Question

If the reported average amount of CO2 (ppm) levels, in parts per million, in 50 kitchen gas
cooking appliances was 600.1 ppm and a sample standard deviation of 95.2 ppm.
a. Find a 95% confidence interval for the mean amount of CO2 in all kitchen gas cooking appliances.

b. If the investigators thought the standard deviation was 130 ppm, what sample size would be neces-
sary to obtain an interval with of 50 ppm for a confidence level of 95/

Answer 1

As per the provided information,

The sample mean is 600.1 ppm

The sample standard deviation (s) is 95.2 ppm

The sample size (n) is 50.

(a).

The degrees of freedom (df) = n – 1 = 50 – 1 = 49

At the significance level 0.05 and the degrees of freedom 49, the two-tailed critical value obtained from t-table is +/- 2.0096.

The 95% confidence interval for the mean amount of CO2 in all kitchen gas cooking appliances can be calculated as,

Thus, the required confidence interval is (573.0 ppm, 627.2 ppm).

(b).

The population standard deviation is 130 ppm

The margin of error (ME) is 50 ppm

At the significance level 0.05, the two tailed critical value obtained from z-table is 1.96.

The sample size is given as,

Therefore, the sample size is 26.