Question

The following molarities were calculated from replicate standardization of a solution: .3036, 0.3039, 0.3033, 0.3031, 0.3035, 0.3042, 0.3027, 0.3036.

-calculate the sample standard diviation and relative deviation

-what range are you 95% certain that the true value of the molarity falls.

-the correct concentration of the standardized solution is 0.3033M. write a null hypothesis and determine if we are 98% confident that the method is appropriate

Answer 1

The relative standard deviation formula is:
100 * s / |x̄|
Where:
s = the sample standard deviation
x̄ = sample mean

Relative deviation =

Relative standard deviation =

95% confidence interval

Since n<30 we will use t critical value for finding 95% confidence interval

xbar = 0.3035
t critical value = 2.36
sM = √((0.000464)^2/8) = 0.00016
μ = M ± t(s_M)

μ = 0.3035 ± 2.36*0.00016
μ = 0.3035 ± 0.000388

M = 0.3035, 95% CI [0.303112, 0.303888].

You can be 95% confident that the population mean (μ) falls between 0.303112 and 0.303888.

Null Hypothesis H0:

Alternative Hypothesis Ha:

Under null hypothesis test statistic is

Degree of freedom= n-1= 8-1=7

alpha= 0.02

tc= 2.998

Conclusion: Since critical value of t is greater than calculated value of t therefore we DO NOT REJECT null hypothesis at 0.02 level of significance.