To determine if the sample size of 35 operators is sufficient at a confidence level of 95%, we can perform a hypothesis test using the given data.
Let's set up the null and alternative hypotheses:
Null hypothesis (H0): The true average time for the operators is equal to or less than the maximum allowed time of 14 minutes.
Alternative hypothesis (H1): The true average time for the operators exceeds the maximum allowed time of 14 minutes.
We can use a one-sample t-test to test these hypotheses, where the test statistic is given by:
t = (sample_mean - hypothesized_mean) / (sample_standard_deviation / sqrt(sample_size))
In this case:
sample_mean = 15.5 minutes (average time for 35 operators)
hypothesized_mean = 14 minutes (maximum allowed time)
sample_standard_deviation = sqrt(sample_variance) = sqrt(1.2 minutes) ≈ 1.0954
sample_size = 35
Calculating the test statistic:
t = (15.5 - 14) / (1.0954 / sqrt(35)) ≈ 3.7561
The critical t-value for a one-tailed test at a 95% confidence level with 34 degrees of freedom (35 - 1) is approximately 1.691 (obtained from a t-table or calculator).
Since the calculated t-value (3.7561) is greater than the critical t-value (1.691), we reject the null hypothesis. This means that there is sufficient evidence to suggest that the true average time for the operators exceeds the maximum allowed time.
Therefore, with a sample size of 35 operators, we can conclude that n=35 is a sufficient sample size at a confidence level of 95% to determine that the average time for the operators is higher than the maximum allowed time.