Question

1. A researcher studying the lifespan of a certain species of bacteria. A preliminary sample of 35 bacteria reveals a sample mean of ¯ x = 64 x ¯ = 64 hours with a standard deviation of s = 5.6 s = 5.6 hours. He would like to estimate the mean lifespan for this species of bacteria to within a margin of error of 0.5 hours at a 98% level of confidence. What sample size should you gather to achieve a 0.5 hour margin of error? He would need to sample bacteria.

2. For a confidence level of 98% with a sample size of 26, find the critical t value. Critical Value = (round answer to 3 decimal places)

3. Assume that a sample is used to estimate a population mean μμ. Find the 80% confidence interval for a sample of size 31 with a mean of 85.9 and a standard deviation of 6.3. Enter your answer accurate to one decimal place. I am 80% confident that the mean μμ is between (blank) and (blank)

4. Assume that a sample is used to estimate a population mean μμ. Find the 95% confidence interval for a sample of size 34 with a mean of 81.3 and a standard deviation of 16.8. Enter your answer accurate to one decimal place.
I am 95% confident that the mean μμ is between (blank) and (blank)

Answer 1

1. Margin of error = E = 0.5

s = sample standard deviation = 5.6

Confidence level = c = 0.98

zc for (1+c)/2 = (1+0.98)/2 = 0.99 is

zc = 2.33             (From statistical table of z values)

Sample size:

n = 681.0012

n ~ 681

Sample size to estimate the mean lifespan of certain species = 681

2. Sample size = n = 26

Confidence level = c = 0.98

t critical value for c = 0.98 and degrees of freedom = n - 1 = 26 - 1 = 25 is

t critical = 2.485                        (From statistical table of t values)

t critical value = 2.485

3. sample size = n = 31

Confidence level = c = 0.80

t critical for c = 0.80 and degrees of freedom = n -1 = 31-1 = 30 is

tc = 1.310    (From statistical table of t values)

80% confidence interval for population mean is

         (Round to one decimal)

80% confidence interval for population mean is (84.4, 87.4)

I am 80% confident that the mean μ is between 84.4 and 87.4

4. Sample size = 34

Confidence level = c = 0.95

t critical for c = 0.95 and degrees of freedom = n -1 = 34-1 = 33 Can be calculated from excel using command:

=T.DISt.2T(0.05,33)

= 2.035              (Round to 3 decimal)

tc = 2.035    (From statistical table of t values)

95% confidence interval for population mean is

            (Round to one decimal)

95% confidence interval for population mean is (75.4,87.1)

I am 95% confident that the mean μ is between 75.4 and 87.1