Question

*UPDATED*
Use the sample you chose to perform a meaningful hypothesis test for the mean. Include all the assumptions, identify clearly the null and alternate hypotheses, show the method used to test the hypothesis and write a clear conclusion

sample x ̅=200 population mean= 266.5

sample n=32 population size=263

S=828.1

Assumptions/Claim: -10 pts-

Null Hypothesis H0: -5 pts-

Alternate Hypothesis H1: -5 pts-

Method to Test the Hypothesis: -5 pts-

Hypothesis conclusion: -10 pts-

Sample is: 19 24 8 21 70 45 4 13 482 4 8 4 16 0 16 52 30 42 4 5 143 7 18 109 22 26 0 4702 38 71 376 17

Answer 1

Solution

Let X be the variable.

Let µ and σ be the mean and standard deviation of X respectively, Then, X ~ N(µ, σ²)

Claim: The population mean is not different from 266.5. Answer 1

Hypotheses:

Null H₀: µ = µ₀ = 266.5  Answer 2

Vs   

Alternative H_A: µ ≠ 266.5 Answer 3

Method to Test the Hypothesis

i) Test statistic:

t = (√n)(Xbar - µ₀)/s, = - 0.4312

where

n = sample size;

Xbar = sample average;

s = sample standard deviation.

Summary of Excel Calculations is given below:

Given, n	32
µ0	266.5
Xbar	199.875
s	828.1209
tcal	- 0.455
Assume α	0.05
tcrit	2.039513
p-value	0.6622

ii) Distribution, Critical Value and p-value

Under H₀, t ~ t_{n – 1} [i.e., t-distribution with (n -1) degrees of freedom)

Critical value = upper (α/2)% point of t_{n – 1}.

p-value = P(t_{n – 1} > |tcal|)

Using Excel Functions: Statistical TINV and TDIST, tcrit and p-value are found to be as shown in the above table

Decision:

Since | tcal | > < tcrit, or equivalently, since p-value > α. H₀ is accepted. Answer 4

Conclusion:

There is sufficient evidence to support the claim and hence we conclude that the population mean is 266.5 Answer 5

DONE