You wish to test the following claim ( H a ) at a significance level of α = 0.01 . H o : μ = 60.8 H a : μ ≠ 60.8 You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size n = 8 with mean M = 66.9 and a standard deviation of S D = 10.7 . What is the p-value for this sample? (Report answer accurate to four decimal places.) p-value = -.1228 Incorrect The p-value is... less than (or equal to) α greater than α Correct This p-value leads to a decision to... reject the null accept the null fail to reject the null Incorrect

Step-by-step explanation:

This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ = 60.8

For the alternative hypothesis,

µ ≠ 60.8

Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 8,

Degrees of freedom, df = n - 1 = 8 - 1 = 7

t = (x - µ)/(s/√n)

Where

x = sample mean = 66.9

µ = population mean = 60.8

s = samples standard deviation = 10.7

t = (66.9 - 60.8)/(10.7/√8) = 1.61

We would determine the p value using the t test calculator. It becomes

p = 0.076

Since alpha, 0.01 < than the p value, 0.076, then we would fail to reject the null hypothesis. Therefore, At a 1 % level of significance, the sample data showed that there is no significant evidence that μ ≠ 60.8

Therefore, this p-value leads to a decision to accept the null hypothesis

step-by-step explanation:

the original amount doesn't ''re just looking for the time it takes for 1/2 of the substance to we have

(1/2) =)     take the ln   of both sides

ln (1/2)   = )   and we can write

ln(1/2) = ) lne     and lne   = 1     so we have

ln(1/2) = -.1092t     divide both sides by -.1092

t = ln(1/2)/-.1092 = about 6.347 days = about 6.3 days (rounded)

step- i d k and i d c : )

you: : 0

bye