Question

Please refer “sampling” data sheet. The general rule to test a claim to see whether it is true, is to find out whether it has a very small chance of occurring. In other words, if assuming the claim is true, the result of sampling should not be a rare result. in this case, the supplier claims that the weight of its product will not be less than 343.5 oz.
Column C is the sample I have drawn to test the claim. Column E and F contain descriptive information regarding the sampling result.
Use an appropriate Excel formula to find out what is the probability of obtaining this particular sampling result? (hint: (1) if the claim is true, a random sample should also have a sample average that is somewhat similar to what the supplier is claiming; (2) In this case, mean value is 343.5, and standard deviation is 6.80, determine what is the value of X by yourself.)

Sample 1
Mean	342.83
Standard Error	6.80
Median	349.92
Mode	#N/A
Standard Deviation	37.24
Sample Variance	1386.49
Kurtosis	-0.36
Skewness	0.02
Range	145.56
Minimum	276.31
Maximum	421.87
Sum	10284.82
Count	30.00

Answer 1

One-Sample T: Sample 1(column C)

Descriptive Statistics

N	Mean	StDev	SE Mean	95% Lower Bound for μ
30	342.83	37.24	6.80	331.28

μ: mean of Sample 1(column C)

Test statistic =

Null hypothesis	H₀: μ = 343.5
Alternative hypothesis	H₁: μ > 343.5

T VALUE	P-Value
-0.10	0.539

If the claim is true the sample mean should be 343.5 OZ. in that case the probability = 0.5 which can be calculated by using (=T.DIST.RT(0,29))

But as per the t-test for the given sample test static = -0.1 for this the probability = 0.539 since it is a right tailed test.

Since p-value is greater than alpha 0.05 we fail to reject null hypothesis and conclude that samples weight approximately 343.5 OZ. that the claim is true.