to compare the indoor air quality in offices where smoking was permitted with that in offices where smoking was not permitted. Measurements were made of carbon monoxide (CO) at 1:20 p.m. in 40 work areas where smoking was permitted and in 40 work areas where smoking was not permitted. Where smoking was permitted, the mean CO level was 11.6 parts per million (ppm) and the standard deviation CO was 7.3 ppm. Where smoking was not permitted, the mean CO was 6.9 ppm and the standard deviation CO was 2.7 ppm.
To test for whether or not the mean CO is significantly different in the two types of working environments, perform a t-test for unequal variance and report the p-value

Answer 1

Answer:

The null hypothesis is not rejected.

There is no enough evidence to support the claim that the CO level is lower  in non-smoking working areas compared to smoking work areas.

P-value = 0.07.

Step-by-step explanation:

We have to perform a test on the difference of means.

The claim that we want to test is that CO is less present in no-smoking work areas.

Then, the null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2 > 0$

being μ1: mean CO level in smoking work areas, and μ2: mean CO level in no-smoking work areas.

The significance level is assumed to be 0.05.

Smoking areas sample

Sample size n1=40.

Sample mean M1=11.6

Sample standard deviation s1=7.3

No-smoking areas sample

Sample size n2=40

Sample mean M2=6.9

Sample standard deviation s2=2.7

First, we calculate the difference between means:

$M_d=M_1-M_2=11.6-7.3=4.3$

Second, we calculate the standard error for the difference between means:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{7.3^2}{40}+\dfrac{2.7^2}{40}}=\sqrt{\dfrac{53.29+7.29}{40}}=\sqrt{\dfrac{60.58}{40}}\\\\\\s_{M_d}=\sqrt{1.5145}=1.23$

Now, we can calculate the t-statistic:

$t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{4.3-0}{1.23}=3.5$

The degrees of freedom are calculated with the Welch–Satterthwaite equation:

$df=\dfrac{(\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2})^2}{\dfrac{s_1^4}{n_1(n_1-1)}+\dfrac{s_2^4}{n_2(n_2-1)}} \\\\\\\\df=\dfrac{(\dfrac{7.3^2}{40}+\dfrac{2.7^2}{40})^2}{\dfrac{7.3^4}{40(39)}+\dfrac{2.7^4}{40(39)}} =\dfrac{(\dfrac{53.29}{40}+\dfrac{7.29}{40})^2}{\dfrac{2839.82}{1560}+\dfrac{53.14}{1560}} \\\\\\\\df=\dfrac{1.5145^2}{1.8545}=\dfrac{2.2937}{1.8545}=1.237$

The P-value for this right tail test, with 1.237 degrees of freedom and t=3.5 is:

$P-value=P(t>3.5)=0.07$

The P-value is bigger than the significance level, so the effect is not significant. The null hypothesis is not rejected.

There is no enough evidence to support the claim that the CO level is lower  in non-smoking working areas compared to smoking work areas.