In a study of factors affecting job satisfaction,320 employed individuals were measured for an index of satisfaction and the sample mean was 28.8 and the sample standard deviation was 7.3.Following,the 98% confidence interval for the population mean would be _____.
A)[27.85,29.75]
B)[27.96,29.64]
C)[11.82,45.78]
D)[28.75,28.85]
E)[28.60,29.00]

Question

Quizplus · Accepted Answer

The 98% confidence interval for the population mean can be calculated using the formula for the confidence interval: mean ± (z*standard deviation/√n), where z is the z-score corresponding to the confidence level (for 98%, it's approximately 2.33), the standard deviation is given as 7.3, and n is the sample size (320). Plugging in the values: 28.8 ± (2.33*7.3/√320) gives an interval of approximately [27.85, 29.75].

In a Study of Factors Affecting Job Satisfaction,320 Employed Individuals