A recent poll of a random sample of 400 students at a public 4-year college showed the average (mean) amount of time spent using email was 49 minutes per day with a standard deviation of 44 minutes.
a. find a 95% confidence interval fot the population mean
It would help if you proofread your questions before you posted them. The information below only applies to a normal distribution. Your data do not indicate a normal distribution. Anything over 1.2 SD below the mean would be negative.
95% = mean ± 1.96 SEm
SEm = SD/√n
To find a 95% confidence interval for the population mean, you can use the following formula:
Confidence Interval = Sample Mean ± (Critical Value × Standard Error)
Where:
Sample Mean = 49 minutes (given in the question)
Standard Error = Standard Deviation / √(Sample Size)
Critical Value = Dependent on the desired confidence level and the sample size
Let's calculate each part step by step:
Sample Mean = 49 minutes
Standard Error = 44 minutes / √(400)
To calculate the standard error, you divide the standard deviation by the square root of the sample size.
Standard Error = 44 / √(400)
Standard Error = 44 / 20
Standard Error ≈ 2.2 minutes
Now we need to find the critical value at a 95% confidence level. For a sample size of 400, the critical value can be found using a t-distribution table or a statistical calculator.
Assuming a t-distribution, with a sample size of 400 and a confidence level of 95%, the critical value (t) would be approximately 1.96.
Confidence Interval = 49 ± (1.96 × 2.2)
Calculating the confidence interval:
Lower Bound = 49 - (1.96 × 2.2)
Lower Bound ≈ 44.08 minutes
Upper Bound = 49 + (1.96 × 2.2)
Upper Bound ≈ 53.92 minutes
Therefore, the 95% confidence interval for the population mean of time spent using email is approximately 44.08 minutes to 53.92 minutes.